Chapter 13 Limits And Derivatives

We need to evaluate the limit: $\lim\limits_{x \to 2} \left( \frac{1}{x − 2} − \frac{2(2x − 3)}{x^3 − 3x^2 + 2x} \right)$.

First, let's simplify the expression inside the limit. The denominator of the second term is $x^3 - 3x^2 + 2x$. We can factor out $x$:

$x^3 - 3x^2 + 2x = x(x^2 - 3x + 2)$

Now, factor the quadratic $x^2 - 3x + 2$. We look for two numbers that multiply to 2 and add to -3. These numbers are -1 and -2.

$x^2 - 3x + 2 = (x - 1)(x - 2)$

So, the denominator of the second term is $x(x - 1)(x - 2)$.

The expression becomes:

$\frac{1}{x − 2} − \frac{2(2x − 3)}{x(x − 1)(x − 2)}$

To combine these fractions, we find a common denominator, which is $x(x - 1)(x - 2)$.

Multiply the numerator and denominator of the first fraction by $x(x - 1)$:

$\frac{1}{x − 2} \times \frac{x(x - 1)}{x(x - 1)} = \frac{x(x - 1)}{x(x - 1)(x - 2)}$

Now, combine the two fractions:

$\frac{x(x - 1)}{x(x - 1)(x - 2)} - \frac{2(2x - 3)}{x(x - 1)(x - 2)} = \frac{x(x - 1) - 2(2x - 3)}{x(x - 1)(x - 2)}$

Simplify the numerator:

$x(x - 1) - 2(2x - 3) = (x^2 - x) - (4x - 6)$

$= x^2 - x - 4x + 6$

$= x^2 - 5x + 6$

Factor the quadratic numerator $x^2 - 5x + 6$. We look for two numbers that multiply to 6 and add to -5. These numbers are -2 and -3.

$x^2 - 5x + 6 = (x - 2)(x - 3)$

Substitute the factored numerator back into the expression:

$\frac{(x - 2)(x - 3)}{x(x - 1)(x - 2)}$

For $x \ne 2$ (which is the case when evaluating the limit as $x \to 2$), we can cancel the $(x - 2)$ term from the numerator and the denominator:

$\frac{\cancel{(x - 2)}(x - 3)}{x(x - 1)\cancel{(x - 2)}} = \frac{x - 3}{x(x - 1)}$

Now, evaluate the limit of the simplified expression as $x \to 2$:

$\lim\limits_{x \to 2} \frac{x - 3}{x(x - 1)}$

Since the function $\frac{x - 3}{x(x - 1)}$ is continuous at $x = 2$, we can substitute $x = 2$ directly into the expression:

$\frac{2 - 3}{2(2 - 1)} = \frac{-1}{2(1)} = \frac{-1}{2}$

Thus, the value of the limit is $-\frac{1}{2}$.

$\lim\limits_{x \to 2} \frac{1}{x − 2} − \frac{2(2x − 3)}{x^3 − 3x^2 + 2x} = \mathbf{-\frac{1}{2}}$

We need to evaluate the limit: $\lim\limits_{x \to 0} \frac{\sqrt{2 + x} − \sqrt{2}}{x}$.

This limit is of the indeterminate form $\frac{0}{0}$ as $x \to 0$.

To evaluate this, we can multiply the numerator and the denominator by the conjugate of the numerator, which is $(\sqrt{2 + x} + \sqrt{2})$.

Multiply the expression by $\frac{\sqrt{2 + x} + \sqrt{2}}{\sqrt{2 + x} + \sqrt{2}}$:

$\lim\limits_{x \to 0} \frac{\sqrt{2 + x} − \sqrt{2}}{x} \times \frac{\sqrt{2 + x} + \sqrt{2}}{\sqrt{2 + x} + \sqrt{2}}$

Using the difference of squares formula, $(a-b)(a+b) = a^2 - b^2$, the numerator becomes:

$(\sqrt{2 + x})^2 - (\sqrt{2})^2 = (2 + x) - 2 = x$

So, the expression inside the limit becomes:

$\frac{x}{x(\sqrt{2 + x} + \sqrt{2})}$

For $x \ne 0$ (which is relevant for the limit as $x \to 0$), we can cancel the $x$ term from the numerator and the denominator:

$\frac{\cancel{x}}{\cancel{x}(\sqrt{2 + x} + \sqrt{2})} = \frac{1}{\sqrt{2 + x} + \sqrt{2}}$

Now, evaluate the limit of the simplified expression as $x \to 0$:

$\lim\limits_{x \to 0} \frac{1}{\sqrt{2 + x} + \sqrt{2}}$

Since the function $\frac{1}{\sqrt{2 + x} + \sqrt{2}}$ is continuous at $x = 0$, we can substitute $x = 0$ directly into the expression:

$\frac{1}{\sqrt{2 + 0} + \sqrt{2}} = \frac{1}{\sqrt{2} + \sqrt{2}} = \frac{1}{2\sqrt{2}}$

We can rationalize the denominator by multiplying the numerator and denominator by $\sqrt{2}$:

$\frac{1}{2\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2 \times 2} = \frac{\sqrt{2}}{4}$

Thus, the value of the limit is $\frac{1}{2\sqrt{2}}$ or $\frac{\sqrt{2}}{4}$.

$\lim\limits_{x \to 0} \frac{\sqrt{2 + x} − \sqrt{2}}{x} = \mathbf{\frac{1}{2\sqrt{2}}}$ or $\mathbf{\frac{\sqrt{2}}{4}}$

We are given the equation involving a limit:

We know the standard limit formula:

Comparing the given limit (i) with the standard formula (ii), we can see that $a = 3$.

Applying the formula (ii) to the left side of equation (i), we get:

Now, substitute this back into equation (i):

We are looking for a positive integer value of $n$ that satisfies equation (iii).

Let's test positive integer values for $n$:

• If $n = 1$: $1 \cdot 3^{1-1} = 1 \cdot 3^0 = 1 \cdot 1 = 1$. $1 \ne 108$.
• If $n = 2$: $2 \cdot 3^{2-1} = 2 \cdot 3^1 = 2 \cdot 3 = 6$. $6 \ne 108$.
• If $n = 3$: $3 \cdot 3^{3-1} = 3 \cdot 3^2 = 3 \cdot 9 = 27$. $27 \ne 108$.
• If $n = 4$: $4 \cdot 3^{4-1} = 4 \cdot 3^3 = 4 \cdot 27 = 108$. $108 = 108$.

The equation $n \cdot 3^{n-1} = 108$ is satisfied when $n = 4$.

Alternatively, we can prime factorize 108:

$\begin{array}{c|cc} 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, $108 = 2^2 \cdot 3^3 = 4 \cdot 3^3$.

Equation (iii) is $n \cdot 3^{n-1} = 108$.

Substituting the prime factorization of 108:

By comparing the structure of both sides, we can see that if $n=4$, the equation holds:

This confirms that $n=4$ is the positive integer solution.

The positive integer $n$ is $\mathbf{4}$.

We need to evaluate the limit: $\lim\limits_{x \to \frac{π}{2}} (\sec x − \tan x)$.

As $x \to \frac{π}{2}$, $\sec x = \frac{1}{\cos x} \to \frac{1}{0}$, which approaches $\infty$.

Also, as $x \to \frac{π}{2}$, $\tan x = \frac{\sin x}{\cos x} \to \frac{1}{0}$, which approaches $\infty$.

The limit is of the indeterminate form $\infty - \infty$.

We can rewrite the expression in terms of sine and cosine:

$\sec x − \tan x = \frac{1}{\cos x} − \frac{\sin x}{\cos x} = \frac{1 - \sin x}{\cos x}$

Now, the limit becomes: $\lim\limits_{x \to \frac{π}{2}} \frac{1 - \sin x}{\cos x}$.

As $x \to \frac{π}{2}$, the numerator $1 - \sin x \to 1 - \sin(\frac{π}{2}) = 1 - 1 = 0$.

The denominator $\cos x \to \cos(\frac{π}{2}) = 0$.

The limit is now of the indeterminate form $\frac{0}{0}$.

We can apply L'Hopital's Rule since we have the indeterminate form $\frac{0}{0}$. We take the derivative of the numerator and the denominator:

Applying L'Hopital's Rule:

Now, substitute $x = \frac{π}{2}$ into the simplified expression, as the resulting function is continuous at $x = \frac{π}{2}$:

Thus, the value of the limit is 0.

$\lim\limits_{x \to \frac{π}{2}} (\sec x − \tan x) = \mathbf{0}$.

We need to evaluate the limit: $\lim\limits_{x \to 0} \frac{\sin (2 + x) − \sin (2 − x)}{x}$.

As $x \to 0$, the numerator $\sin (2 + x) − \sin (2 − x) \to \sin(2) - \sin(2) = 0$.

The denominator $x \to 0$.

This is an indeterminate form of type $\frac{0}{0}$.

We can use the trigonometric identity for the difference of sines: $\sin A - \sin B = 2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)$.

Let $A = 2 + x$ and $B = 2 - x$.

Using the identity, the numerator becomes:

Now, substitute this back into the limit expression:

We can factor out the constant term $2 \cos(2)$ from the limit:

We know the standard limit: $\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$.

Substituting this value:

Thus, the value of the limit is $2 \cos(2)$.

$\lim\limits_{x \to 0} \frac{\sin (2 + x) − \sin (2 − x)}{x} = \mathbf{2 \cos(2)}$.

Alternate Method using L'Hopital's Rule:

Since the limit is in the indeterminate form $\frac{0}{0}$ as $x \to 0$, we can apply L'Hopital's Rule.

Let $f(x) = \sin (2 + x) − \sin (2 − x)$ and $g(x) = x$.

Find the derivatives of $f(x)$ and $g(x)$ with respect to $x$:

Apply L'Hopital's Rule:

Now, substitute $x = 0$ into the resulting expression:

Both methods yield the same result.

We are asked to find the derivative of the function $f(x) = ax + b$ using the first principle.

The definition of the derivative of a function $f(x)$ by the first principle is given by:

First, we find $f(x + h)$ by replacing $x$ with $x + h$ in the function $f(x) = ax + b$:

Next, we calculate the difference $f(x + h) - f(x)$:

Now, substitute this difference into the limit formula for the derivative:

Since we are considering the limit as $h \to 0$, $h$ is never exactly zero, so we can cancel $h$ from the numerator and the denominator:

The limit of a constant ($a$) as $h$ approaches any value is the constant itself.

Thus, the derivative of $f(x) = ax + b$ by the first principle is $\mathbf{a}$.

We are asked to find the derivative of the function $f(x) = ax^2 + bx + c$ using the first principle.

The definition of the derivative of a function $f(x)$ by the first principle is given by:

First, we find $f(x + h)$ by replacing $x$ with $x + h$ in the function $f(x) = ax^2 + bx + c$:

Expand the terms:

Next, we calculate the difference $f(x + h) - f(x)$:

Remove the parentheses and combine like terms:

The terms $ax^2$, $bx$, and $c$ cancel out:

Now, substitute this difference into the limit formula for the derivative:

Factor out the common term $h$ from the numerator:

Since we are considering the limit as $h \to 0$, $h$ is never exactly zero, so we can cancel $h$ from the numerator and the denominator:

Now, evaluate the limit as $h \to 0$. Substitute $h=0$ into the expression $(2ax + ah + b)$:

Thus, the derivative of $f(x) = ax^2 + bx + c$ by the first principle is $\mathbf{2ax + b}$.

We are asked to find the derivative of the function $f(x) = x^3$ using the first principle.

The definition of the derivative of a function $f(x)$ by the first principle is given by:

First, we find $f(x + h)$ by replacing $x$ with $x + h$ in the function $f(x) = x^3$:

Expand $(x + h)^3$ using the binomial expansion $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$:

Next, we calculate the difference $f(x + h) - f(x)$:

The $x^3$ terms cancel out:

Now, substitute this difference into the limit formula for the derivative:

Factor out the common term $h$ from the numerator:

Since we are considering the limit as $h \to 0$, $h$ is never exactly zero, so we can cancel $h$ from the numerator and the denominator:

Now, evaluate the limit as $h \to 0$. Substitute $h=0$ into the expression $(3x^2 + 3xh + h^2)$:

Thus, the derivative of $f(x) = x^3$ by the first principle is $\mathbf{3x^2}$.

We are asked to find the derivative of the function $f(x) = \frac{1}{x}$ using the first principle.

The definition of the derivative of a function $f(x)$ by the first principle is given by:

First, we find $f(x + h)$ by replacing $x$ with $x + h$ in the function $f(x) = \frac{1}{x}$:

Next, we calculate the difference $f(x + h) - f(x)$:

To subtract the fractions, find a common denominator, which is $x(x+h)$:

Now, substitute this difference into the limit formula for the derivative:

Divide by $h$ (which is the same as multiplying by $\frac{1}{h}$):

Since we are considering the limit as $h \to 0$, $h$ is never exactly zero, so we can cancel $h$ from the numerator and the denominator:

Now, evaluate the limit as $h \to 0$. Substitute $h=0$ into the expression $\frac{-1}{x(x + h)}$:

Thus, the derivative of $f(x) = \frac{1}{x}$ by the first principle is $\mathbf{-\frac{1}{x^2}}$.

We are asked to find the derivative of the function $f(x) = \sin x$ using the first principle.

The definition of the derivative of a function $f(x)$ by the first principle is given by:

First, we find $f(x + h)$ by replacing $x$ with $x + h$ in the function $f(x) = \sin x$:

Next, we calculate the difference $f(x + h) - f(x)$:

Use the trigonometric identity $\sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$. Let $A = x+h$ and $B = x$.

So, the difference is:

Now, substitute this difference into the limit formula for the derivative:

Rearrange the terms and factor out the constant 2:

To use the standard limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$, we need $\frac{h}{2}$ in the denominator. We can achieve this by multiplying the numerator and denominator of the last term by $\frac{1}{2}$.

Now, we can evaluate the limit term by term:

Combine the limits:

Thus, the derivative of $f(x) = \sin x$ by the first principle is $\mathbf{\cos x}$.

We are asked to find the derivative of the function $f(x) = x^n$, where $n$ is a positive integer, using the first principle.

The definition of the derivative of a function $f(x)$ by the first principle is given by:

First, we find $f(x + h)$ by replacing $x$ with $x + h$ in the function $f(x) = x^n$:

We use the binomial theorem to expand $(x + h)^n$:

where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ are the binomial coefficients.

Note that $\binom{n}{1} = n$ and $\binom{n}{n} = 1$.

Next, we calculate the difference $f(x + h) - f(x)$:

The $x^n$ terms cancel out:

Now, substitute this difference into the limit formula for the derivative:

Factor out the common term $h$ from the numerator:

Since we are considering the limit as $h \to 0$, $h$ is never exactly zero, so we can cancel $h$ from the numerator and the denominator:

Now, evaluate the limit as $h \to 0$. As $h$ approaches 0, all terms containing $h$ as a factor (which are all terms except the first one, $nx^{n-1}$) will go to zero:

Thus, the derivative of $f(x) = x^n$ by the first principle, where $n$ is a positive integer, is $\mathbf{nx^{n-1}}$.

We need to find the derivative of the function $f(x) = 2x^4 + x$.

We can find the derivative using the standard rules of differentiation:

The derivative of a sum of functions is the sum of their derivatives (Sum Rule):

The derivative of a constant times a function is the constant times the derivative of the function (Constant Multiple Rule):

The derivative of $x^n$ is $nx^{n-1}$ (Power Rule):

Apply the Sum Rule to $f(x) = 2x^4 + x$:

Apply the Constant Multiple Rule to the first term:

Apply the Power Rule with $n=4$ to $\frac{d}{dx}(x^4)$:

So, the derivative of the first term is:

Apply the Power Rule with $n=1$ to the second term $\frac{d}{dx}(x)$:

Combine the derivatives of the two terms:

Thus, the derivative of $f(x) = 2x^4 + x$ is $\mathbf{8x^3 + 1}$.

We need to find the derivative of the function $f(x) = x^2 \cos x$.

This function is a product of two functions, $u(x) = x^2$ and $v(x) = \cos x$. We will use the Product Rule for differentiation:

First, find the derivatives of $u(x)$ and $v(x)$:

Derivative of $u(x) = x^2$ (using the Power Rule):

Derivative of $v(x) = \cos x$:

Now, apply the Product Rule:

We can factor out $x$ from the expression:

Thus, the derivative of $f(x) = x^2 \cos x$ is $\mathbf{2x \cos x - x^2 \sin x}$ or $\mathbf{x(2 \cos x - x \sin x)}$.

We need to evaluate the limit: $\lim\limits_{x \to \frac{π}{6}} \frac{2\sin^2 x + \sin x − 1}{2 \sin^2 x − 3\sin x + 1}$.

Let's substitute $x = \frac{π}{6}$ into the expression. We know that $\sin(\frac{π}{6}) = \frac{1}{2}$.

Numerator: $2\sin^2 (\frac{π}{6}) + \sin (\frac{π}{6}) − 1 = 2\left(\frac{1}{2}\right)^2 + \frac{1}{2} − 1 = 2\left(\frac{1}{4}\right) + \frac{1}{2} − 1 = \frac{1}{2} + \frac{1}{2} − 1 = 1 − 1 = 0$.

Denominator: $2 \sin^2 (\frac{π}{6}) − 3\sin (\frac{π}{6}) + 1 = 2\left(\frac{1}{2}\right)^2 − 3\left(\frac{1}{2}\right) + 1 = 2\left(\frac{1}{4}\right) − \frac{3}{2} + 1 = \frac{1}{2} − \frac{3}{2} + 1 = \frac{1 - 3}{2} + 1 = -1 + 1 = 0$.

Since the limit is of the indeterminate form $\frac{0}{0}$, we can factor the numerator and the denominator.

Let $y = \sin x$. As $x \to \frac{π}{6}$, $y \to \frac{1}{2}$. The expression can be written as a rational function of $y$:

Factor the numerator $2y^2 + y - 1$. We look for factors of the form $(2y + a)(y + b)$. The factors are $(2y - 1)(y + 1)$.

Factor the denominator $2y^2 - 3y + 1$. We look for factors of the form $(2y + a)(y + b)$. The factors are $(2y - 1)(y - 1)$.

Substitute the factored forms back into the expression:

For $y \ne \frac{1}{2}$ (which is the case as $y$ approaches $\frac{1}{2}$), we can cancel the common factor $(2y - 1)$:

Now, substitute back $y = \sin x$:

Evaluate the limit as $x \to \frac{π}{6}$ by substituting $\sin(\frac{π}{6}) = \frac{1}{2}$ into the simplified expression:

Thus, the value of the limit is $\mathbf{-3}$.

Alternate Method using L'Hopital's Rule:

Since the limit is of the indeterminate form $\frac{0}{0}$ as $x \to \frac{π}{6}$, we can apply L'Hopital's Rule.

Let $f(x) = 2\sin^2 x + \sin x − 1$ and $g(x) = 2 \sin^2 x − 3\sin x + 1$.

Find the derivatives of $f(x)$ and $g(x)$ with respect to $x$:

Apply L'Hopital's Rule:

Factor out $\cos x$ from the numerator and denominator:

For $x$ approaching $\frac{π}{6}$ (but not equal to $\frac{π}{6}$), $\cos x \ne 0$, so we can cancel $\cos x$:

Now, substitute $x = \frac{π}{6}$ into the simplified expression:

Both methods yield the same result.

We need to evaluate the limit: $\lim\limits_{x \to 0} \frac{\tan x − \sin x}{\sin^3 x}$.

As $x \to 0$, $\tan x \to \tan(0) = 0$, $\sin x \to \sin(0) = 0$, and $\sin^3 x \to (\sin(0))^3 = 0$.

The limit is of the indeterminate form $\frac{0}{0}$.

We can rewrite the expression by expressing $\tan x$ in terms of $\sin x$ and $\cos x$:

Find a common denominator in the numerator:

Rewrite the complex fraction:

For $x \ne 0$ (which is the case as $x \to 0$), $\sin x \ne 0$, so we can cancel one power of $\sin x$ from the numerator and denominator:

Now, use the identity $\sin^2 x = 1 - \cos^2 x$:

Use the difference of squares factorization in the denominator: $1 - \cos^2 x = (1 - \cos x)(1 + \cos x)$:

For $x \ne 0$ (and near 0), $1 - \cos x \ne 0$, so we can cancel the term $(1 - \cos x)$ from the numerator and the denominator:

Now, evaluate the limit of the simplified expression as $x \to 0$. Substitute $x=0$ into the expression:

Since $\cos(0) = 1$:

Thus, the value of the limit is $\frac{1}{2}$.

$\lim\limits_{x \to 0} \frac{\tan x − \sin x}{\sin^3 x} = \mathbf{\frac{1}{2}}$.

We need to evaluate the limit: $\lim\limits_{x \to a} \frac{\sqrt{a + 2x} − \sqrt{3 x}}{\sqrt{3a + x} − 2\sqrt{x}}$.

Substitute $x = a$ into the expression. The numerator becomes $\sqrt{a + 2a} - \sqrt{3a} = \sqrt{3a} - \sqrt{3a} = 0$. The denominator becomes $\sqrt{3a + a} - 2\sqrt{a} = \sqrt{4a} - 2\sqrt{a} = 2\sqrt{a} - 2\sqrt{a} = 0$.

This is an indeterminate form of type $\frac{0}{0}$ (assuming $a \ge 0$ for the terms under the square root to be real).

To evaluate this limit, we multiply the numerator and the denominator by the conjugates of both the numerator and the denominator.

The conjugate of the numerator $(\sqrt{a + 2x} − \sqrt{3 x})$ is $(\sqrt{a + 2x} + \sqrt{3 x})$.

The conjugate of the denominator $(\sqrt{3a + x} − 2\sqrt{x})$ is $(\sqrt{3a + x} + 2\sqrt{x})$.

Multiply the expression by $\frac{\sqrt{a + 2x} + \sqrt{3 x}}{\sqrt{a + 2x} + \sqrt{3 x}}$ and $\frac{\sqrt{3a + x} + 2\sqrt{x}}{\sqrt{3a + x} + 2\sqrt{x}}$:

$\lim\limits_{x \to a} \frac{(\sqrt{a + 2x} − \sqrt{3 x})(\sqrt{a + 2x} + \sqrt{3 x})}{(\sqrt{3a + x} − 2\sqrt{x})(\sqrt{3a + x} + 2\sqrt{x})} \cdot \frac{\sqrt{3a + x} + 2\sqrt{x}}{\sqrt{a + 2x} + \sqrt{3 x}}$

Using the difference of squares formula $(A-B)(A+B) = A^2 - B^2$ on the numerator terms and denominator terms separately:

Numerator part: $(\sqrt{a + 2x})^2 − (\sqrt{3 x})^2 = (a + 2x) - 3x = a - x$

Denominator part: $(\sqrt{3a + x})^2 − (2\sqrt{x})^2 = (3a + x) - 4x = 3a - 3x = 3(a - x)$

Substitute these back into the expression:

$\lim\limits_{x \to a} \frac{a - x}{3(a - x)} \cdot \frac{\sqrt{3a + x} + 2\sqrt{x}}{\sqrt{a + 2x} + \sqrt{3 x}}$

For $x \ne a$ (which is the case when evaluating the limit as $x \to a$), $a - x \ne 0$, so we can cancel the term $(a - x)$:

$\lim\limits_{x \to a} \frac{\cancel{a - x}}{3\cancel{(a - x)}} \cdot \frac{\sqrt{3a + x} + 2\sqrt{x}}{\sqrt{a + 2x} + \sqrt{3 x}}$

$\lim\limits_{x \to a} \frac{1}{3} \cdot \frac{\sqrt{3a + x} + 2\sqrt{x}}{\sqrt{a + 2x} + \sqrt{3 x}}$

$\lim\limits_{x \to a} \frac{\sqrt{3a + x} + 2\sqrt{x}}{3(\sqrt{a + 2x} + \sqrt{3 x})}$

Now, evaluate the limit by substituting $x = a$ into the simplified expression (assuming $a > 0$ for the denominator to be non-zero):

$\frac{\sqrt{3a + a} + 2\sqrt{a}}{3(\sqrt{a + 2a} + \sqrt{3 a})} = \frac{\sqrt{4a} + 2\sqrt{a}}{3(\sqrt{3a} + \sqrt{3a})}$

$= \frac{2\sqrt{a} + 2\sqrt{a}}{3(2\sqrt{a})}$

$= \frac{4\sqrt{a}}{6\sqrt{a}}$

Assuming $a > 0$, we can cancel $\sqrt{a}$:

$= \frac{4}{6} = \frac{2}{3}$

Thus, the value of the limit is $\mathbf{\frac{2}{3}}$.

We need to evaluate the limit: $\lim\limits_{x \to 0} \frac{\cos ax − \cos bx}{\cos cx − 1}$.

As $x \to 0$, the numerator $\cos(a \cdot 0) - \cos(b \cdot 0) = \cos(0) - \cos(0) = 1 - 1 = 0$.

The denominator $\cos(c \cdot 0) - 1 = \cos(0) - 1 = 1 - 1 = 0$.

This is an indeterminate form of type $\frac{0}{0}$.

We can use the trigonometric identity for the difference of cosines: $\cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$.

Apply this to the numerator with $A = ax$ and $B = bx$:

For the denominator, we can use the identity $1 - \cos \theta = 2 \sin^2(\frac{\theta}{2})$.

Substitute these expressions back into the limit:

Cancel the factor of -2:

We use the standard limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$. We rewrite the expression by dividing and multiplying by appropriate terms:

As $x \to 0$, we have $\lim\limits_{x \to 0} \frac{\sin(kx)}{kx} = 1$ for any constant $k$. So, the terms of the form $\frac{\sin(...)}{...}$ approach 1.

Assuming $c \ne 0$, for $x \ne 0$, we can cancel $x^2$:

The limit of a constant is the constant itself:

This result holds provided $c \ne 0$.

Thus, the value of the limit is $\mathbf{\frac{a^2 - b^2}{c^2}}$.

Alternate Method using L'Hopital's Rule:

Since the limit is of the indeterminate form $\frac{0}{0}$ as $x \to 0$, we can apply L'Hopital's Rule twice.

Let $f(x) = \cos ax − \cos bx$ and $g(x) = \cos cx − 1$.

First derivative:

The new limit is $\lim\limits_{x \to 0} \frac{b \sin bx - a \sin ax}{-c \sin cx}$. This is still $\frac{0}{0}$.

Second derivative:

Apply L'Hopital's Rule again:

Substitute $x = 0$ into this expression (assuming $c \ne 0$):

Both methods give the same result.

We need to evaluate the limit: $\lim\limits_{ℎ \to 0} \frac{(a + ℎ)^2 \sin (a + ℎ) − a^2 \sin a}{ℎ}$.

This limit is in the form of the definition of the derivative of a function $f(x)$ at a point $x=a$, which is given by:

Comparing the given limit with the definition, we can identify the function $f(x)$ as $f(x) = x^2 \sin x$. The limit is the derivative of $f(x)$ evaluated at $x = a$.

Now, we need to find the derivative of $f(x) = x^2 \sin x$. This function is a product of two functions, $u(x) = x^2$ and $v(x) = \sin x$. We use the Product Rule:

Find the derivatives of $u(x)$ and $v(x)$:

Apply the Product Rule to find $f'(x)$:

The given limit is $f'(a)$. Substitute $x=a$ into $f'(x)$:

Thus, the value of the limit is $\mathbf{2a \sin a + a^2 \cos a}$.

We are asked to find the derivative of the function $f(x) = \tan(ax + b)$ using the first principle.

The definition of the derivative of a function $f(x)$ by the first principle is given by:

First, we find $f(x + h)$ by replacing $x$ with $x + h$ in the function $f(x) = \tan(ax + b)$:

Next, we calculate the difference $f(x + h) - f(x)$:

We use the trigonometric identity for the difference of tangents: $\tan A - \tan B = \frac{\sin(A - B)}{\cos A \cos B}$.

Let $A = ax + ah + b$ and $B = ax + b$. Then $A - B = (ax + ah + b) - (ax + b) = ah$.

Now, substitute this difference into the limit formula for the derivative:

Rewrite the expression:

We can rewrite this by separating the terms to use the standard limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$. We need $ah$ in the denominator for the $\sin(ah)$ term. We can multiply the numerator and denominator by $a$ (assuming $a \ne 0$):

Rearrange the terms:

Now, evaluate the limit term by term:

Combine the limits:

Using the identity $\frac{1}{\cos \theta} = \sec \theta$:

Thus, the derivative of $f(x) = \tan(ax + b)$ by the first principle is $\mathbf{a \sec^2(ax + b)}$.

We are asked to find the derivative of the function $f(x) = \sqrt{\sin x}$ using the first principle.

The definition of the derivative of a function $f(x)$ by the first principle is given by:

First, we find $f(x + h)$ by replacing $x$ with $x + h$ in the function $f(x) = \sqrt{\sin x}$:

Next, we calculate the difference $f(x + h) - f(x)$:

Now, substitute this difference into the limit formula for the derivative:

This limit is of the indeterminate form $\frac{0}{0}$. To evaluate it, multiply the numerator and denominator by the conjugate of the numerator, which is $(\sqrt{\sin(x + h)} + \sqrt{\sin x})$:

Use the difference of squares formula $(A-B)(A+B) = A^2 - B^2$ in the numerator:

We can split this limit into two parts:

The first part is the definition of the derivative of $\sin x$ with respect to $x$:

Evaluate the second part by substituting $h=0$ (since the denominator is continuous and non-zero for $\sin x > 0$):

Multiply the results of the two limits:

Thus, the derivative of $f(x) = \sqrt{\sin x}$ by the first principle is $\mathbf{\frac{\cos x}{2\sqrt{\sin x}}}$.

We need to find the derivative of the function $f(x) = \frac{\cos x}{1 + \sin x}$.

This function is a quotient of two functions. We will use the Quotient Rule for differentiation:

Let $u(x) = \cos x$ and $v(x) = 1 + \sin x$.

Find the derivatives of $u(x)$ and $v(x)$:

Apply the Quotient Rule formula:

Simplify the numerator:

Use the Pythagorean identity $\sin^2 x + \cos^2 x = 1$:

Substitute the simplified numerator back into the derivative expression:

Assuming $1 + \sin x \ne 0$, we can cancel one factor of $(1 + \sin x)$ from the numerator and denominator:

Thus, the derivative of $f(x) = \frac{\cos x}{1 + \sin x}$ is $\mathbf{-\frac{1}{1 + \sin x}}$.

We need to evaluate the limit: $\lim\limits_{x \to 0} \frac{\sin x}{x(1 + \cos x)}$.

As $x \to 0$, the numerator $\sin x \to \sin(0) = 0$.

The denominator $x(1 + \cos x) \to 0(1 + \cos(0)) = 0(1 + 1) = 0(2) = 0$.

This is an indeterminate form of type $\frac{0}{0}$.

We can rewrite the expression and use the standard limit $\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$.

Now, evaluate the limit of each part:

Multiply the limits of the two parts:

The value of the limit is $\frac{1}{2}$.

Comparing with the given options, the correct answer is (B).

The correct answer is $\mathbf{(B)\ \frac{1}{2}}$.

We need to evaluate the limit: $\lim\limits_{x \to \frac{π}{2}} \frac{1 − \sin x}{\cos x}$.

Substitute $x = \frac{π}{2}$ into the expression.

Numerator: $1 - \sin(\frac{π}{2}) = 1 - 1 = 0$.

Denominator: $\cos(\frac{π}{2}) = 0$.

This is an indeterminate form of type $\frac{0}{0}$.

Let $y = x - \frac{π}{2}$. As $x \to \frac{π}{2}$, $y \to 0$. Then $x = \frac{π}{2} + y$. Substitute this into the limit:

Use the trigonometric identities: $\sin(\frac{π}{2} + y) = \cos y$ and $\cos(\frac{π}{2} + y) = -\sin y$.

Now, use the half-angle identities: $1 - \cos y = 2 \sin^2(\frac{y}{2})$ and $\sin y = 2 \sin(\frac{y}{2}) \cos(\frac{y}{2})$.

Cancel the term $2 \sin(\frac{y}{2})$ (since $y \to 0$, $\sin(\frac{y}{2}) \ne 0$ for $y$ near 0):

Now, substitute $y = 0$ into the simplified expression:

Thus, the value of the limit is 0.

Comparing with the given options, the correct answer is (A).

The correct answer is $\mathbf{(A)\ 0}$.

We need to evaluate the limit: $\lim\limits_{x \to 0} \frac{|x|}{x}$.

The absolute value function $|x|$ is defined as:

$|x| = \begin{cases} x & , & x > 0 \\ -x & , & x < 0 \\ 0 & , & x = 0 \end{cases}$

To determine if the limit exists as $x \to 0$, we need to check the left-hand limit and the right-hand limit.

Right-hand limit ($\lim\limits_{x \to 0^+}$):

As $x \to 0$ from the right side ($x > 0$), $|x| = x$.

For $x > 0$, $\frac{x}{x} = 1$.

Left-hand limit ($\lim\limits_{x \to 0^-}$):

As $x \to 0$ from the left side ($x < 0$), $|x| = -x$.

For $x < 0$, $\frac{-x}{x} = -1$.

The limit of a function at a point exists if and only if the left-hand limit and the right-hand limit at that point exist and are equal.

In this case, the right-hand limit is 1, and the left-hand limit is -1.

Since the left-hand limit is not equal to the right-hand limit ($1 \ne -1$), the limit $\lim\limits_{x \to 0} \frac{|x|}{x}$ does not exist.

Comparing with the given options, the correct answer is (D).

The correct answer is $\mathbf{(D)\ does\ not\ exists}$.

We need to evaluate the limit: $\lim\limits_{x \to 1} [x − 1]$.

The notation $[y]$ represents the greatest integer less than or equal to $y$.

To evaluate the limit as $x \to 1$, we consider the left-hand limit and the right-hand limit.

Let $y = x - 1$. As $x \to 1$, $y \to 0$. The limit can be rewritten as $\lim\limits_{y \to 0} [y]$.

Right-hand limit ($\lim\limits_{x \to 1^+}$):

As $x$ approaches 1 from the right side ($x > 1$), $x - 1$ approaches 0 from the right side ($x - 1 > 0$).

For a small positive value of $h$, let $x = 1 + h$, where $h \to 0^+$.

Then $x - 1 = (1 + h) - 1 = h$.

So, the limit becomes:

For any positive number $h$ that is very close to 0 (e.g., 0.1, 0.01, 0.001), the greatest integer less than or equal to $h$ is 0.

Left-hand limit ($\lim\limits_{x \to 1^-}$):

As $x$ approaches 1 from the left side ($x < 1$), $x - 1$ approaches 0 from the left side ($x - 1 < 0$).

For a small positive value of $h$, let $x = 1 - h$, where $h \to 0^+$.

Then $x - 1 = (1 - h) - 1 = -h$.

So, the limit becomes:

For any positive number $h$ that is very close to 0, $-h$ is a small negative number (e.g., -0.1, -0.01, -0.001). The greatest integer less than or equal to $-h$ is -1.

The limit of a function at a point exists if and only if the left-hand limit and the right-hand limit at that point exist and are equal.

In this case, we have:

Since the left-hand limit is not equal to the right-hand limit ($0 \ne -1$), the limit $\lim\limits_{x \to 1} [x - 1]$ does not exist.

Comparing with the given options, the correct answer is (D).

The correct answer is $\mathbf{(D)\ does\ not\ exists}$.

We need to evaluate the limit: $\lim\limits_{x \to 0} x \sin \frac{1}{x}$.

As $x \to 0$, the term $x$ approaches 0. The term $\sin \frac{1}{x}$ behaves differently.

As $x \to 0$, $\frac{1}{x}$ approaches $\pm \infty$. The value of $\sin \frac{1}{x}$ oscillates between -1 and 1.

So, the term $\sin \frac{1}{x}$ does not approach a single value as $x \to 0$; it oscillates infinitely often between -1 and 1.

However, the sine function is bounded, which means:

We can use the Squeeze Theorem to evaluate the limit.

Multiply the inequality $-1 \leq \sin \left(\frac{1}{x}\right) \leq 1$ by $|x|$.

If $x > 0$, $|x| = x$, so $-x \leq x \sin \left(\frac{1}{x}\right) \leq x$.

If $x < 0$, $|x| = -x$. When multiplying by a negative number, the inequality signs flip: $(-x)(-1) \geq (-x) \sin \left(\frac{1}{x}\right) \geq (-x)(1)$, which is $x \geq -x \sin \left(\frac{1}{x}\right) \geq -x$. This can be written as $-x \leq -x \sin \left(\frac{1}{x}\right) \leq x$. Since $-x \sin(\frac{1}{x}) = -|x| \sin(\frac{1}{x})$, this approach is a bit tricky. It's easier to multiply the absolute value inequality.

Multiply the inequality $-1 \leq \sin \left(\frac{1}{x}\right) \leq 1$ by $|x| \ge 0$:

Since $|x| \sin(\frac{1}{x})$ can be equal to $x \sin(\frac{1}{x})$ or $-x \sin(\frac{1}{x})$, it is better to work with the absolute value of the expression:

Since $0 \leq \left|\sin \frac{1}{x}\right| \leq 1$ for $x \ne 0$, multiplying by $|x|$ gives:

This inequality is equivalent to:

Now, consider the limits of the lower and upper bounds as $x \to 0$:

By the Squeeze Theorem, since the function $x \sin \frac{1}{x}$ is squeezed between $-|x|$ and $|x|$, and both $-|x|$ and $|x|$ approach 0 as $x \to 0$, the function $x \sin \frac{1}{x}$ must also approach 0.

Thus, the value of the limit is 0.

Comparing with the given options, the correct answer is (A).

The correct answer is $\mathbf{(A)\ 0}$.

We need to evaluate the limit: $\lim\limits_{n \to ∞} \frac{1 + 2 + 3 + … + n}{n^2}$, where $n \in \mathbb{N}$.

The sum of the first $n$ positive integers is given by the formula:

Substitute this formula into the numerator of the limit expression:

Simplify the expression:

To evaluate the limit of this rational expression as $n \to \infty$, divide both the numerator and the denominator by the highest power of $n$ in the denominator, which is $n^2$:

As $n \to \infty$, the term $\frac{1}{n}$ approaches 0.

The value of the limit is $\frac{1}{2}$.

Comparing with the given options, the correct answer is (C).

The correct answer is $\mathbf{(C)\ \frac{1}{2}}$.

We are given the function $f(x) = x \sin x$ and need to find its derivative $f'(x)$ and then evaluate $f'(\frac{π}{2})$.

The function $f(x)$ is a product of two functions: $u(x) = x$ and $v(x) = \sin x$. We will use the Product Rule for differentiation:

Find the derivatives of $u(x)$ and $v(x)$:

Apply the Product Rule to find $f'(x)$:

Now, we need to evaluate $f'(\frac{π}{2})$. Substitute $x = \frac{π}{2}$ into the expression for $f'(x)$:

We know that $\sin(\frac{π}{2}) = 1$ and $\cos(\frac{π}{2}) = 0$.

Thus, the value of $f'(\frac{π}{2})$ is 1.

Comparing with the given options, the correct answer is (B).

The correct answer is $\mathbf{(B)\ 1}$.

Solution:

We are asked to evaluate the limit:

$\lim\limits_{x \to 3} \frac{x^2 − 9}{x − 3}$

When we substitute $x=3$ directly into the expression, we get $\frac{3^2 - 9}{3 - 3} = \frac{9-9}{0} = \frac{0}{0}$. This is an indeterminate form.

To evaluate this limit, we need to simplify the expression. The numerator $x^2 - 9$ is a difference of squares, which can be factored as $x^2 - 3^2 = (x-3)(x+3)$.

Substituting the factored form into the limit expression, we get:

$\lim\limits_{x \to 3} \frac{(x-3)(x+3)}{x − 3}$

Since the limit is taken as $x$ approaches 3 (but $x \neq 3$), the term $(x-3)$ is not zero. Therefore, we can cancel the common factor $(x-3)$ from the numerator and the denominator:

$\frac{\cancel{(x-3)}(x+3)}{\cancel{(x-3)}} = x+3$, for $x \neq 3$.

Now, the limit of the simplified expression is:

$\lim\limits_{x \to 3} (x+3)$

We can evaluate this limit by directly substituting $x=3$ into the simplified expression:

$3+3 = 6$

Therefore, the value of the limit is 6.

Final Answer: $6$

Solution:

We are asked to evaluate the limit:

$\lim\limits_{x \to \frac{1}{2}} \frac{4x^2 − 1}{2x − 1}$

If we substitute $x = \frac{1}{2}$ directly into the expression, we get $\frac{4(\frac{1}{2})^2 - 1}{2(\frac{1}{2}) - 1} = \frac{4(\frac{1}{4}) - 1}{1 - 1} = \frac{1 - 1}{0} = \frac{0}{0}$. This is an indeterminate form of type $\frac{0}{0}$.

To evaluate the limit, we can factor the numerator. The numerator $4x^2 - 1$ is a difference of squares, which can be written as $(2x)^2 - 1^2$.

So, $4x^2 - 1 = (2x - 1)(2x + 1)$.

Now, substitute this factored form back into the limit expression:

$\lim\limits_{x \to \frac{1}{2}} \frac{(2x − 1)(2x + 1)}{2x − 1}$

Since $x \to \frac{1}{2}$, $x$ is approaching $\frac{1}{2}$ but is not equal to $\frac{1}{2}$. Therefore, $2x - 1 \neq 0$. We can cancel the common factor $(2x - 1)$ from the numerator and the denominator.

The expression simplifies to:

$\frac{\cancel{(2x − 1)}(2x + 1)}{\cancel{2x − 1}} = 2x + 1$, for $x \neq \frac{1}{2}$.

Now, evaluate the limit of the simplified expression by substituting $x = \frac{1}{2}$:

$\lim\limits_{x \to \frac{1}{2}} (2x + 1) = 2(\frac{1}{2}) + 1 = 1 + 1 = 2$.

Thus, the value of the limit is 2.

Final Answer: $2$

Solution:

We are asked to evaluate the limit:

$\lim\limits_{x \to 0} \frac{(x + 2)^{\frac{1}{3}} − 2^{\frac{1}{3}}}{x}$

When we substitute $x=0$ directly into the expression, we get $\frac{(0+2)^{\frac{1}{3}} - 2^{\frac{1}{3}}}{0} = \frac{2^{\frac{1}{3}} - 2^{\frac{1}{3}}}{0} = \frac{0}{0}$. This is an indeterminate form of type $\frac{0}{0}$.

To evaluate this limit, we can use the standard limit formula which is a direct consequence of L'Hopital's rule or algebraic manipulation:

$\lim\limits_{y \to a} \frac{y^n - a^n}{y - a} = n a^{n-1}$

Let us rewrite the expression to match this form. Let $y = x + 2$. As $x \to 0$, $y \to 0 + 2 = 2$. Also, from the substitution $y = x+2$, we have $x = y - 2$.

Substitute these into the given limit expression:

$\lim\limits_{y \to 2} \frac{y^{\frac{1}{3}} − 2^{\frac{1}{3}}}{y - 2}$

This limit is now in the standard form $\lim\limits_{y \to a} \frac{y^n - a^n}{y - a}$, where:

$y$ is the variable

$a = 2$

$n = \frac{1}{3}$

Using the formula $\lim\limits_{y \to a} \frac{y^n - a^n}{y - a} = n a^{n-1}$, we can evaluate the limit:

Limit value $= n \times a^{n-1}$

Limit value $= \frac{1}{3} \times 2^{\frac{1}{3} - 1}$

Calculate the exponent: $\frac{1}{3} - 1 = \frac{1}{3} - \frac{3}{3} = -\frac{2}{3}$.

So the limit value is:

$\frac{1}{3} \times 2^{-\frac{2}{3}}$

This can also be written as:

$\frac{1}{3 \cdot 2^{\frac{2}{3}}}$

or

$\frac{1}{3 \sqrt[3]{2^2}} = \frac{1}{3 \sqrt[3]{4}}$

Therefore, the value of the limit is $\frac{1}{3 \cdot 2^{2/3}}$.

Final Answer: $\frac{1}{3 \cdot 2^{2/3}}$

Solution:

We are asked to evaluate the limit:

$\lim\limits_{x \to 1} \frac{(1 + x)^6 − 1}{(1 + x)^2 − 1}$

Let $y = 1 + x$. As $x \to 1$, the value of $y$ approaches $1 + 1 = 2$. So, we can rewrite the limit in terms of $y$ as $y \to 2$.

The expression becomes:

$\lim\limits_{y \to 2} \frac{y^6 − 1}{y^2 − 1}$

We can factor the numerator and the denominator as differences of squares repeatedly or recognize them as polynomials.

Numerator: $y^6 - 1 = (y^3)^2 - 1^2 = (y^3 - 1)(y^3 + 1)$.

Denominator: $y^2 - 1 = y^2 - 1^2 = (y - 1)(y + 1)$.

So the limit expression is:

$\lim\limits_{y \to 2} \frac{(y^3 - 1)(y^3 + 1)}{(y - 1)(y + 1)}$

We can further factor the cubic terms in the numerator using the formulas $a^3 - b^3 = (a-b)(a^2+ab+b^2)$ and $a^3 + b^3 = (a+b)(a^2-ab+b^2)$.

$y^3 - 1 = (y - 1)(y^2 + y + 1)$

$y^3 + 1 = (y + 1)(y^2 - y + 1)$

Substitute these factored forms into the expression:

$\lim\limits_{y \to 2} \frac{(y - 1)(y^2 + y + 1)(y + 1)(y^2 - y + 1)}{(y - 1)(y + 1)}$

As $y \to 2$, $y$ is close to 2 but not equal to 2. Therefore, $y - 1 \neq 0$ and $y + 1 \neq 0$. We can cancel the common factors $(y - 1)$ and $(y + 1)$ from the numerator and the denominator.

The simplified expression is:

$\lim\limits_{y \to 2} (y^2 + y + 1)(y^2 - y + 1)$

Now, we can evaluate the limit by substituting $y = 2$ into the simplified expression, as it is a polynomial function which is continuous everywhere.

Limit value $= (2^2 + 2 + 1)(2^2 - 2 + 1)$

Limit value $= (4 + 2 + 1)(4 - 2 + 1)$

Limit value $= (7)(3)$

Limit value $= 21$

Alternatively, since the denominator $y^2 - 1$ is non-zero at $y=2$, we could have evaluated the limit $\lim\limits_{y \to 2} \frac{y^6 - 1}{y^2 - 1}$ by direct substitution:

$\frac{2^6 - 1}{2^2 - 1} = \frac{64 - 1}{4 - 1} = \frac{63}{3} = 21$.

Therefore, the value of the limit is 21.

Final Answer: $21$

Solution:

We are asked to evaluate the limit:

$\lim\limits_{x \to a} \frac{(2 + x)^{\frac{5}{2}} − (a + 2)^{\frac{5}{2}}}{x − a}$

This limit is in the form of $\lim\limits_{x \to a} \frac{f(x) - f(a)}{x - a}$, which is the definition of the derivative of the function $f(x)$ at $x=a$.

Let $f(x) = (2+x)^{\frac{5}{2}}$. Then $f(a) = (2+a)^{\frac{5}{2}} = (a+2)^{\frac{5}{2}}$.

The limit is equal to $f'(a)$.

We need to find the derivative of $f(x)$ with respect to $x$.

$f'(x) = \frac{d}{dx} \left( (2+x)^{\frac{5}{2}} \right)$

Using the power rule and chain rule, where the outer function is $(\cdot)^{\frac{5}{2}}$ and the inner function is $2+x$:

$f'(x) = \frac{5}{2} (2+x)^{\frac{5}{2} - 1} \cdot \frac{d}{dx}(2+x)$

$f'(x) = \frac{5}{2} (2+x)^{\frac{3}{2}} \cdot (1)$

$f'(x) = \frac{5}{2} (2+x)^{\frac{3}{2}}$

Now, evaluate $f'(a)$ by substituting $x=a$ into the expression for $f'(x)$:

$f'(a) = \frac{5}{2} (2+a)^{\frac{3}{2}}$

So, the value of the limit is $\frac{5}{2} (a+2)^{\frac{3}{2}}$.

Alternatively, we can use the standard limit formula:

$\lim\limits_{y \to c} \frac{y^n - c^n}{y - c} = n c^{n-1}$

In our limit, let $y = 2+x$ and $c = 2+a$. As $x \to a$, $y = 2+x \to 2+a = c$.

The denominator $x-a$ can be rewritten in terms of $y$ and $c$. Since $y = 2+x$, $x = y-2$. Since $c = 2+a$, $a = c-2$.

So, $x-a = (y-2) - (c-2) = y - 2 - c + 2 = y - c$.

The limit expression becomes:

$\lim\limits_{y \to c} \frac{y^{\frac{5}{2}} - c^{\frac{5}{2}}}{y - c}$

This matches the standard formula form with $n = \frac{5}{2}$ and the limit taken as $y \to c$.

Using the formula, the limit is $n c^{n-1}$.

Limit value $= \frac{5}{2} c^{\frac{5}{2} - 1}$

Limit value $= \frac{5}{2} c^{\frac{3}{2}}$

Substitute back $c = a+2$:

Limit value $= \frac{5}{2} (a+2)^{\frac{3}{2}}$

Both methods yield the same result. The value of the limit is $\frac{5}{2} (a+2)^{\frac{3}{2}}$.

Final Answer: $\frac{5}{2} (a+2)^{\frac{3}{2}}$

Solution:

We are asked to evaluate the limit:

$\lim\limits_{x \to a} \frac{(2 + x)^{\frac{5}{2}} − (a + 2)^{\frac{5}{2}}}{x − a}$

When we substitute $x=a$ directly into the expression, we get $\frac{(2 + a)^{\frac{5}{2}} − (a + 2)^{\frac{5}{2}}}{a − a} = \frac{0}{0}$. This is an indeterminate form.

We can evaluate this limit by recognizing its form. The limit is in the form $\lim\limits_{x \to a} \frac{f(x) - f(a)}{x - a}$, which is the definition of the derivative of the function $f(x)$ at the point $x=a$.

Let $f(x) = (2 + x)^{\frac{5}{2}}$. Then $f(a) = (2 + a)^{\frac{5}{2}}$.

The limit is equal to the derivative of $f(x)$ evaluated at $x=a$, i.e., $f'(a)$.

We need to find the derivative of $f(x) = (2 + x)^{\frac{5}{2}}$ with respect to $x$.

Using the chain rule and the power rule for differentiation, $\frac{d}{dx} (u^n) = n u^{n-1} \frac{du}{dx}$ where $u = 2+x$ and $n = \frac{5}{2}$:

$f'(x) = \frac{d}{dx} (2 + x)^{\frac{5}{2}}$

$f'(x) = \frac{5}{2} (2 + x)^{\frac{5}{2} - 1} \cdot \frac{d}{dx}(2 + x)$

$f'(x) = \frac{5}{2} (2 + x)^{\frac{3}{2}} \cdot (1)$

$f'(x) = \frac{5}{2} (2 + x)^{\frac{3}{2}}$

Now, substitute $x=a$ into the expression for $f'(x)$ to find $f'(a)$:

$f'(a) = \frac{5}{2} (2 + a)^{\frac{3}{2}}$

Therefore, the value of the limit is $\frac{5}{2} (a + 2)^{\frac{3}{2}}$.

Alternate Solution:

We can use the standard limit formula:

$\lim\limits_{y \to c} \frac{y^n - c^n}{y - c} = n c^{n-1}$

Let the variable in our limit be $y = 2+x$. As $x \to a$, the value of $y$ approaches $2+a$. So, $c = 2+a$.

From the substitution $y = 2+x$, we have $x = y-2$. Also, $a = c-2$.

The denominator $x-a$ becomes $(y-2) - (c-2) = y - 2 - c + 2 = y - c$.

The limit expression can be rewritten in terms of $y$ and $c$:

$\lim\limits_{y \to c} \frac{y^{\frac{5}{2}} - c^{\frac{5}{2}}}{y - c}$

This limit is exactly in the form of the standard limit formula, with $n = \frac{5}{2}$ and the limit taken as $y$ approaches $c$.

Using the formula, the value of the limit is $n c^{n-1}$.

Limit value $= \frac{5}{2} c^{\frac{5}{2} - 1}$

Limit value $= \frac{5}{2} c^{\frac{3}{2}}$

Substitute back $c = a+2$:

Limit value $= \frac{5}{2} (a+2)^{\frac{3}{2}}$

Both methods give the same result.

Final Answer: $\frac{5}{2} (a + 2)^{\frac{3}{2}}$

Solution:

We are asked to evaluate the limit:

$\lim\limits_{x \to 1} \frac{x^4 − \sqrt{x}}{\sqrt{x} − 1}$

When we substitute $x=1$ directly into the expression, we get $\frac{1^4 − \sqrt{1}}{\sqrt{1} − 1} = \frac{1 - 1}{1 - 1} = \frac{0}{0}$. This is an indeterminate form of type $\frac{0}{0}$.

To evaluate this limit, we can use a substitution. Let $y = \sqrt{x}$.

Since $y = \sqrt{x}$, squaring both sides gives $y^2 = x$. Raising to the power of 4 gives $(y^2)^4 = x^4$, so $y^8 = x^4$.

As $x$ approaches 1, $\sqrt{x}$ approaches $\sqrt{1}$, which is 1. Therefore, as $x \to 1$, $y \to 1$.

Substitute $y = \sqrt{x}$, $x = y^2$, and $x^4 = y^8$ into the limit expression. The limit becomes:

$\lim\limits_{y \to 1} \frac{y^8 − y}{y − 1}$

Now, we can factor the numerator $y^8 - y$ by taking out the common factor $y$:

$y^8 - y = y(y^7 - 1)$

The limit expression becomes:

$\lim\limits_{y \to 1} \frac{y(y^7 - 1)}{y − 1}$

We can rewrite this as a product of two limits:

$\lim\limits_{y \to 1} y \cdot \lim\limits_{y \to 1} \frac{y^7 - 1}{y − 1}$

The first part is $\lim\limits_{y \to 1} y$. This is a simple direct substitution:

$\lim\limits_{y \to 1} y = 1$

The second part is $\lim\limits_{y \to 1} \frac{y^7 - 1}{y − 1}$. This limit is in the standard form $\lim\limits_{y \to a} \frac{y^n - a^n}{y - a}$, where $a = 1$ and $n = 7$.

Using the standard limit formula: $\lim\limits_{y \to a} \frac{y^n - a^n}{y - a} = n a^{n-1}$

We have:

$\lim\limits_{y \to 1} \frac{y^7 - 1^7}{y − 1} = 7 \cdot 1^{7-1} = 7 \cdot 1^6 = 7 \cdot 1 = 7$

Now, combine the results of the two limits:

$\lim\limits_{y \to 1} y \cdot \lim\limits_{y \to 1} \frac{y^7 - 1}{y − 1} = 1 \cdot 7 = 7$

Therefore, the value of the limit is 7.

Final Answer: $7$

Solution:

We are asked to evaluate the limit:

$\lim\limits_{x \to 2} \frac{x^2 − 4}{\sqrt{3x − 2} − \sqrt{x + 2}}$

When we substitute $x=2$ directly into the expression, we get $\frac{2^2 - 4}{\sqrt{3(2) - 2} - \sqrt{2 + 2}} = \frac{4 - 4}{\sqrt{6 - 2} - \sqrt{4}} = \frac{0}{\sqrt{4} - 2} = \frac{0}{2 - 2} = \frac{0}{0}$. This is an indeterminate form of type $\frac{0}{0}$.

To evaluate this limit, we can rationalize the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator, which is $\sqrt{3x - 2} + \sqrt{x + 2}$.

Multiply the expression by $\frac{\sqrt{3x − 2} + \sqrt{x + 2}}{\sqrt{3x − 2} + \sqrt{x + 2}}$:

$\lim\limits_{x \to 2} \frac{x^2 − 4}{\sqrt{3x − 2} − \sqrt{x + 2}} \times \frac{\sqrt{3x − 2} + \sqrt{x + 2}}{\sqrt{3x − 2} + \sqrt{x + 2}}$

Simplify the denominator using the difference of squares formula $(a-b)(a+b) = a^2 - b^2$:

Denominator $= (\sqrt{3x - 2})^2 - (\sqrt{x + 2})^2 = (3x - 2) - (x + 2)$

Denominator $= 3x - 2 - x - 2 = 2x - 4$

The numerator becomes $(x^2 - 4)(\sqrt{3x - 2} + \sqrt{x + 2})$.

The limit expression is now:

$\lim\limits_{x \to 2} \frac{(x^2 − 4)(\sqrt{3x − 2} + \sqrt{x + 2})}{2x - 4}$

Factor the numerator ($x^2 - 4 = (x-2)(x+2)$) and the denominator ($2x - 4 = 2(x-2)$):

$\lim\limits_{x \to 2} \frac{(x-2)(x+2)(\sqrt{3x − 2} + \sqrt{x + 2})}{2(x-2)}$

Since $x \to 2$, $x$ is approaching 2 but is not equal to 2. Therefore, $x-2 \neq 0$. We can cancel the common factor $(x-2)$ from the numerator and the denominator.

The simplified expression is:

$\lim\limits_{x \to 2} \frac{(x+2)(\sqrt{3x − 2} + \sqrt{x + 2})}{2}$

Now, substitute $x=2$ into the simplified expression, as it is a continuous function at $x=2$:

Limit value $= \frac{(2+2)(\sqrt{3(2) − 2} + \sqrt{2 + 2})}{2}$

Limit value $= \frac{(4)(\sqrt{6 − 2} + \sqrt{4})}{2}$

Limit value $= \frac{(4)(\sqrt{4} + \sqrt{4})}{2}$

Limit value $= \frac{(4)(2 + 2)}{2}$

Limit value $= \frac{4 \times 4}{2}$

Limit value $= \frac{16}{2}$

Limit value $= 8$

Therefore, the value of the limit is 8.

Final Answer: $8$

Solution:

We are asked to evaluate the limit:

$\lim\limits_{x \to \sqrt{2}} \frac{x^4 − 4}{x^2 + 3\sqrt{2}x − 8}$

When we substitute $x = \sqrt{2}$ directly into the expression:

Numerator: $(\sqrt{2})^4 - 4 = 4 - 4 = 0$.

Denominator: $(\sqrt{2})^2 + 3\sqrt{2}(\sqrt{2}) - 8 = 2 + 3(2) - 8 = 2 + 6 - 8 = 0$.

Since we get the indeterminate form $\frac{0}{0}$, we need to simplify the expression.

Let's factor the numerator and the denominator.

The numerator is a difference of squares:

$x^4 - 4 = (x^2)^2 - 2^2 = (x^2 - 2)(x^2 + 2)$

The term $(x^2 - 2)$ is also a difference of squares:

$x^2 - 2 = x^2 - (\sqrt{2})^2 = (x - \sqrt{2})(x + \sqrt{2})$

So, the numerator is $(x - \sqrt{2})(x + \sqrt{2})(x^2 + 2)$.

For the denominator, $x^2 + 3\sqrt{2}x - 8$, since substituting $x = \sqrt{2}$ gives 0, $(x - \sqrt{2})$ must be a factor. We can find the other factor using polynomial division or by observation.

Using polynomial division:

$\begin{array}{r} x + 4\sqrt{2}\phantom{)} \\ x-\sqrt{2}{\overline{\smash{\big)}\,x^2+3\sqrt{2}x-8\phantom{)}}} \\ \underline{-~\phantom{(}(x^2-\sqrt{2}x)\phantom{-b)}} \\ 0+4\sqrt{2}x-8\phantom{)} \\ \underline{-~\phantom{()}(4\sqrt{2}x-8)} \\ 0+0\phantom{)} \end{array}$

So, the denominator factors as $(x - \sqrt{2})(x + 4\sqrt{2})$.

Now, rewrite the limit expression with the factored numerator and denominator:

$\lim\limits_{x \to \sqrt{2}} \frac{(x - \sqrt{2})(x + \sqrt{2})(x^2 + 2)}{(x - \sqrt{2})(x + 4\sqrt{2})}$

Since $x \to \sqrt{2}$, $x$ is approaching $\sqrt{2}$ but is not equal to $\sqrt{2}$. Thus, $(x - \sqrt{2}) \neq 0$. We can cancel the common factor $(x - \sqrt{2})$ from the numerator and the denominator.

The simplified expression is:

$\lim\limits_{x \to \sqrt{2}} \frac{(x + \sqrt{2})(x^2 + 2)}{x + 4\sqrt{2}}$

Now, substitute $x = \sqrt{2}$ into the simplified expression, as the denominator $x + 4\sqrt{2}$ is non-zero when $x = \sqrt{2}$ ($ \sqrt{2} + 4\sqrt{2} = 5\sqrt{2} \neq 0$).

Limit value $= \frac{(\sqrt{2} + \sqrt{2})((\sqrt{2})^2 + 2)}{\sqrt{2} + 4\sqrt{2}}$

Limit value $= \frac{(2\sqrt{2})(2 + 2)}{5\sqrt{2}}$

Limit value $= \frac{(2\sqrt{2})(4)}{5\sqrt{2}}$

Limit value $= \frac{8\sqrt{2}}{5\sqrt{2}}$

Cancel the $\sqrt{2}$ term:

Limit value $= \frac{8}{5}$

Therefore, the value of the limit is $\frac{8}{5}$.

Final Answer: $\frac{8}{5}$

Solution:

We are asked to evaluate the limit:

$\lim\limits_{x \to 1} \frac{x^7 − 2x^5 + 1}{x^3 − 3x^2 + 2}$

When we substitute $x=1$ directly into the expression, we get:

Numerator: $1^7 - 2(1)^5 + 1 = 1 - 2 + 1 = 0$

Denominator: $1^3 - 3(1)^2 + 2 = 1 - 3 + 2 = 0$

This gives the indeterminate form $\frac{0}{0}$, which indicates that $(x-1)$ is a common factor in both the numerator and the denominator.

Let's factor the numerator $N(x) = x^7 − 2x^5 + 1$. Since $N(1) = 0$, $(x-1)$ is a factor. We can use polynomial long division or synthetic division.

By grouping terms:

$x^7 − 2x^5 + 1 = x^7 - x^5 - x^5 + 1$

$= x^5(x^2 - 1) - (x^5 - 1)$

$= x^5(x-1)(x+1) - (x-1)(x^4 + x^3 + x^2 + x + 1)$

$= (x-1) [x^5(x+1) - (x^4 + x^3 + x^2 + x + 1)]$

$= (x-1) [x^6 + x^5 - x^4 - x^3 - x^2 - x - 1]$

So, $x^7 − 2x^5 + 1 = (x-1)(x^6 + x^5 - x^4 - x^3 - x^2 - x - 1)$.

Let's factor the denominator $D(x) = x^3 − 3x^2 + 2$. Since $D(1) = 0$, $(x-1)$ is a factor. We can use polynomial long division or synthetic division.

Using synthetic division with root 1:

$\begin{array}{c|cccc} 1 & 1 & -3 & 0 & 2 \\ & & 1 & -2 & -2 \\ \hline & 1 & -2 & -2 & 0 \\ \end{array}$

The coefficients of the quotient are $1, -2, -2$. So, $x^3 − 3x^2 + 2 = (x-1)(x^2 - 2x - 2)$.

Substitute the factored forms into the limit expression:

$\lim\limits_{x \to 1} \frac{(x-1)(x^6 + x^5 - x^4 - x^3 - x^2 - x - 1)}{(x-1)(x^2 - 2x - 2)}$

Since $x \to 1$, $x$ is approaching 1 but is not equal to 1. Thus, $x-1 \neq 0$. We can cancel the common factor $(x-1)$ from the numerator and the denominator.

The simplified expression is:

$\lim\limits_{x \to 1} \frac{x^6 + x^5 - x^4 - x^3 - x^2 - x - 1}{x^2 - 2x - 2}$

Now, substitute $x=1$ into the simplified expression, as the denominator is non-zero at $x=1$ ($1^2 - 2(1) - 2 = 1 - 2 - 2 = -3 \neq 0$).

Numerator: $1^6 + 1^5 - 1^4 - 1^3 - 1^2 - 1 - 1 = 1 + 1 - 1 - 1 - 1 - 1 - 1 = 2 - 5 = -3$

Denominator: $1^2 - 2(1) - 2 = 1 - 2 - 2 = -3$

Limit value $= \frac{-3}{-3} = 1$

Therefore, the value of the limit is 1.

Final Answer: $1$

Solution:

We are asked to evaluate the limit:

$\lim\limits_{x \to 0} \frac{\sqrt{1 + x^3} − \sqrt{1 − x^3}}{x^2}$

When we substitute $x=0$ directly into the expression, we get:

Numerator: $\sqrt{1 + 0^3} − \sqrt{1 − 0^3} = \sqrt{1} − \sqrt{1} = 1 - 1 = 0$

Denominator: $0^2 = 0$

This gives the indeterminate form $\frac{0}{0}$.

To evaluate this limit, we will rationalize the numerator by multiplying the numerator and the denominator by the conjugate of the numerator, which is $\sqrt{1 + x^3} + \sqrt{1 − x^3}$.

Multiply the expression by $\frac{\sqrt{1 + x^3} + \sqrt{1 − x^3}}{\sqrt{1 + x^3} + \sqrt{1 − x^3}}$:

$\lim\limits_{x \to 0} \frac{\sqrt{1 + x^3} − \sqrt{1 − x^3}}{x^2} \times \frac{\sqrt{1 + x^3} + \sqrt{1 − x^3}}{\sqrt{1 + x^3} + \sqrt{1 − x^3}}$

Now, simplify the numerator using the difference of squares formula $(a-b)(a+b) = a^2 - b^2$, where $a = \sqrt{1 + x^3}$ and $b = \sqrt{1 − x^3}$.

Numerator $= (\sqrt{1 + x^3})^2 - (\sqrt{1 − x^3})^2$

Numerator $= (1 + x^3) - (1 − x^3)$

Numerator $= 1 + x^3 - 1 + x^3$

Numerator $= 2x^3$

The denominator becomes $x^2 (\sqrt{1 + x^3} + \sqrt{1 − x^3})$.

The limit expression is now:

$\lim\limits_{x \to 0} \frac{2x^3}{x^2 (\sqrt{1 + x^3} + \sqrt{1 − x^3})}$

Since $x \to 0$, $x$ is approaching 0 but is not equal to 0. Thus, $x^2 \neq 0$. We can cancel the common factor $x^2$ from the numerator and the denominator.

$\frac{2x^{\cancel{3}^{1}}}{\cancel{x^2} (\sqrt{1 + x^3} + \sqrt{1 − x^3})} = \frac{2x}{\sqrt{1 + x^3} + \sqrt{1 − x^3}}$

The simplified expression is:

$\lim\limits_{x \to 0} \frac{2x}{\sqrt{1 + x^3} + \sqrt{1 − x^3}}$

Now, substitute $x=0$ into the simplified expression. The denominator $\sqrt{1 + x^3} + \sqrt{1 − x^3}$ is non-zero at $x=0$ ($\sqrt{1} + \sqrt{1} = 2$).

Numerator $= 2(0) = 0$

Denominator $= \sqrt{1 + 0^3} + \sqrt{1 − 0^3} = \sqrt{1} + \sqrt{1} = 1 + 1 = 2$

Limit value $= \frac{0}{2} = 0$

Therefore, the value of the limit is 0.

Final Answer: $0$

Solution:

We are asked to evaluate the limit:

$\lim\limits_{x \to −3} \frac{x^3 + 27}{x^5 + 243}$

When we substitute $x=-3$ directly into the expression, we get:

Numerator: $(-3)^3 + 27 = -27 + 27 = 0$

Denominator: $(-3)^5 + 243 = -243 + 243 = 0$

This is an indeterminate form of type $\frac{0}{0}$.

To evaluate this limit, we can use the standard limit formula:

$\lim\limits_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}$

First, rewrite the numerator and the denominator to fit this form. Note that $27 = 3^3$ and $243 = 3^5$. Also, the limit is $x \to -3$. So, we need terms in the form of $(x - (-3))$ in the denominator.

Numerator: $x^3 + 27 = x^3 - (-27) = x^3 - (-3)^3$.

Denominator: $x^5 + 243 = x^5 - (-243) = x^5 - (-3)^5$.

Divide both the numerator and the denominator by $(x - (-3)) = (x + 3)$:

$\lim\limits_{x \to −3} \frac{\frac{x^3 - (-3)^3}{x - (-3)}}{\frac{x^5 - (-3)^5}{x - (-3)}}$

We can use the property of limits $\lim \frac{f(x)}{g(x)} = \frac{\lim f(x)}{\lim g(x)}$, provided the limits of the numerator and denominator exist and the limit of the denominator is non-zero.

Numerator limit: $\lim\limits_{x \to −3} \frac{x^3 - (-3)^3}{x - (-3)}$

This is in the form $\lim\limits_{x \to a} \frac{x^n - a^n}{x - a}$ with $a = -3$ and $n = 3$.

Using the formula, the numerator limit is $3 \cdot (-3)^{3-1} = 3 \cdot (-3)^2 = 3 \cdot 9 = 27$.

Denominator limit: $\lim\limits_{x \to −3} \frac{x^5 - (-3)^5}{x - (-3)}$

This is in the form $\lim\limits_{x \to a} \frac{x^n - a^n}{x - a}$ with $a = -3$ and $n = 5$.

Using the formula, the denominator limit is $5 \cdot (-3)^{5-1} = 5 \cdot (-3)^4 = 5 \cdot 81 = 405$.

Now, the limit of the original expression is the ratio of these limits:

$\lim\limits_{x \to −3} \frac{x^3 + 27}{x^5 + 243} = \frac{\lim\limits_{x \to −3} \frac{x^3 - (-3)^3}{x - (-3)}}{\lim\limits_{x \to −3} \frac{x^5 - (-3)^5}{x - (-3)}}$

Limit value $= \frac{27}{405}$

Simplify the fraction by dividing both numerator and denominator by their greatest common divisor, which is 27.

$\frac{27}{405} = \frac{\cancel{27}^1}{405 \div 27}$

$405 \div 27 = 15$.

So, $\frac{27}{405} = \frac{1}{15}$.

Therefore, the value of the limit is $\frac{1}{15}$.

Final Answer: $\frac{1}{15}$

Solution:

We are asked to evaluate the limit:

$\lim\limits_{x \to \frac{1}{2}} \left( \frac{8x − 3}{2x − 1} − \frac{4x^2 + 1}{4x^2 − 1} \right)$

First, let's examine the expression as $x \to \frac{1}{2}$.

For the first term, as $x \to \frac{1}{2}$, the numerator approaches $8(\frac{1}{2}) - 3 = 4 - 3 = 1$, and the denominator approaches $2(\frac{1}{2}) - 1 = 1 - 1 = 0$. So, the first term approaches $\frac{1}{0}$, which tends to $\pm \infty$.

For the second term, as $x \to \frac{1}{2}$, the numerator approaches $4(\frac{1}{2})^2 + 1 = 4(\frac{1}{4}) + 1 = 1 + 1 = 2$, and the denominator approaches $4(\frac{1}{2})^2 - 1 = 4(\frac{1}{4}) - 1 = 1 - 1 = 0$. So, the second term approaches $\frac{2}{0}$, which also tends to $\pm \infty$.

The limit is of the indeterminate form $\infty - \infty$. We need to combine the terms into a single fraction.

Find a common denominator for the two fractions. The denominator of the second term is $4x^2 - 1$, which can be factored as a difference of squares: $4x^2 - 1 = (2x)^2 - 1^2 = (2x - 1)(2x + 1)$.

The common denominator is $4x^2 - 1 = (2x - 1)(2x + 1)$.

Combine the fractions:

$\frac{8x − 3}{2x − 1} − \frac{4x^2 + 1}{4x^2 − 1} = \frac{(8x − 3)(2x + 1)}{(2x − 1)(2x + 1)} − \frac{4x^2 + 1}{(2x − 1)(2x + 1)}$

$= \frac{(8x − 3)(2x + 1) - (4x^2 + 1)}{(2x − 1)(2x + 1)}$

Expand the numerator:

$(8x − 3)(2x + 1) - (4x^2 + 1) = (16x^2 + 8x - 6x - 3) - (4x^2 + 1)$

$= 16x^2 + 2x - 3 - 4x^2 - 1$

$= 12x^2 + 2x - 4$

The combined expression is $\frac{12x^2 + 2x - 4}{4x^2 - 1}$. Now, we evaluate the limit of this single fraction:

$\lim\limits_{x \to \frac{1}{2}} \frac{12x^2 + 2x - 4}{4x^2 - 1}$

Substitute $x = \frac{1}{2}$ into this new expression:

Numerator: $12(\frac{1}{2})^2 + 2(\frac{1}{2}) - 4 = 12(\frac{1}{4}) + 1 - 4 = 3 + 1 - 4 = 0$

Denominator: $4(\frac{1}{2})^2 - 1 = 4(\frac{1}{4}) - 1 = 1 - 1 = 0$

We still have the indeterminate form $\frac{0}{0}$. This means $(x - \frac{1}{2})$ or $(2x - 1)$ is a common factor in the numerator and the denominator.

Factor the numerator $12x^2 + 2x - 4$. We know $(2x - 1)$ is a factor. We can use polynomial division:

$\begin{array}{r} 6x + 4\phantom{)} \\ 2x-1{\overline{\smash{\big)}\,12x^2+2x-4}} \\ \underline{-~\phantom{(}(12x^2-6x)\phantom{-b)}} \\ 0+8x-4\phantom{)} \\ \underline{-~\phantom{()}(8x-4)} \\ 0+0\phantom{)} \end{array}$

So, $12x^2 + 2x - 4 = (2x - 1)(6x + 4)$.

The denominator is $4x^2 - 1 = (2x - 1)(2x + 1)$.

Substitute the factored forms into the limit expression:

$\lim\limits_{x \to \frac{1}{2}} \frac{(2x - 1)(6x + 4)}{(2x - 1)(2x + 1)}$

Since $x \to \frac{1}{2}$, $x \neq \frac{1}{2}$, so $2x - 1 \neq 0$. We can cancel the common factor $(2x - 1)$.

The simplified expression is:

$\lim\limits_{x \to \frac{1}{2}} \frac{6x + 4}{2x + 1}$

Now, substitute $x = \frac{1}{2}$ into the simplified expression. The denominator $2x+1$ is non-zero at $x = \frac{1}{2}$.

Limit value $= \frac{6(\frac{1}{2}) + 4}{2(\frac{1}{2}) + 1}$

Limit value $= \frac{3 + 4}{1 + 1}$

Limit value $= \frac{7}{2}$

Therefore, the value of the limit is $\frac{7}{2}$.

Final Answer: $\frac{7}{2}$

Solution:

We are given the equation involving a limit:

$\lim\limits_{x \to 2} \frac{x^n − 2^n}{x − 2} = 80$

where $n \in N$ (n is a natural number).

We can use the standard limit formula:

$\lim\limits_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}$

Comparing the given limit with the standard formula, we have:

$a = 2$

The form of the limit matches the standard formula.

Applying the formula to the given limit:

$\lim\limits_{x \to 2} \frac{x^n − 2^n}{x − 2} = n \cdot 2^{n-1}$

We are given that this limit is equal to 80. So, we have the equation:

$n \cdot 2^{n-1} = 80$

We need to find the natural number $n$ that satisfies this equation.

Let's find the prime factorization of 80:

$\begin{array}{c|cc} 2 & 80 \\ \hline 2 & 40 \\ \hline 2 & 20 \\ \hline 2 & 10 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

So, $80 = 2 \times 2 \times 2 \times 2 \times 5 = 2^4 \times 5 = 5 \times 2^4$.

Now, we equate the two forms:

$n \cdot 2^{n-1} = 5 \cdot 2^4$

Since $n$ is a natural number, $2^{n-1}$ is a power of 2 (for $n=1$, $2^0=1$; for $n>1$, $2^{n-1}$ is an even integer). $n$ is the other factor.

By comparing the two sides of the equation $n \cdot 2^{n-1} = 5 \cdot 2^4$, we can equate the non-power-of-2 factors and the power-of-2 factors.

Equating the non-power-of-2 factors, we get:

$n = 5$

Equating the power-of-2 factors, we get:

$2^{n-1} = 2^4$

From the exponents, we have $n-1 = 4$.

Solving for $n$: $n = 4 + 1 = 5$.

Both comparisons give $n=5$. Since $n=5$ is a natural number, this is the required value of $n$.

Let's verify the result for $n=5$:

$\lim\limits_{x \to 2} \frac{x^5 − 2^5}{x − 2} = 5 \cdot 2^{5-1} = 5 \cdot 2^4 = 5 \cdot 16 = 80$

This matches the given value.

Final Answer: $n = 5$

Solution:

We are asked to evaluate the limit:

$\lim\limits_{x \to a} \frac{\sin 3x}{\sin 7x}$

The value of the limit depends on the value of $a$.

Case 1: $\sin 7a \neq 0$.

In this case, the denominator is non-zero at $x=a$. The function $f(x) = \frac{\sin 3x}{\sin 7x}$ is continuous at $x=a$ because $\sin 3x$ and $\sin 7x$ are continuous functions, and the denominator is non-zero.

Therefore, the limit can be found by direct substitution:

$\lim\limits_{x \to a} \frac{\sin 3x}{\sin 7x} = \frac{\sin 3a}{\sin 7a}$

Case 2: $\sin 7a = 0$.

This occurs when $7a = k\pi$ for some integer $k$, i.e., $a = \frac{k\pi}{7}$ for $k \in \mathbb{Z}$.

If $\sin 3a \neq 0$ when $\sin 7a = 0$, the limit is of the form $\frac{\text{non-zero}}{0}$, which means the limit does not exist (approaches $\pm \infty$). This happens if $a = \frac{k\pi}{7}$ and $3a$ is not an integer multiple of $\pi$, i.e., $3k/7$ is not an integer.

If $\sin 3a = 0$ when $\sin 7a = 0$, this means $3a$ is also an integer multiple of $\pi$. This occurs when $a$ is a common multiple of $\pi/3$ and $\pi/7$, which means $a$ is an integer multiple of $\pi$. Let $a = p\pi$ for some integer $p$. In this case, the limit is of the indeterminate form $\frac{0}{0}$.

Evaluating the limit in the indeterminate case ($a = p\pi$ for $p \in \mathbb{Z}$):

When $a = p\pi$, the limit is $\lim\limits_{x \to p\pi} \frac{\sin 3x}{\sin 7x}$. This is an indeterminate form $\frac{0}{0}$.

Let $x = p\pi + h$. As $x \to p\pi$, $h \to 0$.

Substitute this into the limit expression:

$\lim\limits_{x \to p\pi} \frac{\sin 3x}{\sin 7x} = \lim\limits_{h \to 0} \frac{\sin(3(p\pi + h))}{\sin(7(p\pi + h))}$

$= \lim\limits_{h \to 0} \frac{\sin(3p\pi + 3h)}{\sin(7p\pi + 7h)}$

Using the identity $\sin(n\pi + \theta) = (-1)^n \sin \theta$ for integer $n$:

$= \lim\limits_{h \to 0} \frac{(-1)^{3p} \sin(3h)}{(-1)^{7p} \sin(7h)}$

Since $3p$ and $7p$ are integers, $(-1)^{3p} = (-1)^p$ and $(-1)^{7p} = (-1)^p$.

$= \lim\limits_{h \to 0} \frac{(-1)^p \sin(3h)}{(-1)^p \sin(7h)} = \lim\limits_{h \to 0} \frac{\sin(3h)}{\sin(7h)}$

This is a standard limit of the form $\lim\limits_{\theta \to 0} \frac{\sin k\theta}{\sin m\theta}$. We evaluate this using the fundamental limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$.

$\lim\limits_{h \to 0} \frac{\sin(3h)}{\sin(7h)} = \lim\limits_{h \to 0} \frac{\frac{\sin(3h)}{3h} \cdot 3h}{\frac{\sin(7h)}{7h} \cdot 7h}$

$= \lim\limits_{h \to 0} \frac{\frac{\sin(3h)}{3h}}{\frac{\sin(7h)}{7h}} \cdot \frac{3h}{7h}$

Cancel $h$ (since $h \to 0$, $h \neq 0$):

$= \lim\limits_{h \to 0} \frac{\frac{\sin(3h)}{3h}}{\frac{\sin(7h)}{7h}} \cdot \frac{3}{7}$

Using limit properties and the standard limit:

$= \frac{\lim\limits_{h \to 0} \frac{\sin(3h)}{3h}}{\lim\limits_{h \to 0} \frac{\sin(7h)}{7h}} \cdot \lim\limits_{h \to 0} \frac{3}{7}$

$= \frac{1}{1} \cdot \frac{3}{7}$

$= \frac{3}{7}$

Therefore, if $a = p\pi$ for some integer $p$, the limit is $\frac{3}{7}$. Given that this question is likely testing the evaluation of indeterminate forms using standard trigonometric limits, the intended answer is most probably for the case where $a$ makes the limit indeterminate, i.e., $a = p\pi$.

Final Answer: $\frac{3}{7}$ (assuming the indeterminate case $a = p\pi$ is intended)

Solution:

We are asked to evaluate the limit:

$\lim\limits_{x \to 0} \frac{\sin^2 2x}{\sin^2 4x}$

When we substitute $x=0$ directly into the expression, we get $\frac{\sin^2 (2 \cdot 0)}{\sin^2 (4 \cdot 0)} = \frac{\sin^2 0}{\sin^2 0} = \frac{0^2}{0^2} = \frac{0}{0}$. This is an indeterminate form of type $\frac{0}{0}$.

To evaluate this limit, we can use the fundamental trigonometric limit:

$\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$

We can rewrite the given expression by multiplying and dividing by appropriate terms to match the form of the standard limit:

$\frac{\sin^2 2x}{\sin^2 4x} = \frac{(\sin 2x)^2}{(\sin 4x)^2}$

$= \frac{\left(\frac{\sin 2x}{2x} \cdot 2x\right)^2}{\left(\frac{\sin 4x}{4x} \cdot 4x\right)^2}$

$= \frac{\left(\frac{\sin 2x}{2x}\right)^2 \cdot (2x)^2}{\left(\frac{\sin 4x}{4x}\right)^2 \cdot (4x)^2}$

$= \frac{\left(\frac{\sin 2x}{2x}\right)^2 \cdot 4x^2}{\left(\frac{\sin 4x}{4x}\right)^2 \cdot 16x^2}$

Now, consider the limit as $x \to 0$. Since $x \to 0$, $x \neq 0$, and therefore $x^2 \neq 0$. We can cancel the $x^2$ term from the numerator and denominator:

$\lim\limits_{x \to 0} \frac{\left(\frac{\sin 2x}{2x}\right)^2 \cdot \cancel{4x^2}}{\left(\frac{\sin 4x}{4x}\right)^2 \cdot \cancel{16x^2}} = \lim\limits_{x \to 0} \frac{\left(\frac{\sin 2x}{2x}\right)^2 \cdot 4}{\left(\frac{\sin 4x}{4x}\right)^2 \cdot 16}$

$= \lim\limits_{x \to 0} \frac{4}{16} \cdot \frac{\left(\frac{\sin 2x}{2x}\right)^2}{\left(\frac{\sin 4x}{4x}\right)^2}$

$= \frac{1}{4} \cdot \lim\limits_{x \to 0} \frac{\left(\frac{\sin 2x}{2x}\right)^2}{\left(\frac{\sin 4x}{4x}\right)^2}$

As $x \to 0$, $2x \to 0$ and $4x \to 0$. Using the standard limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$, we have:

$\lim\limits_{x \to 0} \frac{\sin 2x}{2x} = 1$

$\lim\limits_{x \to 0} \frac{\sin 4x}{4x} = 1$

Substitute these values back into the limit expression:

Limit value $= \frac{1}{4} \cdot \frac{(1)^2}{(1)^2}$

Limit value $= \frac{1}{4} \cdot \frac{1}{1} = \frac{1}{4}$

Therefore, the value of the limit is $\frac{1}{4}$.

Final Answer: $\frac{1}{4}$

Solution:

We are asked to evaluate the limit:

$\lim\limits_{x \to 0} \frac{1 − \cos 2x}{x^2}$

When we substitute $x=0$ directly into the expression, we get $\frac{1 − \cos (2 \cdot 0)}{0^2} = \frac{1 − \cos 0}{0} = \frac{1 − 1}{0} = \frac{0}{0}$. This is an indeterminate form of type $\frac{0}{0}$.

To evaluate this limit, we can use the trigonometric identity $1 − \cos 2x = 2 \sin^2 x$.

Substitute this identity into the limit expression:

$\lim\limits_{x \to 0} \frac{2 \sin^2 x}{x^2}$

We can rewrite the expression as:

$\lim\limits_{x \to 0} 2 \left(\frac{\sin x}{x}\right)^2$

We know the fundamental trigonometric limit:

$\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$

Using the properties of limits, the limit of a constant times a function is the constant times the limit of the function, and the limit of a power of a function is the power of the limit:

$\lim\limits_{x \to 0} 2 \left(\frac{\sin x}{x}\right)^2 = 2 \left(\lim\limits_{x \to 0} \frac{\sin x}{x}\right)^2$

Substitute the value of the fundamental limit:

Limit value $= 2 (1)^2 = 2 \cdot 1 = 2$

Therefore, the value of the limit is 2.

Final Answer: $2$

Solution:

We are asked to evaluate the limit:

$\lim\limits_{x \to 0} \frac{2 \sin x − \sin 2x}{x^3}$

When we substitute $x=0$ directly into the expression, we get $\frac{2 \sin 0 − \sin (2 \cdot 0)}{0^3} = \frac{2(0) − \sin 0}{0} = \frac{0 - 0}{0} = \frac{0}{0}$. This is an indeterminate form of type $\frac{0}{0}$.

To evaluate this limit, we can use the double angle formula for sine: $\sin 2x = 2 \sin x \cos x$.

Substitute this into the numerator:

Numerator $= 2 \sin x − (2 \sin x \cos x)$

Factor out $2 \sin x$ from the numerator:

Numerator $= 2 \sin x (1 − \cos x)$

The limit expression becomes:

$\lim\limits_{x \to 0} \frac{2 \sin x (1 − \cos x)}{x^3}$

We can use another trigonometric identity: $1 − \cos x = 2 \sin^2 \left(\frac{x}{2}\right)$.

Substitute this into the numerator:

Numerator $= 2 \sin x \left(2 \sin^2 \left(\frac{x}{2}\right)\right)$

Numerator $= 4 \sin x \sin^2 \left(\frac{x}{2}\right)$

The limit expression is now:

$\lim\limits_{x \to 0} \frac{4 \sin x \sin^2 \left(\frac{x}{2}\right)}{x^3}$

We can rearrange the terms and use the fundamental trigonometric limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$.

$\lim\limits_{x \to 0} 4 \cdot \frac{\sin x}{x} \cdot \frac{\sin^2 \left(\frac{x}{2}\right)}{x^2}$

$\lim\limits_{x \to 0} 4 \cdot \frac{\sin x}{x} \cdot \left(\frac{\sin \left(\frac{x}{2}\right)}{x}\right)^2$

For the term $\left(\frac{\sin \left(\frac{x}{2}\right)}{x}\right)^2$, we need a $\frac{x}{2}$ in the denominator to match the standard limit form. We can rewrite the denominator $x$ as $2 \cdot \frac{x}{2}$:

$\frac{\sin \left(\frac{x}{2}\right)}{x} = \frac{\sin \left(\frac{x}{2}\right)}{2 \cdot \frac{x}{2}} = \frac{1}{2} \cdot \frac{\sin \left(\frac{x}{2}\right)}{\frac{x}{2}}$

So, $\left(\frac{\sin \left(\frac{x}{2}\right)}{x}\right)^2 = \left(\frac{1}{2} \cdot \frac{\sin \left(\frac{x}{2}\right)}{\frac{x}{2}}\right)^2 = \frac{1}{4} \left(\frac{\sin \left(\frac{x}{2}\right)}{\frac{x}{2}}\right)^2$

Substitute this back into the limit expression:

$\lim\limits_{x \to 0} 4 \cdot \frac{\sin x}{x} \cdot \frac{1}{4} \left(\frac{\sin \left(\frac{x}{2}\right)}{\frac{x}{2}}\right)^2$

Cancel the factor of 4:

$\lim\limits_{x \to 0} \frac{\sin x}{x} \cdot \left(\frac{\sin \left(\frac{x}{2}\right)}{\frac{x}{2}}\right)^2$

As $x \to 0$, $x \to 0$ and $\frac{x}{2} \to 0$. Using the standard limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:

$\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$

$\lim\limits_{x \to 0} \frac{\sin \left(\frac{x}{2}\right)}{\frac{x}{2}} = 1$

Substitute these limit values:

Limit value $= 1 \cdot (1)^2 = 1 \cdot 1 = 1$

Therefore, the value of the limit is 1.

Final Answer: $1$

Solution:

We are asked to evaluate the limit:

$\lim\limits_{x \to 0} \frac{1 − \cos mx}{1 − \cos nx}$

When we substitute $x=0$ directly into the expression, we get $\frac{1 − \cos (m \cdot 0)}{1 − \cos (n \cdot 0)} = \frac{1 − \cos 0}{1 − \cos 0} = \frac{1 − 1}{1 − 1} = \frac{0}{0}$. This is an indeterminate form of type $\frac{0}{0}$.

To evaluate this limit, we can use the trigonometric identity $1 − \cos 2\theta = 2 \sin^2 \theta$.

Applying this identity to the numerator with $2\theta = mx$, so $\theta = \frac{mx}{2}$:

$1 − \cos mx = 2 \sin^2 \left(\frac{mx}{2}\right)$

Applying this identity to the denominator with $2\theta = nx$, so $\theta = \frac{nx}{2}$:

$1 − \cos nx = 2 \sin^2 \left(\frac{nx}{2}\right)$

Substitute these into the limit expression:

$\lim\limits_{x \to 0} \frac{2 \sin^2 \left(\frac{mx}{2}\right)}{2 \sin^2 \left(\frac{nx}{2}\right)}$

Cancel the constant factor 2:

$\lim\limits_{x \to 0} \frac{\sin^2 \left(\frac{mx}{2}\right)}{\sin^2 \left(\frac{nx}{2}\right)}$

We can rewrite the expression using the property $(\sin \theta)^2 = \sin^2 \theta$:

$\lim\limits_{x \to 0} \frac{\left(\sin \left(\frac{mx}{2}\right)\right)^2}{\left(\sin \left(\frac{nx}{2}\right)\right)^2}$

Now, we will use the fundamental trigonometric limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$. To do this, we multiply and divide each sine term by its argument:

$\lim\limits_{x \to 0} \frac{\left(\frac{\sin \left(\frac{mx}{2}\right)}{\frac{mx}{2}} \cdot \frac{mx}{2}\right)^2}{\left(\frac{\sin \left(\frac{nx}{2}\right)}{\frac{nx}{2}} \cdot \frac{nx}{2}\right)^2}$

$= \lim\limits_{x \to 0} \frac{\left(\frac{\sin \left(\frac{mx}{2}\right)}{\frac{mx}{2}}\right)^2 \cdot \left(\frac{mx}{2}\right)^2}{\left(\frac{\sin \left(\frac{nx}{2}\right)}{\frac{nx}{2}}\right)^2 \cdot \left(\frac{nx}{2}\right)^2}$

$= \lim\limits_{x \to 0} \frac{\left(\frac{\sin \left(\frac{mx}{2}\right)}{\frac{mx}{2}}\right)^2 \cdot \frac{m^2 x^2}{4}}{\left(\frac{\sin \left(\frac{nx}{2}\right)}{\frac{nx}{2}}\right)^2 \cdot \frac{n^2 x^2}{4}}$

Since $x \to 0$, $x$ is approaching 0 but $x \neq 0$. Therefore, $x^2 \neq 0$. We can cancel the common factor $\frac{x^2}{4}$ from the numerator and the denominator:

$\lim\limits_{x \to 0} \frac{\left(\frac{\sin \left(\frac{mx}{2}\right)}{\frac{mx}{2}}\right)^2 \cdot m^2 \cancel{\frac{x^2}{4}}}{\left(\frac{\sin \left(\frac{nx}{2}\right)}{\frac{nx}{2}}\right)^2 \cdot n^2 \cancel{\frac{x^2}{4}}} = \lim\limits_{x \to 0} \frac{\left(\frac{\sin \left(\frac{mx}{2}\right)}{\frac{mx}{2}}\right)^2 m^2}{\left(\frac{\sin \left(\frac{nx}{2}\right)}{\frac{nx}{2}}\right)^2 n^2}$

Rearrange the terms:

$= \frac{m^2}{n^2} \cdot \lim\limits_{x \to 0} \frac{\left(\frac{\sin \left(\frac{mx}{2}\right)}{\frac{mx}{2}}\right)^2}{\left(\frac{\sin \left(\frac{nx}{2}\right)}{\frac{nx}{2}}\right)^2}$

As $x \to 0$, we have $\frac{mx}{2} \to 0$ and $\frac{nx}{2} \to 0$. Using the standard limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:

$\lim\limits_{x \to 0} \frac{\sin \left(\frac{mx}{2}\right)}{\frac{mx}{2}} = 1$

$\lim\limits_{x \to 0} \frac{\sin \left(\frac{nx}{2}\right)}{\frac{nx}{2}} = 1$

Substitute these values into the expression:

Limit value $= \frac{m^2}{n^2} \cdot \frac{(1)^2}{(1)^2}$

Limit value $= \frac{m^2}{n^2} \cdot \frac{1}{1} = \frac{m^2}{n^2}$

Therefore, the value of the limit is $\frac{m^2}{n^2}$.

Final Answer: $\frac{m^2}{n^2}$

Solution:

We are asked to evaluate the limit:

$\lim\limits_{x \to \frac{π}{3}} \frac{\sqrt{1 − \cos 6x}}{\sqrt{2} (\frac{π}{3} − x)}$

When we substitute $x = \frac{\pi}{3}$ into the expression, we get:

Numerator: $\sqrt{1 − \cos (6 \cdot \frac{\pi}{3})} = \sqrt{1 − \cos 2\pi} = \sqrt{1 − 1} = \sqrt{0} = 0$

Denominator: $\sqrt{2} (\frac{\pi}{3} − \frac{\pi}{3}) = \sqrt{2} \cdot 0 = 0$

This is an indeterminate form of type $\frac{0}{0}$.

To simplify the expression, we use the trigonometric identity $1 − \cos 2\theta = 2 \sin^2 \theta$. Let $2\theta = 6x$, so $\theta = 3x$.

The numerator becomes:

$\sqrt{1 − \cos 6x} = \sqrt{2 \sin^2 3x} = \sqrt{2} |\sin 3x|$

The limit expression is now:

$\lim\limits_{x \to \frac{π}{3}} \frac{\sqrt{2} |\sin 3x|}{\sqrt{2} (\frac{π}{3} − x)}$

We can cancel the factor $\sqrt{2}$ (since $\sqrt{2} \neq 0$):

$\lim\limits_{x \to \frac{π}{3}} \frac{|\sin 3x|}{\frac{π}{3} − x}$

To evaluate this limit, let's make a substitution. Let $x = \frac{\pi}{3} + h$. As $x \to \frac{\pi}{3}$, $h$ approaches 0. The expression becomes:

Numerator: $|\sin (3(\frac{\pi}{3} + h))| = |\sin (\pi + 3h)|$

Using the identity $\sin(\pi + \theta) = -\sin\theta$, we have $|\sin (\pi + 3h)| = |-\sin 3h| = |\sin 3h|$.

Denominator: $\frac{\pi}{3} − x = \frac{\pi}{3} − (\frac{\pi}{3} + h) = -h$.

The limit is transformed into:

$\lim\limits_{h \to 0} \frac{|\sin 3h|}{-h}$

Because of the absolute value in the numerator and the term $-h$ in the denominator, we need to evaluate the one-sided limits as $h$ approaches 0.

Left-hand limit (as $h \to 0^-$, meaning $h < 0$):

If $h < 0$, then $3h < 0$. For $h$ values close to 0 but negative, $3h$ is in the fourth quadrant (or negative small angle), so $\sin 3h < 0$.

Thus, $|\sin 3h| = -(\sin 3h)$.

The limit becomes:

$\lim\limits_{h \to 0^-} \frac{-\sin 3h}{-h} = \lim\limits_{h \to 0^-} \frac{\sin 3h}{h}$

Multiply and divide by 3 to use the standard limit form $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:

$\lim\limits_{h \to 0^-} \frac{\sin 3h}{3h} \cdot 3$

As $h \to 0^-$, $3h \to 0^-$. Using the standard limit:

$= 1 \cdot 3 = 3$

Right-hand limit (as $h \to 0^+$, meaning $h > 0$):

If $h > 0$, then $3h > 0$. For $h$ values close to 0 but positive, $3h$ is in the first quadrant (or positive small angle), so $\sin 3h > 0$.

Thus, $|\sin 3h| = \sin 3h$.

The limit becomes:

$\lim\limits_{h \to 0^+} \frac{\sin 3h}{-h} = \lim\limits_{h \to 0^+} - \frac{\sin 3h}{h}$

Multiply and divide by 3:

$\lim\limits_{h \to 0^+} - \frac{\sin 3h}{3h} \cdot 3$

As $h \to 0^+$, $3h \to 0^+$. Using the standard limit:

$= -1 \cdot 3 = -3$

Since the left-hand limit (3) and the right-hand limit (-3) are not equal, the two-sided limit does not exist.

Final Answer: The limit does not exist.

Solution:

We are asked to evaluate the limit:

$\lim\limits_{x \to \frac{π}{4}} \frac{\sin x − \cos x}{x − \frac{π}{4}}$

When we substitute $x = \frac{\pi}{4}$ directly into the expression, we get:

Numerator: $\sin(\frac{\pi}{4}) - \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = 0$

Denominator: $\frac{\pi}{4} - \frac{\pi}{4} = 0$

This gives the indeterminate form $\frac{0}{0}$.

This limit is in the form of the definition of the derivative of a function at a point.

The definition of the derivative of a function $f(x)$ at $x=a$ is given by:

$f'(a) = \lim\limits_{x \to a} \frac{f(x) - f(a)}{x - a}$

Comparing the given limit with the definition, we have $a = \frac{\pi}{4}$.

Let $f(x) = \sin x − \cos x$.

Then $f(a) = f(\frac{\pi}{4}) = \sin(\frac{\pi}{4}) − \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} − \frac{\sqrt{2}}{2} = 0$.

So the numerator is $f(x) - f(\frac{\pi}{4}) = (\sin x - \cos x) - 0 = \sin x - \cos x$.

The given limit is exactly in the form of the derivative of $f(x)$ at $x = \frac{\pi}{4}$.

We need to find the derivative of $f(x) = \sin x - \cos x$.

$f'(x) = \frac{d}{dx}(\sin x - \cos x)$

$f'(x) = \frac{d}{dx}(\sin x) - \frac{d}{dx}(\cos x)$

$f'(x) = \cos x - (-\sin x)$

$f'(x) = \cos x + \sin x$

Now, evaluate $f'(\frac{\pi}{4})$ by substituting $x = \frac{\pi}{4}$ into the expression for $f'(x)$:

$f'(\frac{\pi}{4}) = \cos(\frac{\pi}{4}) + \sin(\frac{\pi}{4})$

$f'(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}$

$f'(\frac{\pi}{4}) = \frac{2\sqrt{2}}{2}$

$f'(\frac{\pi}{4}) = \sqrt{2}$

Thus, the value of the limit is $\sqrt{2}$.

Alternate Solution:

We can use trigonometric identities and the standard limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$.

Consider the numerator $\sin x - \cos x$. We can rewrite this expression using the auxiliary angle method or trigonometric identities.

$\sin x - \cos x = \sqrt{2} \left( \frac{1}{\sqrt{2}}\sin x - \frac{1}{\sqrt{2}}\cos x \right)$

Using $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$ and $\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$:

$\sin x - \cos x = \sqrt{2} \left( \cos(\frac{\pi}{4})\sin x - \sin(\frac{\pi}{4})\cos x \right)$

Using the identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$, with $A = x$ and $B = \frac{\pi}{4}$:

$\sin x - \cos x = \sqrt{2} \sin(x - \frac{\pi}{4})$

Substitute this into the limit expression:

$\lim\limits_{x \to \frac{π}{4}} \frac{\sqrt{2} \sin(x − \frac{π}{4})}{x − \frac{π}{4}}$

Let $h = x - \frac{\pi}{4}$. As $x \to \frac{\pi}{4}$, $h$ approaches 0.

The limit becomes:

$\lim\limits_{h \to 0} \frac{\sqrt{2} \sin h}{h}$

Using the property of limits, $\lim\limits_{h \to 0} c \cdot g(h) = c \cdot \lim\limits_{h \to 0} g(h)$:

$= \sqrt{2} \lim\limits_{h \to 0} \frac{\sin h}{h}$

Using the fundamental trigonometric limit $\lim\limits_{h \to 0} \frac{\sin h}{h} = 1$:

Limit value $= \sqrt{2} \cdot 1 = \sqrt{2}$

Both methods give the same result. The value of the limit is $\sqrt{2}$.

Final Answer: $\sqrt{2}$

Solution:

We are asked to evaluate the limit:

$\lim\limits_{x \to \frac{π}{6}} \frac{\sqrt{3} \sin x − \cos x}{x − \frac{π}{6}}$

When we substitute $x = \frac{\pi}{6}$ directly into the expression, we get:

Numerator: $\sqrt{3} \sin(\frac{\pi}{6}) - \cos(\frac{\pi}{6}) = \sqrt{3} (\frac{1}{2}) - \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} = 0$

Denominator: $\frac{\pi}{6} - \frac{\pi}{6} = 0$

This gives the indeterminate form $\frac{0}{0}$.

This limit is in the form of the definition of the derivative of a function at a point.

The definition of the derivative of a function $f(x)$ at $x=a$ is given by:

$f'(a) = \lim\limits_{x \to a} \frac{f(x) - f(a)}{x - a}$

Comparing the given limit with the definition, we have $a = \frac{\pi}{6}$.

Let $f(x) = \sqrt{3} \sin x − \cos x$.

Then $f(a) = f(\frac{\pi}{6}) = \sqrt{3} \sin(\frac{\pi}{6}) − \cos(\frac{\pi}{6}) = \sqrt{3} (\frac{1}{2}) - \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} = 0$.

So the numerator is $f(x) - f(\frac{\pi}{6}) = (\sqrt{3} \sin x - \cos x) - 0 = \sqrt{3} \sin x - \cos x$.

The given limit is exactly in the form of the derivative of $f(x)$ at $x = \frac{\pi}{6}$.

We need to find the derivative of $f(x) = \sqrt{3} \sin x - \cos x$.

$f'(x) = \frac{d}{dx}(\sqrt{3} \sin x - \cos x)$

$f'(x) = \sqrt{3} \frac{d}{dx}(\sin x) - \frac{d}{dx}(\cos x)$

$f'(x) = \sqrt{3} \cos x - (-\sin x)$

$f'(x) = \sqrt{3} \cos x + \sin x$

Now, evaluate $f'(\frac{\pi}{6})$ by substituting $x = \frac{\pi}{6}$ into the expression for $f'(x)$:

$f'(\frac{\pi}{6}) = \sqrt{3} \cos(\frac{\pi}{6}) + \sin(\frac{\pi}{6})$

$f'(\frac{\pi}{6}) = \sqrt{3} (\frac{\sqrt{3}}{2}) + \frac{1}{2}$

$f'(\frac{\pi}{6}) = \frac{3}{2} + \frac{1}{2}$

$f'(\frac{\pi}{6}) = \frac{4}{2} = 2$

Thus, the value of the limit is 2.

Alternate Solution:

We can use trigonometric identities and the standard limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$.

Consider the numerator $\sqrt{3} \sin x - \cos x$. We can express this in the form $R \sin(x - \alpha)$.

$R \sin(x - \alpha) = R (\sin x \cos \alpha - \cos x \sin \alpha) = (R \cos \alpha)\sin x - (R \sin \alpha)\cos x$

Comparing with $\sqrt{3} \sin x - \cos x$, we have:

Square and add equations (i) and (ii):

$(R \cos \alpha)^2 + (R \sin \alpha)^2 = (\sqrt{3})^2 + 1^2$

$R^2 \cos^2 \alpha + R^2 \sin^2 \alpha = 3 + 1$

$R^2 (\cos^2 \alpha + \sin^2 \alpha) = 4$

$R^2 (1) = 4 \implies R = 2$ (since $R > 0$)

Divide equation (ii) by equation (i):

$\frac{R \sin \alpha}{R \cos \alpha} = \frac{1}{\sqrt{3}}$

$\tan \alpha = \frac{1}{\sqrt{3}}$

Since $\cos \alpha = \frac{\sqrt{3}}{2} > 0$ and $\sin \alpha = \frac{1}{2} > 0$, $\alpha$ is in the first quadrant. So, $\alpha = \frac{\pi}{6}$.

Thus, $\sqrt{3} \sin x - \cos x = 2 \sin(x - \frac{\pi}{6})$.

Substitute this into the limit expression:

$\lim\limits_{x \to \frac{π}{6}} \frac{2 \sin(x − \frac{π}{6})}{x − \frac{π}{6}}$

Let $h = x - \frac{\pi}{6}$. As $x \to \frac{\pi}{6}$, $h$ approaches 0.

The limit becomes:

$\lim\limits_{h \to 0} \frac{2 \sin h}{h}$

Using the property of limits, $\lim\limits_{h \to 0} c \cdot g(h) = c \cdot \lim\limits_{h \to 0} g(h)$:

$= 2 \lim\limits_{h \to 0} \frac{\sin h}{h}$

Using the fundamental trigonometric limit $\lim\limits_{h \to 0} \frac{\sin h}{h} = 1$:

Limit value $= 2 \cdot 1 = 2$

Both methods give the same result. The value of the limit is 2.

Final Answer: $2$

Solution:

We are asked to evaluate the limit:

$\lim\limits_{x \to 0} \frac{\sin 2x + 3x}{2x + \tan 3x}$

When we substitute $x=0$ directly into the expression, we get $\frac{\sin (2 \cdot 0) + 3(0)}{2(0) + \tan (3 \cdot 0)} = \frac{\sin 0 + 0}{0 + \tan 0} = \frac{0 + 0}{0 + 0} = \frac{0}{0}$. This is an indeterminate form of type $\frac{0}{0}$.

To evaluate this limit, we can divide both the numerator and the denominator by $x$. This is allowed because as $x \to 0$, $x \neq 0$.

$\lim\limits_{x \to 0} \frac{\frac{\sin 2x + 3x}{x}}{\frac{2x + \tan 3x}{x}} = \lim\limits_{x \to 0} \frac{\frac{\sin 2x}{x} + \frac{3x}{x}}{\frac{2x}{x} + \frac{\tan 3x}{x}}$

$= \lim\limits_{x \to 0} \frac{\frac{\sin 2x}{x} + 3}{2 + \frac{\tan 3x}{x}}$

Now, we use the standard trigonometric limits $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$ and $\lim\limits_{\theta \to 0} \frac{\tan \theta}{\theta} = 1$.

For the term $\frac{\sin 2x}{x}$, multiply the denominator by 2 and the whole term by 2:

$\frac{\sin 2x}{x} = \frac{\sin 2x}{2x} \cdot 2$

As $x \to 0$, $2x \to 0$, so $\lim\limits_{x \to 0} \frac{\sin 2x}{2x} = 1$. Thus, $\lim\limits_{x \to 0} \frac{\sin 2x}{x} = \lim\limits_{x \to 0} \frac{\sin 2x}{2x} \cdot 2 = 1 \cdot 2 = 2$.

For the term $\frac{\tan 3x}{x}$, multiply the denominator by 3 and the whole term by 3:

$\frac{\tan 3x}{x} = \frac{\tan 3x}{3x} \cdot 3$

As $x \to 0$, $3x \to 0$, so $\lim\limits_{x \to 0} \frac{\tan 3x}{3x} = 1$. Thus, $\lim\limits_{x \to 0} \frac{\tan 3x}{x} = \lim\limits_{x \to 0} \frac{\tan 3x}{3x} \cdot 3 = 1 \cdot 3 = 3$.

Substitute these limit values back into the expression after dividing by $x$:

$\lim\limits_{x \to 0} \frac{\frac{\sin 2x}{x} + 3}{2 + \frac{\tan 3x}{x}} = \frac{\lim\limits_{x \to 0} (\frac{\sin 2x}{x} + 3)}{\lim\limits_{x \to 0} (2 + \frac{\tan 3x}{x})}$

$= \frac{\lim\limits_{x \to 0} \frac{\sin 2x}{x} + \lim\limits_{x \to 0} 3}{\lim\limits_{x \to 0} 2 + \lim\limits_{x \to 0} \frac{\tan 3x}{x}}$

$= \frac{2 + 3}{2 + 3}$

$= \frac{5}{5}$

$= 1$

Therefore, the value of the limit is 1.

Final Answer: $1$

Solution:

We are asked to evaluate the limit:

$\lim\limits_{x \to a} \frac{\sin x − \sin a}{\sqrt{x} − \sqrt{a}}$

When we substitute $x=a$ directly into the expression, we get $\frac{\sin a − \sin a}{\sqrt{a} − \sqrt{a}} = \frac{0}{0}$. This is an indeterminate form of type $\frac{0}{0}$. We assume $a \ge 0$ for the square roots to be defined.

To evaluate this limit, we can use a substitution. Let $y = \sqrt{x}$. Since $x \to a$, assuming $a \ge 0$, $y \to \sqrt{a}$. Also, $x = y^2$.

The limit expression can be rewritten in terms of $y$. Note that $a = (\sqrt{a})^2$.

The limit becomes:

$\lim\limits_{y \to \sqrt{a}} \frac{\sin(y^2) − \sin((\sqrt{a})^2)}{y − \sqrt{a}}$

This limit is in the form of the definition of the derivative of a function. The definition of the derivative of a function $f(y)$ at a point $c$ is given by $\lim\limits_{y \to c} \frac{f(y) - f(c)}{y - c}$.

Let $f(y) = \sin(y^2)$ and $c = \sqrt{a}$. The limit is $\lim\limits_{y \to c} \frac{f(y) - f(c)}{y - c}$, which is equal to $f'(c)$.

We need to find the derivative of $f(y) = \sin(y^2)$ with respect to $y$. Using the chain rule, $\frac{d}{dy}(\sin(g(y))) = \cos(g(y)) \cdot g'(y)$, where $g(y) = y^2$ and $g'(y) = 2y$:

$f'(y) = \frac{d}{dy}(\sin(y^2)) = \cos(y^2) \cdot \frac{d}{dy}(y^2)$

$f'(y) = \cos(y^2) \cdot (2y)$

$f'(y) = 2y \cos(y^2)$

Now, evaluate $f'(\sqrt{a})$ by substituting $y = \sqrt{a}$ into the expression for $f'(y)$:

$f'(\sqrt{a}) = 2(\sqrt{a}) \cos((\sqrt{a})^2)$

$f'(\sqrt{a}) = 2\sqrt{a} \cos a$

Therefore, the value of the limit is $2\sqrt{a} \cos a$. This result is valid for $a \ge 0$. If $a=0$, the limit is 0, and the formula gives $2\sqrt{0}\cos 0 = 0 \cdot 1 = 0$.

Alternate Solution:

We can use trigonometric identities and algebraic manipulation.

Rewrite the numerator using the sum-to-product identity $\sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$, with $A=x$ and $B=a$:

$\sin x - \sin a = 2 \cos\left(\frac{x+a}{2}\right) \sin\left(\frac{x-a}{2}\right)$

Rationalize the denominator by multiplying the numerator and denominator by the conjugate $\sqrt{x} + \sqrt{a}$ (assuming $x, a \ge 0$):

$\frac{\sin x − \sin a}{\sqrt{x} − \sqrt{a}} = \frac{\sin x − \sin a}{\sqrt{x} − \sqrt{a}} \cdot \frac{\sqrt{x} + \sqrt{a}}{\sqrt{x} + \sqrt{a}}$

$= \frac{(\sin x − \sin a)(\sqrt{x} + \sqrt{a})}{(\sqrt{x})^2 − (\sqrt{a})^2} = \frac{(\sin x − \sin a)(\sqrt{x} + \sqrt{a})}{x − a}$

Substitute the identity for the numerator:

$= \frac{2 \cos\left(\frac{x+a}{2}\right) \sin\left(\frac{x-a}{2}\right) (\sqrt{x} + \sqrt{a})}{x − a}$

Rearrange the terms to use the standard limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:

$= 2 \cos\left(\frac{x+a}{2}\right) \cdot \frac{\sin\left(\frac{x-a}{2}\right)}{x − a} \cdot (\sqrt{x} + \sqrt{a})$

Now, evaluate the limit of each factor as $x \to a$:

$\lim\limits_{x \to a} 2 \cos\left(\frac{x+a}{2}\right) = 2 \cos\left(\frac{a+a}{2}\right) = 2 \cos a$ (by direct substitution, as cosine is continuous).

$\lim\limits_{x \to a} (\sqrt{x} + \sqrt{a}) = \sqrt{a} + \sqrt{a} = 2\sqrt{a}$ (by direct substitution, as square root is continuous for $x \ge 0$).

For the term $\lim\limits_{x \to a} \frac{\sin\left(\frac{x-a}{2}\right)}{x − a}$, let $h = x - a$. As $x \to a$, $h \to 0$. The limit becomes $\lim\limits_{h \to 0} \frac{\sin(h/2)}{h}$.

$\lim\limits_{h \to 0} \frac{\sin(h/2)}{h} = \lim\limits_{h \to 0} \frac{\sin(h/2)}{h/2 \cdot 2} = \frac{1}{2} \lim\limits_{h \to 0} \frac{\sin(h/2)}{h/2}$

Using the standard limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$, with $\theta = h/2$:

$= \frac{1}{2} \cdot 1 = \frac{1}{2}$

Multiply the limits of the factors:

Limit value $= (2 \cos a) \cdot \left(\frac{1}{2}\right) \cdot (2\sqrt{a})$

Limit value $= 2\sqrt{a} \cos a$

Both methods yield the same result.

Final Answer: $2\sqrt{a} \cos a$

Solution:

We are asked to evaluate the limit:

$\lim\limits_{x \to \frac{π}{6}} \frac{\cot^2 x − 3}{\text{cosec}\; x − 2}$

When we substitute $x = \frac{\pi}{6}$ directly into the expression, we get:

Numerator: $\cot^2(\frac{\pi}{6}) − 3 = (\sqrt{3})^2 − 3 = 3 − 3 = 0$

Denominator: $\text{cosec}(\frac{\pi}{6}) − 2 = 2 − 2 = 0$

This is an indeterminate form of type $\frac{0}{0}$.

To evaluate this limit, we can use the trigonometric identity $\cot^2 x = \text{cosec}^2 x - 1$. Substitute this into the numerator:

Numerator $= (\text{cosec}^2 x - 1) - 3 = \text{cosec}^2 x - 4$

The numerator is a difference of squares, $\text{cosec}^2 x - 2^2$, which can be factored as $(\text{cosec}\; x - 2)(\text{cosec}\; x + 2)$.

The limit expression becomes:

$\lim\limits_{x \to \frac{π}{6}} \frac{(\text{cosec}\; x − 2)(\text{cosec}\; x + 2)}{\text{cosec}\; x − 2}$

Since $x \to \frac{\pi}{6}$, $x$ is approaching $\frac{\pi}{6}$ but is not equal to $\frac{\pi}{6}$. Therefore, $\text{cosec}\; x$ is approaching $\text{cosec}(\frac{\pi}{6}) = 2$, but $\text{cosec}\; x \neq 2$. Thus, $(\text{cosec}\; x - 2) \neq 0$. We can cancel the common factor $(\text{cosec}\; x - 2)$ from the numerator and the denominator.

The simplified expression is:

$\lim\limits_{x \to \frac{π}{6}} (\text{cosec}\; x + 2)$

Now, substitute $x = \frac{\pi}{6}$ into the simplified expression, as it is a continuous function at $x = \frac{\pi}{6}$ (where $\text{cosec}\; x$ is defined and continuous):

Limit value $= \text{cosec}(\frac{\pi}{6}) + 2$

Limit value $= 2 + 2$

Limit value $= 4$

Alternate Solution (using L'Hopital's Rule):

Since the limit is of the indeterminate form $\frac{0}{0}$, we can apply L'Hopital's Rule.

Let $f(x) = \cot^2 x - 3$ and $g(x) = \text{cosec}\; x - 2$.

We need to find the derivatives $f'(x)$ and $g'(x)$.

$f'(x) = \frac{d}{dx}(\cot^2 x - 3) = 2 \cot x \cdot \frac{d}{dx}(\cot x) - 0$

$f'(x) = 2 \cot x \cdot (-\text{cosec}^2 x) = -2 \cot x \text{cosec}^2 x$

$g'(x) = \frac{d}{dx}(\text{cosec}\; x - 2) = \frac{d}{dx}(\text{cosec}\; x) - 0$

$g'(x) = -\text{cosec}\; x \cot x$

Now, evaluate the limit of the ratio of the derivatives:

$\lim\limits_{x \to \frac{π}{6}} \frac{f'(x)}{g'(x)} = \lim\limits_{x \to \frac{π}{6}} \frac{-2 \cot x \text{cosec}^2 x}{-\text{cosec}\; x \cot x}$

Since $x \to \frac{\pi}{6}$, $x$ is close to $\frac{\pi}{6}$ but not equal to $\frac{\pi}{6}$. At $x = \frac{\pi}{6}$, $\cot x = \sqrt{3} \neq 0$ and $\text{cosec}\; x = 2 \neq 0$. Thus, for $x$ close to $\frac{\pi}{6}$ but not equal, $\cot x \neq 0$ and $\text{cosec}\; x \neq 0$. We can cancel the common factors $-\cot x$ and $\text{cosec}\; x$ from the numerator and the denominator.

$\frac{\cancel{-}2 \cancel{\cot x} \text{cosec}^{\cancel{2}^1} x}{\cancel{-}\cancel{\text{cosec}\; x} \cancel{\cot x}} = 2 \text{cosec}\; x$

The limit is equal to:

$\lim\limits_{x \to \frac{π}{6}} 2 \text{cosec}\; x$

Substitute $x = \frac{\pi}{6}$:

Limit value $= 2 \text{cosec}(\frac{\pi}{6}) = 2 \cdot 2 = 4$

Both methods yield the same result. The value of the limit is 4.

Final Answer: $4$

Solution:

We are asked to evaluate the limit:

$\lim\limits_{x \to 0} \frac{\sqrt{2} − \sqrt{1 + \cos x}}{\sin^2 x}$

When we substitute $x=0$ directly into the expression, we get:

Numerator: $\sqrt{2} − \sqrt{1 + \cos 0} = \sqrt{2} − \sqrt{1 + 1} = \sqrt{2} − \sqrt{2} = 0$

Denominator: $\sin^2 0 = 0^2 = 0$

This is an indeterminate form of type $\frac{0}{0}$.

To evaluate this limit, we can rationalize the numerator by multiplying the numerator and the denominator by the conjugate of the numerator, which is $\sqrt{2} + \sqrt{1 + \cos x}$.

Multiply the expression by $\frac{\sqrt{2} + \sqrt{1 + \cos x}}{\sqrt{2} + \sqrt{1 + \cos x}}$:

$\lim\limits_{x \to 0} \frac{\sqrt{2} − \sqrt{1 + \cos x}}{\sin^2 x} \times \frac{\sqrt{2} + \sqrt{1 + \cos x}}{\sqrt{2} + \sqrt{1 + \cos x}}$

Simplify the numerator using the difference of squares formula $(a-b)(a+b) = a^2 - b^2$, where $a = \sqrt{2}$ and $b = \sqrt{1 + \cos x}$.

Numerator $= (\sqrt{2})^2 - (\sqrt{1 + \cos x})^2$

Numerator $= 2 - (1 + \cos x)$

Numerator $= 2 - 1 - \cos x = 1 - \cos x$

The denominator becomes $\sin^2 x (\sqrt{2} + \sqrt{1 + \cos x})$.

The limit expression is now:

$\lim\limits_{x \to 0} \frac{1 − \cos x}{\sin^2 x (\sqrt{2} + \sqrt{1 + \cos x})}$

We can use trigonometric identities to simplify further. Use the identity $1 - \cos x = 2 \sin^2 \left(\frac{x}{2}\right)$ and $\sin^2 x = 1 - \cos^2 x = (1 - \cos x)(1 + \cos x)$. Using $\sin^2 x = 2 \sin^2 \frac{x}{2} \cos^2 \frac{x}{2}$ is also possible but the first approach is often simpler here.

Substitute $\sin^2 x = 1 - \cos^2 x$ in the denominator:

$\lim\limits_{x \to 0} \frac{1 − \cos x}{(1 − \cos x)(1 + \cos x) (\sqrt{2} + \sqrt{1 + \cos x})}$

Since $x \to 0$, $x$ is approaching 0 but $x \neq 0$. Therefore, $\cos x$ is approaching $\cos 0 = 1$, but $\cos x \neq 1$. Thus, $1 - \cos x \neq 0$. We can cancel the common factor $(1 - \cos x)$ from the numerator and the denominator.

The simplified expression is:

$\lim\limits_{x \to 0} \frac{1}{(1 + \cos x) (\sqrt{2} + \sqrt{1 + \cos x})}$

Now, substitute $x=0$ into the simplified expression, as the denominator is non-zero at $x=0$:

Denominator at $x=0$: $(1 + \cos 0)(\sqrt{2} + \sqrt{1 + \cos 0}) = (1 + 1)(\sqrt{2} + \sqrt{1 + 1}) = (2)(\sqrt{2} + \sqrt{2}) = 2(2\sqrt{2}) = 4\sqrt{2}$.

Limit value $= \frac{1}{(1 + \cos 0) (\sqrt{2} + \sqrt{1 + \cos 0})}$

Limit value $= \frac{1}{(1 + 1)(\sqrt{2} + \sqrt{1 + 1})}$

Limit value $= \frac{1}{2 (\sqrt{2} + \sqrt{2})}$

Limit value $= \frac{1}{2 (2\sqrt{2})}$

Limit value $= \frac{1}{4\sqrt{2}}$

To rationalize the denominator, multiply numerator and denominator by $\sqrt{2}$:

Limit value $= \frac{1}{4\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{4 \cdot 2} = \frac{\sqrt{2}}{8}$

Therefore, the value of the limit is $\frac{\sqrt{2}}{8}$.

Final Answer: $\frac{\sqrt{2}}{8}$

Solution:

We are asked to evaluate the limit:

$\lim\limits_{x \to 0} \frac{\sin x − 2 \sin 3x + \sin 5x}{x}$

This is an indeterminate form $\frac{0}{0}$ at $x=0$.

We can split the fraction and use the standard limit $\lim\limits_{\theta \to 0} \frac{\sin k\theta}{\theta} = k$:

$\lim\limits_{x \to 0} \left( \frac{\sin x}{x} − \frac{2 \sin 3x}{x} + \frac{\sin 5x}{x} \right)$

$= \lim\limits_{x \to 0} \frac{\sin x}{x} − 2 \lim\limits_{x \to 0} \frac{\sin 3x}{x} + \lim\limits_{x \to 0} \frac{\sin 5x}{x}$

Evaluate each limit using $\lim\limits_{x \to 0} \frac{\sin kx}{x} = k$:

$\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$

$\lim\limits_{x \to 0} \frac{\sin 3x}{x} = 3$

$\lim\limits_{x \to 0} \frac{\sin 5x}{x} = 5$

Substitute these values:

Limit value $= 1 − 2(3) + 5$

Limit value $= 1 − 6 + 5$

Limit value $= 0$

Final Answer: $0$

Solution:

We are given the equation:

$\lim\limits_{x \to 1} \frac{x^4 − 1}{x − 1} = \lim\limits_{x \to k} \frac{x^3 − k^3}{x^2 − k^2}$

Let's evaluate the limit on the left-hand side (LHS).

LHS $= \lim\limits_{x \to 1} \frac{x^4 − 1}{x − 1}$

This limit is in the standard form $\lim\limits_{x \to a} \frac{x^n - a^n}{x - a}$, where $a=1$ and $n=4$.

Using the formula $\lim\limits_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}$, we get:

LHS $= 4 \cdot 1^{4-1} = 4 \cdot 1^3 = 4 \cdot 1 = 4$

Now, let's evaluate the limit on the right-hand side (RHS).

RHS $= \lim\limits_{x \to k} \frac{x^3 − k^3}{x^2 − k^2}$

Substituting $x=k$ gives $\frac{k^3 - k^3}{k^2 - k^2} = \frac{0}{0}$, which is an indeterminate form. We need to factor the numerator and the denominator.

Numerator: $x^3 - k^3 = (x - k)(x^2 + kx + k^2)$ (Difference of Cubes)

Denominator: $x^2 - k^2 = (x - k)(x + k)$ (Difference of Squares)

Substitute the factored forms into the limit expression:

RHS $= \lim\limits_{x \to k} \frac{(x - k)(x^2 + kx + k^2)}{(x - k)(x + k)}$

Since $x \to k$, $x$ is approaching $k$ but is not equal to $k$. Thus, $(x-k) \neq 0$. We can cancel the common factor $(x-k)$ from the numerator and the denominator.

RHS $= \lim\limits_{x \to k} \frac{x^2 + kx + k^2}{x + k}$

Now, substitute $x=k$ into the simplified expression. Note that if $k \neq 0$, the denominator $k+k=2k$ is non-zero.

RHS $= \frac{k^2 + k(k) + k^2}{k + k} = \frac{k^2 + k^2 + k^2}{2k} = \frac{3k^2}{2k}$

Assuming $k \neq 0$, we simplify the expression:

RHS $= \frac{3k^{\cancel{2}^1}}{2\cancel{k}} = \frac{3k}{2}$

If $k=0$, the RHS limit would be $\lim\limits_{x \to 0} \frac{x^3}{x^2} = \lim\limits_{x \to 0} x = 0$. Since the LHS is 4, $k=0$ is not a solution.

We are given that LHS = RHS. So, we equate the values we found:

$4 = \frac{3k}{2}$

Multiply both sides by 2:

$4 \times 2 = 3k$

$8 = 3k$

Divide by 3:

$k = \frac{8}{3}$

The value of $k$ is $\frac{8}{3}$.

Final Answer: $k = \frac{8}{3}$

Given:

The function to be differentiated is given by:

$f(x) = \frac{x^4 + x^3 + x^2 + 1}{1}$

To Differentiate:

We need to find the derivative of $f(x)$ with respect to $x$, denoted as $\frac{d}{dx}(f(x))$.

Solution:

The given function can be simplified by dividing the numerator by 1:

$f(x) = x^4 + x^3 + x^2 + 1$

To find the derivative of $f(x)$ with respect to $x$, we differentiate each term of the polynomial separately using the sum rule of differentiation and the power rule.

The sum rule states that $\frac{d}{dx}(u+v+w+c) = \frac{du}{dx} + \frac{dv}{dx} + \frac{dw}{dx} + \frac{dc}{dx}$, where $u, v, w$ are functions of $x$ and $c$ is a constant.

The power rule states that $\frac{d}{dx}(x^n) = nx^{n-1}$ for any real number $n$. Also, the derivative of a constant is zero, i.e., $\frac{d}{dx}(c) = 0$.

Applying these rules to the function $f(x) = x^4 + x^3 + x^2 + 1$, we get:

$\frac{d}{dx}(f(x)) = \frac{d}{dx}(x^4 + x^3 + x^2 + 1)$

$\frac{d}{dx}(f(x)) = \frac{d}{dx}(x^4) + \frac{d}{dx}(x^3) + \frac{d}{dx}(x^2) + \frac{d}{dx}(1)$

Differentiating each term:

$\frac{d}{dx}(x^4) = 4x^{4-1} = 4x^3$

$\frac{d}{dx}(x^3) = 3x^{3-1} = 3x^2$

$\frac{d}{dx}(x^2) = 2x^{2-1} = 2x^1 = 2x$

$\frac{d}{dx}(1) = 0$

Combining the derivatives of each term:

$\frac{d}{dx}(f(x)) = 4x^3 + 3x^2 + 2x + 0$

Thus, the derivative of the given function with respect to $x$ is:

$\frac{d}{dx}\left(\frac{x^4 + x^3 + x^2 + 1}{1}\right) = 4x^3 + 3x^2 + 2x$

Given:

The function to be differentiated is given by:

$f(x) = \left( x + \frac{1}{x} \right)^3$

To Differentiate:

We need to find the derivative of $f(x)$ with respect to $x$, denoted as $\frac{d}{dx}(f(x))$.

Solution:

We can differentiate this function by first expanding the cubic term and then differentiating the resulting polynomial. The formula for $(a+b)^3$ is $a^3 + 3a^2b + 3ab^2 + b^3$.

Let $a = x$ and $b = \frac{1}{x} = x^{-1}$.

$f(x) = \left( x + \frac{1}{x} \right)^3 = x^3 + 3x^2 \left(\frac{1}{x}\right) + 3x \left(\frac{1}{x}\right)^2 + \left(\frac{1}{x}\right)^3$

Simplify the terms:

$f(x) = x^3 + 3x^2 \cdot x^{-1} + 3x \cdot x^{-2} + x^{-3}$

$f(x) = x^3 + 3x^{2-1} + 3x^{1-2} + x^{-3}$

$f(x) = x^3 + 3x^1 + 3x^{-1} + x^{-3}$

$f(x) = x^3 + 3x + \frac{3}{x} + \frac{1}{x^3}$

Now, differentiate $f(x)$ with respect to $x$ using the sum rule and the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$ and $\frac{d}{dx}(c) = 0$ for a constant $c$.

$\frac{d}{dx}(f(x)) = \frac{d}{dx}(x^3 + 3x + 3x^{-1} + x^{-3})$

$\frac{d}{dx}(f(x)) = \frac{d}{dx}(x^3) + \frac{d}{dx}(3x) + \frac{d}{dx}(3x^{-1}) + \frac{d}{dx}(x^{-3})$

Apply the power rule:

$\frac{d}{dx}(x^3) = 3x^{3-1} = 3x^2$

$\frac{d}{dx}(3x) = 3 \cdot \frac{d}{dx}(x^1) = 3 \cdot 1x^{1-1} = 3 \cdot x^0 = 3 \cdot 1 = 3$

$\frac{d}{dx}(3x^{-1}) = 3 \cdot \frac{d}{dx}(x^{-1}) = 3 \cdot (-1)x^{-1-1} = -3x^{-2} = -\frac{3}{x^2}$

$\frac{d}{dx}(x^{-3}) = -3x^{-3-1} = -3x^{-4} = -\frac{3}{x^4}$

Combine the derivatives:

$\frac{d}{dx}(f(x)) = 3x^2 + 3 - \frac{3}{x^2} - \frac{3}{x^4}$

Thus, the derivative of the given function with respect to $x$ is:

$\frac{d}{dx}\left( \left( x + \frac{1}{x} \right)^3 \right) = 3x^2 + 3 - \frac{3}{x^2} - \frac{3}{x^4}$

Alternate Solution (Using Chain Rule):

We can also use the chain rule. Let $u = x + \frac{1}{x}$. Then the function is $f(x) = u^3$.

According to the chain rule, $\frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx}$.

First, find $\frac{df}{du}$:

$f(u) = u^3$

$\frac{df}{du} = \frac{d}{du}(u^3) = 3u^{3-1} = 3u^2$

Substitute back $u = x + \frac{1}{x}$:

$\frac{df}{du} = 3\left(x + \frac{1}{x}\right)^2$

Next, find $\frac{du}{dx}$:

$u = x + \frac{1}{x} = x + x^{-1}$

$\frac{du}{dx} = \frac{d}{dx}(x + x^{-1}) = \frac{d}{dx}(x) + \frac{d}{dx}(x^{-1})$

$\frac{du}{dx} = 1x^{1-1} + (-1)x^{-1-1} = 1x^0 - 1x^{-2} = 1 - x^{-2} = 1 - \frac{1}{x^2}$

Now, multiply $\frac{df}{du}$ and $\frac{du}{dx}$:

$\frac{df}{dx} = 3\left(x + \frac{1}{x}\right)^2 \cdot \left(1 - \frac{1}{x^2}\right)$

Expand the terms:

$\left(x + \frac{1}{x}\right)^2 = x^2 + 2x\left(\frac{1}{x}\right) + \left(\frac{1}{x}\right)^2 = x^2 + 2 + \frac{1}{x^2}$

So, $\frac{df}{dx} = 3\left(x^2 + 2 + \frac{1}{x^2}\right) \left(1 - \frac{1}{x^2}\right)$

Multiply the two binomials:

$\left(x^2 + 2 + \frac{1}{x^2}\right) \left(1 - \frac{1}{x^2}\right) = x^2(1) + x^2\left(-\frac{1}{x^2}\right) + 2(1) + 2\left(-\frac{1}{x^2}\right) + \frac{1}{x^2}(1) + \frac{1}{x^2}\left(-\frac{1}{x^2}\right)$

$= x^2 - 1 + 2 - \frac{2}{x^2} + \frac{1}{x^2} - \frac{1}{x^4}$

$= x^2 + 1 - \frac{1}{x^2} - \frac{1}{x^4}$

Multiply by 3:

$\frac{df}{dx} = 3\left(x^2 + 1 - \frac{1}{x^2} - \frac{1}{x^4}\right)$

$\frac{df}{dx} = 3x^2 + 3 - \frac{3}{x^2} - \frac{3}{x^4}$

Both methods yield the same result.

$\frac{d}{dx}\left( \left( x + \frac{1}{x} \right)^3 \right) = 3x^2 + 3 - \frac{3}{x^2} - \frac{3}{x^4}$

Given:

The function to be differentiated is given by:

$f(x) = (3x + 5)(1 + \tan x)$

To Differentiate:

We need to find the derivative of $f(x)$ with respect to $x$, denoted as $\frac{d}{dx}(f(x))$.

Solution:

The given function is a product of two functions. Let $u(x) = 3x + 5$ and $v(x) = 1 + \tan x$.

We will use the product rule for differentiation, which states that if $f(x) = u(x)v(x)$, then its derivative is given by:

$\frac{d}{dx}(u(x)v(x)) = u'(x)v(x) + u(x)v'(x)$

First, find the derivative of $u(x)$ with respect to $x$:

$u(x) = 3x + 5$

$u'(x) = \frac{d}{dx}(3x + 5)$

$u'(x) = \frac{d}{dx}(3x) + \frac{d}{dx}(5)$

$u'(x) = 3(1) + 0$

$u'(x) = 3$

Next, find the derivative of $v(x)$ with respect to $x$:

$v(x) = 1 + \tan x$

$v'(x) = \frac{d}{dx}(1 + \tan x)$

$v'(x) = \frac{d}{dx}(1) + \frac{d}{dx}(\tan x)$

$v'(x) = 0 + \sec^2 x$

$v'(x) = \sec^2 x$

Now, apply the product rule:

$\frac{d}{dx}(f(x)) = u'(x)v(x) + u(x)v'(x)$

$\frac{d}{dx}(f(x)) = (3)(1 + \tan x) + (3x + 5)(\sec^2 x)$

Expand the expression:

$\frac{d}{dx}(f(x)) = 3 \cdot 1 + 3 \cdot \tan x + (3x + 5)\sec^2 x$

$\frac{d}{dx}(f(x)) = 3 + 3\tan x + (3x + 5)\sec^2 x$

Thus, the derivative of the given function with respect to $x$ is:

$\frac{d}{dx}((3x + 5)(1 + \tan x)) = 3 + 3\tan x + (3x + 5)\sec^2 x$

Given:

The function to be differentiated is given by:

$f(x) = (\sec x - 1)(\sec x + 1)$

To Differentiate:

We need to find the derivative of $f(x)$ with respect to $x$, denoted as $\frac{d}{dx}(f(x))$.

Solution:

We can first simplify the given function using the algebraic identity $(a-b)(a+b) = a^2 - b^2$.

Here, $a = \sec x$ and $b = 1$.

$f(x) = (\sec x)^2 - (1)^2$

$f(x) = \sec^2 x - 1$

Using the trigonometric identity $\sec^2 x - \tan^2 x = 1$, we can rewrite $\sec^2 x - 1$ as $\tan^2 x$.

$f(x) = \tan^2 x$

Now, we need to differentiate $f(x) = \tan^2 x$ with respect to $x$. We can use the chain rule for this.

Let $y = \tan^2 x = (\tan x)^2$. Let $u = \tan x$. Then $y = u^2$.

The chain rule states that $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

First, find $\frac{dy}{du}$:

$\frac{dy}{du} = \frac{d}{du}(u^2) = 2u^{2-1} = 2u$

Substitute back $u = \tan x$:

$\frac{dy}{du} = 2\tan x$

Next, find $\frac{du}{dx}$:

$\frac{du}{dx} = \frac{d}{dx}(\tan x) = \sec^2 x$

Now, multiply $\frac{dy}{du}$ and $\frac{du}{dx}$:

$\frac{dy}{dx} = (2\tan x)(\sec^2 x)$

Thus, the derivative of the given function with respect to $x$ is:

$\frac{d}{dx}((\sec x - 1)(\sec x + 1)) = 2\tan x \sec^2 x$

Alternate Solution:

Alternatively, one could apply the product rule directly to the function $f(x) = (\sec x - 1)(\sec x + 1)$. Using the product rule $\frac{d}{dx}(uv) = u'v + uv'$ with $u = \sec x - 1$ and $v = \sec x + 1$, the derivative can be found. This would involve calculating the derivatives of $u$ and $v$ (which is $\sec x \tan x$ for both) and then applying the product rule formula, which simplifies to the same result: $2\sec x \tan x \sec x = 2\sec^2 x \tan x$.

Given:

The function to be differentiated is given by:

$f(x) = \frac{3x + 4}{5x^2 − 7x + 9}$

To Differentiate:

We need to find the derivative of $f(x)$ with respect to $x$, denoted as $\frac{d}{dx}(f(x))$.

Solution:

The given function is a quotient of two functions. We will use the quotient rule for differentiation, which states that if $f(x) = \frac{u(x)}{v(x)}$, then its derivative is given by:

$\frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$

Let the numerator be $u(x)$ and the denominator be $v(x)$.

$u(x) = 3x + 4$

$v(x) = 5x^2 - 7x + 9$

First, find the derivative of $u(x)$ with respect to $x$:

$u'(x) = \frac{d}{dx}(3x + 4) = \frac{d}{dx}(3x) + \frac{d}{dx}(4) = 3(1) + 0 = 3$

Next, find the derivative of $v(x)$ with respect to $x$:

$v'(x) = \frac{d}{dx}(5x^2 - 7x + 9) = \frac{d}{dx}(5x^2) - \frac{d}{dx}(7x) + \frac{d}{dx}(9)$

$v'(x) = 5(2x^{2-1}) - 7(1x^{1-1}) + 0 = 10x - 7x^0 = 10x - 7$

Now, apply the quotient rule formula:

$\frac{d}{dx}(f(x)) = \frac{(3)(5x^2 - 7x + 9) - (3x + 4)(10x - 7)}{(5x^2 - 7x + 9)^2}$

Simplify the numerator:

Numerator $= 3(5x^2 - 7x + 9) - (3x + 4)(10x - 7)$

Numerator $= (15x^2 - 21x + 27) - (3x \cdot 10x + 3x \cdot (-7) + 4 \cdot 10x + 4 \cdot (-7))$

Numerator $= 15x^2 - 21x + 27 - (30x^2 - 21x + 40x - 28)$

Numerator $= 15x^2 - 21x + 27 - (30x^2 + 19x - 28)$

Numerator $= 15x^2 - 21x + 27 - 30x^2 - 19x + 28$

Combine like terms in the numerator:

Numerator $= (15x^2 - 30x^2) + (-21x - 19x) + (27 + 28)$

Numerator $= -15x^2 - 40x + 55$

Substitute the simplified numerator back into the derivative expression:

$\frac{d}{dx}(f(x)) = \frac{-15x^2 - 40x + 55}{(5x^2 - 7x + 9)^2}$

Thus, the derivative of the given function with respect to $x$ is:

$\frac{d}{dx}\left(\frac{3x + 4}{5x^2 − 7x + 9}\right) = \frac{-15x^2 - 40x + 55}{(5x^2 - 7x + 9)^2}$

Given:

The function to be differentiated is given by:

$f(x) = \frac{x^5 - \cos x}{\sin x}$

To Differentiate:

We need to find the derivative of $f(x)$ with respect to $x$, denoted as $\frac{d}{dx}(f(x))$.

Solution:

The given function is a quotient of two functions. We will use the quotient rule for differentiation, which states that if $f(x) = \frac{u(x)}{v(x)}$, then its derivative is given by:

$\frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$

Let the numerator be $u(x)$ and the denominator be $v(x)$.

$u(x) = x^5 - \cos x$

$v(x) = \sin x$

First, find the derivative of $u(x)$ with respect to $x$:

$u'(x) = \frac{d}{dx}(x^5 - \cos x)$

$u'(x) = \frac{d}{dx}(x^5) - \frac{d}{dx}(\cos x)$

$u'(x) = 5x^{5-1} - (-\sin x) = 5x^4 + \sin x$

Next, find the derivative of $v(x)$ with respect to $x$:

$v'(x) = \frac{d}{dx}(\sin x) = \cos x$

Now, apply the quotient rule formula:

$\frac{d}{dx}(f(x)) = \frac{(5x^4 + \sin x)(\sin x) - (x^5 - \cos x)(\cos x)}{(\sin x)^2}$

Simplify the numerator:

Numerator $= (5x^4 \sin x + \sin^2 x) - (x^5 \cos x - \cos^2 x)$

Numerator $= 5x^4 \sin x + \sin^2 x - x^5 \cos x + \cos^2 x$

Rearrange and use the identity $\sin^2 x + \cos^2 x = 1$:

Numerator $= 5x^4 \sin x - x^5 \cos x + (\sin^2 x + \cos^2 x)$

Numerator $= 5x^4 \sin x - x^5 \cos x + 1$

Substitute the simplified numerator back into the derivative expression:

$\frac{d}{dx}(f(x)) = \frac{5x^4 \sin x - x^5 \cos x + 1}{\sin^2 x}$

Thus, the derivative of the given function with respect to $x$ is:

$\frac{d}{dx}\left(\frac{x^5 − \cos x}{\sin x}\right) = \frac{5x^4 \sin x - x^5 \cos x + 1}{\sin^2 x}$

Alternate form of the Solution:

The derivative can also be written by separating the terms in the numerator:

$\frac{d}{dx}(f(x)) = \frac{5x^4 \sin x}{\sin^2 x} - \frac{x^5 \cos x}{\sin^2 x} + \frac{1}{\sin^2 x}$

Using $\frac{1}{\sin x} = \text{cosec } x$ and $\frac{\cos x}{\sin x} = \cot x$:

$\frac{d}{dx}(f(x)) = 5x^4 \left(\frac{1}{\sin x}\right) - x^5 \left(\frac{\cos x}{\sin x}\right) \left(\frac{1}{\sin x}\right) + \left(\frac{1}{\sin x}\right)^2$

$\frac{d}{dx}(f(x)) = 5x^4 \text{cosec } x - x^5 \cot x \text{ cosec } x + \text{cosec}^2 x$

$\frac{d}{dx}\left(\frac{x^5 − \cos x}{\sin x}\right) = 5x^4 \text{cosec } x - x^5 \cot x \text{ cosec } x + \text{cosec}^2 x$

Given:

The function to be differentiated is given by:

$f(x) = \frac{x^2 \cos \frac{π}{4}}{\sin x}$

To Differentiate:

We need to find the derivative of $f(x)$ with respect to $x$, denoted as $\frac{d}{dx}(f(x))$.

Solution:

First, we recognize that $\cos \frac{π}{4}$ is a constant value, which is $\frac{\sqrt{2}}{2}$.

So, the function can be written as:

$f(x) = \frac{x^2 \cdot \frac{\sqrt{2}}{2}}{\sin x} = \frac{\frac{\sqrt{2}}{2} x^2}{\sin x}$

This is a quotient of two functions. Let $u(x) = \frac{\sqrt{2}}{2} x^2$ and $v(x) = \sin x$.

We will use the quotient rule for differentiation:

$\frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$

First, find the derivative of $u(x)$ with respect to $x$:

$u(x) = \frac{\sqrt{2}}{2} x^2$

$u'(x) = \frac{d}{dx}\left(\frac{\sqrt{2}}{2} x^2\right) = \frac{\sqrt{2}}{2} \frac{d}{dx}(x^2)$

$u'(x) = \frac{\sqrt{2}}{2} (2x) = \sqrt{2}x$

Next, find the derivative of $v(x)$ with respect to $x$:

$v(x) = \sin x$

$v'(x) = \frac{d}{dx}(\sin x) = \cos x$

Now, apply the quotient rule formula:

$\frac{d}{dx}(f(x)) = \frac{(\sqrt{2}x)(\sin x) - \left(\frac{\sqrt{2}}{2} x^2\right)(\cos x)}{(\sin x)^2}$

Simplify the numerator:

Numerator $= \sqrt{2}x \sin x - \frac{\sqrt{2}}{2} x^2 \cos x$

Substitute the simplified numerator back into the derivative expression:

$\frac{d}{dx}(f(x)) = \frac{\sqrt{2}x \sin x - \frac{\sqrt{2}}{2} x^2 \cos x}{\sin^2 x}$

We can factor out $\frac{\sqrt{2}}{2}x$ from the numerator:

$\frac{d}{dx}(f(x)) = \frac{\frac{\sqrt{2}}{2} x (2\sin x - x \cos x)}{\sin^2 x}$

Alternatively, we can split the fraction and use trigonometric identities:

$\frac{d}{dx}(f(x)) = \frac{\sqrt{2}x \sin x}{\sin^2 x} - \frac{\frac{\sqrt{2}}{2} x^2 \cos x}{\sin^2 x}$

$\frac{d}{dx}(f(x)) = \frac{\sqrt{2}x}{\sin x} - \frac{\sqrt{2}}{2} x^2 \frac{\cos x}{\sin x} \frac{1}{\sin x}$

Using $\frac{1}{\sin x} = \text{cosec } x$ and $\frac{\cos x}{\sin x} = \cot x$:

$\frac{d}{dx}(f(x)) = \sqrt{2}x \text{ cosec } x - \frac{\sqrt{2}}{2} x^2 \cot x \text{ cosec } x$

Thus, the derivative of the given function with respect to $x$ is:

$\frac{d}{dx}\left(\frac{x^2 \cos \frac{π}{4}}{\sin x}\right) = \frac{\sqrt{2}x \sin x - \frac{\sqrt{2}}{2} x^2 \cos x}{\sin^2 x}$

or

$\frac{d}{dx}\left(\frac{x^2 \cos \frac{π}{4}}{\sin x}\right) = \sqrt{2}x \text{ cosec } x - \frac{\sqrt{2}}{2} x^2 \cot x \text{ cosec } x$

Given:

The function to be differentiated is given by:

$f(x) = (ax^2 + \cot x)(p + q \cos x)$

To Differentiate:

We need to find the derivative of $f(x)$ with respect to $x$, denoted as $\frac{d}{dx}(f(x))$.

Solution:

The given function is a product of two functions. Let $u(x) = ax^2 + \cot x$ and $v(x) = p + q \cos x$, where $a$, $p$, and $q$ are constants.

We will use the product rule for differentiation, which states that if $f(x) = u(x)v(x)$, then its derivative is given by:

$\frac{d}{dx}(u(x)v(x)) = u'(x)v(x) + u(x)v'(x)$

First, find the derivative of $u(x)$ with respect to $x$:

$u(x) = ax^2 + \cot x$

$u'(x) = \frac{d}{dx}(ax^2 + \cot x)$

$u'(x) = \frac{d}{dx}(ax^2) + \frac{d}{dx}(\cot x)$

$u'(x) = a(2x^{2-1}) + (-\text{cosec}^2 x) = 2ax - \text{cosec}^2 x$

Next, find the derivative of $v(x)$ with respect to $x$:

$v(x) = p + q \cos x$

$v'(x) = \frac{d}{dx}(p + q \cos x)$

$v'(x) = \frac{d}{dx}(p) + \frac{d}{dx}(q \cos x)$

$v'(x) = 0 + q(-\sin x) = -q \sin x$

Now, apply the product rule formula:

$\frac{d}{dx}(f(x)) = (2ax - \text{cosec}^2 x)(p + q \cos x) + (ax^2 + \cot x)(-q \sin x)$

Expand the terms:

$\frac{d}{dx}(f(x)) = (2ax)(p) + (2ax)(q \cos x) + (-\text{cosec}^2 x)(p) + (-\text{cosec}^2 x)(q \cos x) + (ax^2)(-q \sin x) + (\cot x)(-q \sin x)$

$\frac{d}{dx}(f(x)) = 2apx + 2aqx \cos x - p \text{ cosec}^2 x - q \text{ cosec}^2 x \cos x - aqx^2 \sin x - q \cot x \sin x$

Recall that $\cot x = \frac{\cos x}{\sin x}$. Substitute this into the last term:

$- q \cot x \sin x = - q \left(\frac{\cos x}{\sin x}\right) \sin x = -q \cos x$

Substitute this back into the derivative expression:

$\frac{d}{dx}(f(x)) = 2apx + 2aqx \cos x - p \text{ cosec}^2 x - q \text{ cosec}^2 x \cos x - aqx^2 \sin x - q \cos x$

Thus, the derivative of the given function with respect to $x$ is:

$\frac{d}{dx}((ax^2 + \cot x)(p + q \cos x)) = 2apx + 2aqx \cos x - p \text{ cosec}^2 x - q \text{ cosec}^2 x \cos x - aqx^2 \sin x - q \cos x$

Given:

The function to be differentiated is given by:

$f(x) = \frac{a + b \sin x}{c + d \cos x}$

where $a$, $b$, $c$, and $d$ are constants.

To Differentiate:

We need to find the derivative of $f(x)$ with respect to $x$, denoted as $\frac{d}{dx}(f(x))$.

Solution:

The given function is a quotient of two functions. We will use the quotient rule for differentiation, which states that if $f(x) = \frac{u(x)}{v(x)}$, then its derivative is given by:

$\frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$

Let the numerator be $u(x)$ and the denominator be $v(x)$.

$u(x) = a + b \sin x$

$v(x) = c + d \cos x$

First, find the derivative of $u(x)$ with respect to $x$:

$u'(x) = \frac{d}{dx}(a + b \sin x)$

$u'(x) = \frac{d}{dx}(a) + \frac{d}{dx}(b \sin x)$

$u'(x) = 0 + b \frac{d}{dx}(\sin x)$

$u'(x) = b \cos x$

Next, find the derivative of $v(x)$ with respect to $x$:

$v(x) = c + d \cos x$

$v'(x) = \frac{d}{dx}(c + d \cos x)$

$v'(x) = \frac{d}{dx}(c) + \frac{d}{dx}(d \cos x)$

$v'(x) = 0 + d \frac{d}{dx}(\cos x)$

$v'(x) = d (-\sin x)$

$v'(x) = -d \sin x$

Now, apply the quotient rule formula:

$\frac{d}{dx}(f(x)) = \frac{(b \cos x)(c + d \cos x) - (a + b \sin x)(-d \sin x)}{(c + d \cos x)^2}$

Simplify the numerator:

Numerator $= (b \cos x)(c) + (b \cos x)(d \cos x) - [(a)(-d \sin x) + (b \sin x)(-d \sin x)]$

Numerator $= bc \cos x + bd \cos^2 x - [-ad \sin x - bd \sin^2 x]$

Numerator $= bc \cos x + bd \cos^2 x + ad \sin x + bd \sin^2 x$

Group terms with $bd$ and use the identity $\cos^2 x + \sin^2 x = 1$:

Numerator $= bc \cos x + ad \sin x + bd (\cos^2 x + \sin^2 x)$

Numerator $= bc \cos x + ad \sin x + bd(1)$

Numerator $= bc \cos x + ad \sin x + bd$

Substitute the simplified numerator back into the derivative expression:

$\frac{d}{dx}(f(x)) = \frac{bc \cos x + ad \sin x + bd}{(c + d \cos x)^2}$

Thus, the derivative of the given function with respect to $x$ is:

$\frac{d}{dx}\left(\frac{a + b \sin x}{c + d \cos x}\right) = \frac{bc \cos x + ad \sin x + bd}{(c + d \cos x)^2}$

Given:

The function to be differentiated is given by:

$f(x) = (\sin x + \cos x)^2$

To Differentiate:

We need to find the derivative of $f(x)$ with respect to $x$, denoted as $\frac{d}{dx}(f(x))$.

Solution (Method 1: Simplify First):

We can first simplify the given function by expanding the square. Using the identity $(a+b)^2 = a^2 + 2ab + b^2$, with $a = \sin x$ and $b = \cos x$:

$f(x) = \sin^2 x + 2 \sin x \cos x + \cos^2 x$

Using the trigonometric identity $\sin^2 x + \cos^2 x = 1$, we can simplify the expression:

$f(x) = (\sin^2 x + \cos^2 x) + 2 \sin x \cos x$

$f(x) = 1 + 2 \sin x \cos x$

Using the double angle identity $2 \sin x \cos x = \sin(2x)$, we get:

$f(x) = 1 + \sin(2x)$

Now, we differentiate $f(x)$ with respect to $x$ using the sum rule and the chain rule for $\sin(2x)$.

$\frac{d}{dx}(f(x)) = \frac{d}{dx}(1 + \sin(2x))$

$\frac{d}{dx}(f(x)) = \frac{d}{dx}(1) + \frac{d}{dx}(\sin(2x))$

The derivative of a constant is 0:

$\frac{d}{dx}(1) = 0$

For the term $\frac{d}{dx}(\sin(2x))$, let $u = 2x$. Then $\sin(2x) = \sin u$. Using the chain rule $\frac{d}{dx}(\sin u) = \frac{d}{du}(\sin u) \cdot \frac{du}{dx}$:

$\frac{d}{du}(\sin u) = \cos u$

$\frac{du}{dx} = \frac{d}{dx}(2x) = 2$

So, $\frac{d}{dx}(\sin(2x)) = (\cos u)(2) = 2 \cos(2x)$ (substituting $u = 2x$).

Combining the derivatives of the terms:

$\frac{d}{dx}(f(x)) = 0 + 2 \cos(2x)$

$\frac{d}{dx}(f(x)) = 2 \cos(2x)$

Thus, the derivative of the given function with respect to $x$ is:

$\frac{d}{dx}((\sin x + \cos x)^2) = 2 \cos(2x)$

Alternate Solution (Method 2: Using Chain Rule Directly):

Let $y = (\sin x + \cos x)^2$. Let $u = \sin x + \cos x$. Then $y = u^2$.

Using the chain rule $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$:

First, find $\frac{dy}{du}$:

$\frac{dy}{du} = \frac{d}{du}(u^2) = 2u$

Next, find $\frac{du}{dx}$:

$u = \sin x + \cos x$

$\frac{du}{dx} = \frac{d}{dx}(\sin x + \cos x) = \frac{d}{dx}(\sin x) + \frac{d}{dx}(\cos x) = \cos x - \sin x$

Now, multiply $\frac{dy}{du}$ and $\frac{du}{dx}$:

$\frac{dy}{dx} = (2u)(\cos x - \sin x)$

Substitute back $u = \sin x + \cos x$:

$\frac{dy}{dx} = 2(\sin x + \cos x)(\cos x - \sin x)$

Rearrange the second factor as $(\cos x - \sin x)$ to use the difference of squares formula $(a+b)(a-b) = a^2 - b^2$ with $a = \cos x$ and $b = \sin x$:

$\frac{dy}{dx} = 2(\cos x + \sin x)(\cos x - \sin x)$

$\frac{dy}{dx} = 2(\cos^2 x - \sin^2 x)$

Using the double angle identity $\cos(2x) = \cos^2 x - \sin^2 x$:

$\frac{dy}{dx} = 2 \cos(2x)$

Both methods yield the same result.

$\frac{d}{dx}((\sin x + \cos x)^2) = 2 \cos(2x)$

Given:

The function to be differentiated is given by:

$f(x) = (2x - 7)^2 (3x + 5)^3$

To Differentiate:

We need to find the derivative of $f(x)$ with respect to $x$, denoted as $\frac{d}{dx}(f(x))$.

Solution:

The given function is a product of two functions. We will use the product rule for differentiation, which states that if $f(x) = u(x)v(x)$, then its derivative is given by:

$\frac{d}{dx}(u(x)v(x)) = u'(x)v(x) + u(x)v'(x)$

Let $u(x) = (2x - 7)^2$ and $v(x) = (3x + 5)^3$.

First, find the derivative of $u(x)$ with respect to $x$ using the chain rule:

$u'(x) = \frac{d}{dx}((2x - 7)^2)$

Using the chain rule, $\frac{d}{dx}(g(h(x))) = g'(h(x)) \cdot h'(x)$. Here $g(w) = w^2$ and $h(x) = 2x - 7$.

$g'(w) = 2w$

$h'(x) = \frac{d}{dx}(2x - 7) = 2$

So, $u'(x) = 2(2x - 7) \cdot 2 = 4(2x - 7)$.

Next, find the derivative of $v(x)$ with respect to $x$ using the chain rule:

$v'(x) = \frac{d}{dx}((3x + 5)^3)$

Here $g(z) = z^3$ and $h(x) = 3x + 5$.

$g'(z) = 3z^2$

$h'(x) = \frac{d}{dx}(3x + 5) = 3$

So, $v'(x) = 3(3x + 5)^2 \cdot 3 = 9(3x + 5)^2$.

Now, apply the product rule formula:

$\frac{d}{dx}(f(x)) = u'(x)v(x) + u(x)v'(x)$

$\frac{d}{dx}(f(x)) = [4(2x - 7)][(3x + 5)^3] + [(2x - 7)^2][9(3x + 5)^2]$

Factor out the common terms $(2x - 7)$ and $(3x + 5)^2$:

$\frac{d}{dx}(f(x)) = (2x - 7)(3x + 5)^2 [4(3x + 5) + 9(2x - 7)]$

Simplify the expression inside the square brackets:

$4(3x + 5) + 9(2x - 7) = 12x + 20 + 18x - 63$

$= (12x + 18x) + (20 - 63)$

$= 30x - 43$

Substitute this back into the factored expression:

$\frac{d}{dx}(f(x)) = (2x - 7)(3x + 5)^2 (30x - 43)$

Thus, the derivative of the given function with respect to $x$ is:

$\frac{d}{dx}((2x - 7)^2 (3x + 5)^3) = (2x - 7)(3x + 5)^2 (30x - 43)$

Given:

The function to be differentiated is given by:

$f(x) = x^2 \sin x + \cos 2x$

To Differentiate:

We need to find the derivative of $f(x)$ with respect to $x$, denoted as $\frac{d}{dx}(f(x))$.

Solution:

The given function is a sum of two terms: $x^2 \sin x$ and $\cos 2x$. We will use the sum rule for differentiation, which states that $\frac{d}{dx}(u(x) + v(x)) = \frac{du}{dx} + \frac{dv}{dx}$.

So, $\frac{d}{dx}(f(x)) = \frac{d}{dx}(x^2 \sin x) + \frac{d}{dx}(\cos 2x)$.

First, let's differentiate the term $x^2 \sin x$. This is a product of two functions, $u = x^2$ and $v = \sin x$. We use the product rule: $\frac{d}{dx}(uv) = u'v + uv'$.

Let $u = x^2$. Then $u' = \frac{d}{dx}(x^2) = 2x$.

Let $v = \sin x$. Then $v' = \frac{d}{dx}(\sin x) = \cos x$.

So, $\frac{d}{dx}(x^2 \sin x) = (2x)(\sin x) + (x^2)(\cos x) = 2x \sin x + x^2 \cos x$.

Next, let's differentiate the term $\cos 2x$. We use the chain rule for this. Let $w = 2x$. Then $\cos 2x = \cos w$. The chain rule states $\frac{d}{dx}(\cos w) = \frac{d}{dw}(\cos w) \cdot \frac{dw}{dx}$.

$\frac{d}{dw}(\cos w) = -\sin w$

$\frac{dw}{dx} = \frac{d}{dx}(2x) = 2$

So, $\frac{d}{dx}(\cos 2x) = (-\sin w)(2) = -2 \sin(2x)$ (substituting back $w = 2x$).

Now, combine the derivatives of the two terms:

$\frac{d}{dx}(f(x)) = (2x \sin x + x^2 \cos x) + (-2 \sin(2x))$

$\frac{d}{dx}(f(x)) = 2x \sin x + x^2 \cos x - 2 \sin(2x)$

Thus, the derivative of the given function with respect to $x$ is:

$\frac{d}{dx}(x^2 \sin x + \cos 2x) = 2x \sin x + x^2 \cos x - 2 \sin(2x)$

Given:

The function to be differentiated is given by:

$f(x) = \sin^3 x \cos^3 x$

To Differentiate:

We need to find the derivative of $f(x)$ with respect to $x$, denoted as $\frac{d}{dx}(f(x))$.

Solution:

We can first simplify the function using trigonometric identities before differentiating. The function can be written as:

$f(x) = (\sin x \cos x)^3$

We know the double angle identity for sine: $\sin(2x) = 2 \sin x \cos x$. From this, we have $\sin x \cos x = \frac{1}{2} \sin(2x)$.

Substitute this into the expression for $f(x)$:

$f(x) = \left(\frac{1}{2} \sin(2x)\right)^3$

$f(x) = \left(\frac{1}{2}\right)^3 (\sin(2x))^3$

$f(x) = \frac{1}{8} \sin^3(2x)$

Now, we differentiate $f(x) = \frac{1}{8} \sin^3(2x)$ with respect to $x$ using the chain rule. We have a constant factor $\frac{1}{8}$ and a composite function $\sin^3(2x)$.

Let $y = \frac{1}{8} (\sin(2x))^3$. Let $u = \sin(2x)$. Then $y = \frac{1}{8} u^3$.

Using the chain rule, $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

First, find $\frac{dy}{du}$:

$\frac{dy}{du} = \frac{d}{du}\left(\frac{1}{8} u^3\right) = \frac{1}{8} \cdot 3u^{3-1} = \frac{3}{8} u^2$

Substitute back $u = \sin(2x)$:

$\frac{dy}{du} = \frac{3}{8} (\sin(2x))^2 = \frac{3}{8} \sin^2(2x)$

Next, find $\frac{du}{dx}$ where $u = \sin(2x)$. We need to use the chain rule again. Let $w = 2x$. Then $u = \sin w$. Using $\frac{du}{dx} = \frac{du}{dw} \cdot \frac{dw}{dx}$:

$\frac{du}{dw} = \frac{d}{dw}(\sin w) = \cos w$

$\frac{dw}{dx} = \frac{d}{dx}(2x) = 2$

So, $\frac{du}{dx} = (\cos w)(2) = 2 \cos(2x)$ (substituting back $w = 2x$).

Now, multiply $\frac{dy}{du}$ and $\frac{du}{dx}$ to get $\frac{dy}{dx}$:

$\frac{dy}{dx} = \left(\frac{3}{8} \sin^2(2x)\right) (2 \cos(2x))$

$\frac{dy}{dx} = \frac{3}{8} \cdot 2 \cdot \sin^2(2x) \cos(2x)$

$\frac{dy}{dx} = \frac{6}{8} \sin^2(2x) \cos(2x)$

$\frac{dy}{dx} = \frac{3}{4} \sin^2(2x) \cos(2x)$

Thus, the derivative of the given function with respect to $x$ is:

$\frac{d}{dx}(\sin^3 x \cos^3 x) = \frac{3}{4} \sin^2(2x) \cos(2x)$

Given:

The function to be differentiated is given by:

$f(x) = \frac{1}{ax^2 + bx + c}$

where $a$, $b$, and $c$ are constants.

To Differentiate:

We need to find the derivative of $f(x)$ with respect to $x$, denoted as $\frac{d}{dx}(f(x))$.

Solution (Method 1: Using Chain Rule):

We can rewrite the function using a negative exponent:

$f(x) = (ax^2 + bx + c)^{-1}$

We will use the chain rule to differentiate this function. Let $u = ax^2 + bx + c$. Then $f(x) = u^{-1}$.

The chain rule states that $\frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx}$.

First, find the derivative of $f$ with respect to $u$:

$\frac{df}{du} = \frac{d}{du}(u^{-1}) = -1 \cdot u^{-1-1} = -u^{-2} = -\frac{1}{u^2}$

Next, find the derivative of $u$ with respect to $x$:

$u(x) = ax^2 + bx + c$

$\frac{du}{dx} = \frac{d}{dx}(ax^2 + bx + c) = \frac{d}{dx}(ax^2) + \frac{d}{dx}(bx) + \frac{d}{dx}(c)$

$\frac{du}{dx} = a(2x) + b(1) + 0 = 2ax + b$

Now, apply the chain rule formula $\frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx}$:

$\frac{df}{dx} = \left(-\frac{1}{u^2}\right) (2ax + b)$

Substitute back $u = ax^2 + bx + c$:

$\frac{df}{dx} = -\frac{1}{(ax^2 + bx + c)^2} (2ax + b)$

$\frac{df}{dx} = -\frac{2ax + b}{(ax^2 + bx + c)^2}$

Thus, the derivative of the given function with respect to $x$ is:

$\frac{d}{dx}\left(\frac{1}{ax^2 + bx + c}\right) = -\frac{2ax + b}{(ax^2 + bx + c)^2}$

Alternate Solution (Method 2: Using Quotient Rule):

The given function is in the form $f(x) = \frac{u(x)}{v(x)}$, where $u(x) = 1$ and $v(x) = ax^2 + bx + c$.

We use the quotient rule: $\frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$.

First, find the derivative of the numerator $u(x) = 1$:

$u'(x) = \frac{d}{dx}(1) = 0$

Next, find the derivative of the denominator $v(x) = ax^2 + bx + c$:

$v'(x) = \frac{d}{dx}(ax^2 + bx + c) = 2ax + b$

Now, apply the quotient rule formula:

$\frac{d}{dx}(f(x)) = \frac{(0)(ax^2 + bx + c) - (1)(2ax + b)}{(ax^2 + bx + c)^2}$

Simplify the numerator:

Numerator $= 0 - (2ax + b) = -(2ax + b)$

Substitute the simplified numerator back into the derivative expression:

$\frac{d}{dx}(f(x)) = \frac{-(2ax + b)}{(ax^2 + bx + c)^2}$

$\frac{d}{dx}(f(x)) = -\frac{2ax + b}{(ax^2 + bx + c)^2}$

Both methods yield the same result.

$\frac{d}{dx}\left(\frac{1}{ax^2 + bx + c}\right) = -\frac{2ax + b}{(ax^2 + bx + c)^2}$

Given:

The function to be differentiated is $f(x) = \cos(x^2 + 1)$.

To Differentiate using first principle:

We need to find the derivative of $f(x)$ with respect to $x$ using the definition of the derivative (first principle).

Solution:

The derivative of a function $f(x)$ with respect to $x$ using the first principle is given by:

$f'(x) = \lim\limits_{h \to 0} \frac{f(x + h) - f(x)}{h}$

Here, $f(x) = \cos(x^2 + 1)$.

First, we find $f(x + h)$:

$f(x + h) = \cos((x + h)^2 + 1)$

$f(x + h) = \cos(x^2 + 2xh + h^2 + 1)$

Now, we find the difference $f(x + h) - f(x)$:

$f(x + h) - f(x) = \cos(x^2 + 2xh + h^2 + 1) - \cos(x^2 + 1)$

We use the trigonometric identity $\cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$.

Let $A = x^2 + 2xh + h^2 + 1$ and $B = x^2 + 1$.

$\frac{A+B}{2} = \frac{(x^2 + 2xh + h^2 + 1) + (x^2 + 1)}{2} = \frac{2x^2 + 2xh + h^2 + 2}{2} = x^2 + xh + \frac{h^2}{2} + 1$

$\frac{A-B}{2} = \frac{(x^2 + 2xh + h^2 + 1) - (x^2 + 1)}{2} = \frac{2xh + h^2}{2} = xh + \frac{h^2}{2}$

So, $f(x + h) - f(x) = -2 \sin\left(x^2 + xh + \frac{h^2}{2} + 1\right) \sin\left(xh + \frac{h^2}{2}\right)$.

Now, we form the difference quotient:

$\frac{f(x + h) - f(x)}{h} = \frac{-2 \sin\left(x^2 + xh + \frac{h^2}{2} + 1\right) \sin\left(xh + \frac{h^2}{2}\right)}{h}$

We rearrange the terms to use the standard limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$.

$\frac{f(x + h) - f(x)}{h} = -2 \sin\left(x^2 + xh + \frac{h^2}{2} + 1\right) \cdot \frac{\sin\left(xh + \frac{h^2}{2}\right)}{h}$

To apply the limit, we multiply the numerator and denominator of the second fraction by $xh + \frac{h^2}{2}$:

$\frac{f(x + h) - f(x)}{h} = -2 \sin\left(x^2 + xh + \frac{h^2}{2} + 1\right) \cdot \frac{\sin\left(xh + \frac{h^2}{2}\right)}{xh + \frac{h^2}{2}} \cdot \frac{xh + \frac{h^2}{2}}{h}$

Simplify the last term:

$\frac{xh + \frac{h^2}{2}}{h} = \frac{h(x + \frac{h}{2})}{h} = x + \frac{h}{2}$

So, $\frac{f(x + h) - f(x)}{h} = -2 \sin\left(x^2 + xh + \frac{h^2}{2} + 1\right) \cdot \frac{\sin\left(xh + \frac{h^2}{2}\right)}{xh + \frac{h^2}{2}} \cdot \left(x + \frac{h}{2}\right)$

Now, we take the limit as $h \to 0$:

$f'(x) = \lim\limits_{h \to 0} \left[-2 \sin\left(x^2 + xh + \frac{h^2}{2} + 1\right) \cdot \frac{\sin\left(xh + \frac{h^2}{2}\right)}{xh + \frac{h^2}{2}} \cdot \left(x + \frac{h}{2}\right)\right]$

As $h \to 0$, we have:

$\lim\limits_{h \to 0} \left(x^2 + xh + \frac{h^2}{2} + 1\right) = x^2 + x(0) + \frac{0^2}{2} + 1 = x^2 + 1$

Let $\theta = xh + \frac{h^2}{2}$. As $h \to 0$, $\theta \to 0$. So, $\lim\limits_{h \to 0} \frac{\sin\left(xh + \frac{h^2}{2}\right)}{xh + \frac{h^2}{2}} = \lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$

$\lim\limits_{h \to 0} \left(x + \frac{h}{2}\right) = x + \frac{0}{2} = x$

Substituting these limits:

$f'(x) = -2 \sin(x^2 + 1) \cdot 1 \cdot x$

$f'(x) = -2x \sin(x^2 + 1)$

Thus, the derivative of $\cos(x^2 + 1)$ with respect to $x$ using the first principle is:

$\frac{d}{dx}(\cos(x^2 + 1)) = -2x \sin(x^2 + 1)$

Given:

The function to be differentiated is $f(x) = \frac{ax + b}{cx + d}$, where $a, b, c, d$ are constants and $cx + d \neq 0$.

To Differentiate using first principle:

We need to find the derivative of $f(x)$ with respect to $x$ using the definition of the derivative (first principle).

Solution:

The derivative of a function $f(x)$ with respect to $x$ using the first principle is given by:

$f'(x) = \lim\limits_{h \to 0} \frac{f(x + h) - f(x)}{h}$

Here, $f(x) = \frac{ax + b}{cx + d}$.

First, we find $f(x + h)$ by replacing $x$ with $x+h$:

$f(x + h) = \frac{a(x + h) + b}{c(x + h) + d} = \frac{ax + ah + b}{cx + ch + d}$

Now, we find the difference $f(x + h) - f(x)$:

$f(x + h) - f(x) = \frac{ax + ah + b}{cx + ch + d} - \frac{ax + b}{cx + d}$

To subtract the fractions, we find a common denominator, which is $(cx + ch + d)(cx + d)$.

$f(x + h) - f(x) = \frac{(ax + ah + b)(cx + d) - (ax + b)(cx + ch + d)}{(cx + ch + d)(cx + d)}$

Expand the numerator:

Numerator $= (ax + ah + b)(cx + d) - (ax + b)(cx + ch + d)$

Numerator $= (acx^2 + adx + achx + ahd + bcx + bd) - (acx^2 + achx + adx + bcx + bch + bd)$

Numerator $= acx^2 + adx + achx + ahd + bcx + bd - acx^2 - achx - adx - bcx - bch - bd$

Cancel out the like terms:

Numerator $= ahd - bch$

Numerator $= h(ad - bc)$

Now, we form the difference quotient:

$\frac{f(x + h) - f(x)}{h} = \frac{\frac{h(ad - bc)}{(cx + ch + d)(cx + d)}}{h}$

$\frac{f(x + h) - f(x)}{h} = \frac{h(ad - bc)}{h(cx + ch + d)(cx + d)}$

Cancel the $h$ term in the numerator and denominator (since $h \to 0$, $h \neq 0$):

$\frac{f(x + h) - f(x)}{h} = \frac{ad - bc}{(cx + ch + d)(cx + d)}$

Finally, we take the limit as $h \to 0$:

$f'(x) = \lim\limits_{h \to 0} \frac{ad - bc}{(cx + ch + d)(cx + d)}$

As $h \to 0$, the term $ch$ in the denominator approaches $c(0) = 0$.

$f'(x) = \frac{ad - bc}{(cx + c(0) + d)(cx + d)}$

$f'(x) = \frac{ad - bc}{(cx + d)(cx + d)}$

$f'(x) = \frac{ad - bc}{(cx + d)^2}$

Thus, the derivative of $\frac{ax + b}{cx + d}$ with respect to $x$ using the first principle is:

$\frac{d}{dx}\left(\frac{ax + b}{cx + d}\right) = \frac{ad - bc}{(cx + d)^2}$

Given:

The function to be differentiated is $f(x) = x^{\frac{2}{3}}$.

To Differentiate using first principle:

We need to find the derivative of $f(x)$ with respect to $x$ using the definition of the derivative (first principle).

Solution:

The derivative of a function $f(x)$ with respect to $x$ using the first principle is given by:

$f'(x) = \lim\limits_{h \to 0} \frac{f(x + h) - f(x)}{h}$

Here, $f(x) = x^{2/3}$.

So, $f(x + h) = (x + h)^{2/3}$.

We need to evaluate the limit:

$f'(x) = \lim\limits_{h \to 0} \frac{(x + h)^{2/3} - x^{2/3}}{h}$

Let $A = (x + h)^{1/3}$ and $B = x^{1/3}$. Then $A^3 = x+h$ and $B^3 = x$.

The numerator is $(x+h)^{2/3} - x^{2/3} = (A^2 - B^2) = (A - B)(A + B)$.

Also, $A^3 - B^3 = (A - B)(A^2 + AB + B^2)$. From this, $A - B = \frac{A^3 - B^3}{A^2 + AB + B^2}$.

Substitute the expressions for $A$ and $B$ back:

$A^3 - B^3 = (x+h) - x = h$

$A^2 + AB + B^2 = (x+h)^{2/3} + (x+h)^{1/3} x^{1/3} + x^{2/3}$

So, $A - B = \frac{h}{(x+h)^{2/3} + (x+h)^{1/3} x^{1/3} + x^{2/3}}$.

Now, substitute the expression for the numerator $(A-B)(A+B)$ into the limit:

$\frac{f(x + h) - f(x)}{h} = \frac{(A - B)(A + B)}{h} = \frac{\left(\frac{h}{A^2 + AB + B^2}\right) (A+B)}{h}$

Since $h \to 0$, $h \neq 0$, so we can cancel $h$ from the numerator and denominator:

$\frac{f(x + h) - f(x)}{h} = \frac{A + B}{A^2 + AB + B^2}$

Substitute back $A = (x + h)^{1/3}$ and $B = x^{1/3}$:

$\frac{f(x + h) - f(x)}{h} = \frac{(x + h)^{1/3} + x^{1/3}}{(x + h)^{2/3} + (x + h)^{1/3} x^{1/3} + x^{2/3}}$

Now, take the limit as $h \to 0$:

$f'(x) = \lim\limits_{h \to 0} \frac{(x + h)^{1/3} + x^{1/3}}{(x + h)^{2/3} + (x + h)^{1/3} x^{1/3} + x^{2/3}}$

As $h \to 0$, $(x + h)^{1/3} \to x^{1/3}$ and $(x + h)^{2/3} \to x^{2/3}$.

$f'(x) = \frac{x^{1/3} + x^{1/3}}{x^{2/3} + x^{1/3} x^{1/3} + x^{2/3}}$

$f'(x) = \frac{2x^{1/3}}{x^{2/3} + x^{2/3} + x^{2/3}}$

$f'(x) = \frac{2x^{1/3}}{3x^{2/3}}$

Simplify the expression by subtracting the exponents of $x$:

$f'(x) = \frac{2}{3} x^{\frac{1}{3} - \frac{2}{3}} = \frac{2}{3} x^{-\frac{1}{3}}$

Thus, the derivative of $x^{2/3}$ with respect to $x$ using the first principle is:

$\frac{d}{dx}(x^{\frac{2}{3}}) = \frac{2}{3} x^{-\frac{1}{3}}$

Given:

The function to be differentiated is $f(x) = x \cos x$.

To Differentiate using first principle:

We need to find the derivative of $f(x)$ with respect to $x$ using the definition of the derivative (first principle).

Solution:

The derivative of a function $f(x)$ with respect to $x$ using the first principle is given by:

$f'(x) = \lim\limits_{h \to 0} \frac{f(x + h) - f(x)}{h}$

Here, $f(x) = x \cos x$.

First, we find $f(x + h)$ by replacing $x$ with $(x + h)$:

$f(x + h) = (x + h) \cos(x + h)$

Now, we find the difference $f(x + h) - f(x)$:

$f(x + h) - f(x) = (x + h) \cos(x + h) - x \cos x$

Expand the first term:

$f(x + h) - f(x) = x \cos(x + h) + h \cos(x + h) - x \cos x$

Rearrange the terms to group terms with $x$:

$f(x + h) - f(x) = x (\cos(x + h) - \cos x) + h \cos(x + h)$

Now, we form the difference quotient:

$\frac{f(x + h) - f(x)}{h} = \frac{x (\cos(x + h) - \cos x) + h \cos(x + h)}{h}$

Split the fraction into two parts:

$\frac{f(x + h) - f(x)}{h} = x \frac{\cos(x + h) - \cos x}{h} + \frac{h \cos(x + h)}{h}$

Cancel the $h$ in the second term (since $h \to 0$, $h \neq 0$):

$\frac{f(x + h) - f(x)}{h} = x \frac{\cos(x + h) - \cos x}{h} + \cos(x + h)$

Finally, we take the limit as $h \to 0$:

$f'(x) = \lim\limits_{h \to 0} \left[ x \frac{\cos(x + h) - \cos x}{h} + \cos(x + h) \right]$

Using the property that the limit of a sum is the sum of the limits (if they exist):

$f'(x) = x \lim\limits_{h \to 0} \frac{\cos(x + h) - \cos x}{h} + \lim\limits_{h \to 0} \cos(x + h)$

The first limit, $\lim\limits_{h \to 0} \frac{\cos(x + h) - \cos x}{h}$, is the definition of the derivative of $\cos x$ with respect to $x$. We know that $\frac{d}{dx}(\cos x) = -\sin x$.

So, $\lim\limits_{h \to 0} \frac{\cos(x + h) - \cos x}{h} = -\sin x$.

The second limit is evaluated by direct substitution since $\cos x$ is a continuous function:

$\lim\limits_{h \to 0} \cos(x + h) = \cos(x + 0) = \cos x$

Substitute these limits back into the expression for $f'(x)$:

$f'(x) = x (-\sin x) + \cos x$

$f'(x) = -x \sin x + \cos x$

Thus, the derivative of $x \cos x$ with respect to $x$ using the first principle is:

$\frac{d}{dx}(x \cos x) = \cos x - x \sin x$

Given:

The limit to be evaluated is:

$\lim\limits_{y \to 0} \frac{(x + y) \sec (x + y) − x \sec x}{y}$

To Evaluate:

We need to find the value of the given limit.

Solution:

The limit is in the form of the definition of the derivative of a function $f(x)$ with respect to $x$. The definition of the derivative using the first principle is:

$f'(x) = \lim\limits_{h \to 0} \frac{f(x + h) - f(x)}{h}$

Comparing the given limit with this definition, we can see that $y$ acts as the variable $h$ that approaches 0.

The numerator is $(x + y) \sec (x + y) − x \sec x$. This term can be compared to $f(x + h) - f(x)$.

If we let $f(z) = z \sec z$, then $f(x) = x \sec x$ and $f(x + y) = (x + y) \sec (x + y)$.

So, the given limit is exactly the definition of the derivative of the function $f(x) = x \sec x$ with respect to $x$.

Therefore, evaluating the limit is equivalent to finding the derivative of $f(x) = x \sec x$.

We can use the product rule for differentiation. The product rule states that if $f(x) = u(x)v(x)$, then $f'(x) = u'(x)v(x) + u(x)v'(x)$.

Let $u(x) = x$ and $v(x) = \sec x$.

First, find the derivative of $u(x)$ with respect to $x$:

$u'(x) = \frac{d}{dx}(x) = 1$

Next, find the derivative of $v(x)$ with respect to $x$:

$v'(x) = \frac{d}{dx}(\sec x) = \sec x \tan x$

Now, apply the product rule:

$f'(x) = u'(x)v(x) + u(x)v'(x)$

$f'(x) = (1)(\sec x) + (x)(\sec x \tan x)$

$f'(x) = \sec x + x \sec x \tan x$

Since the given limit is the derivative of $f(x) = x \sec x$, the value of the limit is equal to $f'(x)$.

Therefore, the value of the limit is $\sec x + x \sec x \tan x$.

$\lim\limits_{y \to 0} \frac{(x + y) \sec (x + y) − x \sec x}{y} = \sec x + x \sec x \tan x$

Given:

The limit to be evaluated is:

$L = \lim\limits_{x \to 0} \frac{\sin (α + β) x + \sin (α − β) x + \sin 2α x}{\cos 2βx − \cos 2αx} \cdot x$

To Evaluate:

We need to find the value of the given limit.

Solution:

Let the given expression be $f(x) = \frac{\sin (α + β) x + \sin (α − β) x + \sin 2α x}{\cos 2βx − \cos 2αx} \cdot x$.

We will simplify the numerator and the denominator using trigonometric identities.

Consider the numerator: $N(x) = \sin (α + β) x + \sin (α − β) x + \sin 2α x$.

Using the sum-to-product identity $\sin C + \sin D = 2 \sin\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right)$, apply it to the first two terms with $C = (α + β)x$ and $D = (α − β)x$:

$\sin (α + β) x + \sin (α − β) x = 2 \sin\left(\frac{(α+β)x + (α-β)x}{2}\right) \cos\left(\frac{(α+β)x - (α-β)x}{2}\right)$

$= 2 \sin\left(\frac{2αx}{2}\right) \cos\left(\frac{2βx}{2}\right)$

$= 2 \sin(αx) \cos(βx)$

So, the numerator becomes: $N(x) = 2 \sin(αx) \cos(βx) + \sin 2α x$.

Using the double angle identity $\sin 2A = 2 \sin A \cos A$, apply it to the third term with $A = αx$:

$\sin 2α x = 2 \sin(αx) \cos(αx)$

Substituting this back into the numerator:

$N(x) = 2 \sin(αx) \cos(βx) + 2 \sin(αx) \cos(αx)$

Factor out $2 \sin(αx)$:

$N(x) = 2 \sin(αx) (\cos(βx) + \cos(αx))$

Now consider the denominator: $D(x) = \cos 2βx − \cos 2αx$.

Using the sum-to-product identity $\cos C - \cos D = -2 \sin\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right)$, apply it with $C = 2βx$ and $D = 2αx$:

$\cos 2βx − \cos 2αx = -2 \sin\left(\frac{2βx + 2αx}{2}\right) \sin\left(\frac{2βx - 2αx}{2}\right)$

$= -2 \sin((β + α)x) \sin((β - α)x)$

Using the property $\sin(-z) = -\sin z$:

$= -2 \sin((α + β)x) (-\sin((α - β)x))$

$= 2 \sin((α + β)x) \sin((α - β)x)$

Now, substitute the simplified numerator and denominator back into the original expression for the limit:

$L = \lim\limits_{x \to 0} \frac{2 \sin(αx) (\cos(βx) + \cos(αx))}{2 \sin((α + β)x) \sin((α - β)x)} \cdot x$

Cancel out the factor of 2:

$L = \lim\limits_{x \to 0} \frac{\sin(αx) (\cos(βx) + \cos(αx)) x}{\sin((α + β)x) \sin((α - β)x)}$

To evaluate this limit, we will use the standard limit $\lim\limits_{z \to 0} \frac{\sin z}{z} = 1$. We rearrange the terms by multiplying and dividing by the arguments of the sine functions:

$L = \lim\limits_{x \to 0} \left( \frac{\sin(αx)}{αx} \cdot \frac{(α+β)x}{\sin((α+β)x)} \cdot \frac{(α-β)x}{\sin((α-β)x)} \right) \cdot \left( \frac{αx \cdot (\cos(βx) + \cos(αx)) \cdot x}{(α+β)x \cdot (α-β)x} \right)$

$L = \lim\limits_{x \to 0} \left( \frac{\sin(αx)}{αx} \right) \cdot \lim\limits_{x \to 0} \left( \frac{(α+β)x}{\sin((α+β)x)} \right) \cdot \lim\limits_{x \to 0} \left( \frac{(α-β)x}{\sin((α-β)x)} \right) \cdot \lim\limits_{x \to 0} \left( \frac{α (\cos(βx) + \cos(αx)) x^2}{(α^2 - β^2) x^2} \right)$

(Assuming $α \neq 0$, $α+β \neq 0$, and $α-β \neq 0$, the terms in the first parenthesis approach 1)

For $x \neq 0$, we can cancel the $x^2$ terms in the last part:

$L = 1 \cdot 1 \cdot 1 \cdot \lim\limits_{x \to 0} \frac{α (\cos(βx) + \cos(αx))}{α^2 - β^2}$

Now, evaluate the limit of the remaining expression as $x \to 0$:

$\lim\limits_{x \to 0} (\cos(βx) + \cos(αx)) = \cos(0) + \cos(0) = 1 + 1 = 2$

So, the limit becomes:

$L = \frac{α \cdot 2}{α^2 - β^2}$

Assuming $α^2 - β^2 \neq 0$, the value of the limit is:

$L = \frac{2α}{α^2 - β^2}$

Discussion on Special Cases:

The result $\frac{2α}{α^2 - β^2}$ is valid provided $α^2 - β^2 \neq 0$. This means $α \neq β$ and $α \neq -β$.

If $α = β \neq 0$ or $α = -β \neq 0$, the denominator $\cos 2βx − \cos 2αx$ becomes identically zero for all $x$. In these cases, the expression takes the form $\frac{\text{Numerator}}{\text{0}} \cdot x$. Since the numerator $2 \sin(2αx)$ is not identically zero for $α \neq 0$, the expression is undefined for $x$ near 0, and the limit does not exist.

If $α = 0$ and $β \neq 0$, the numerator is identically zero. The expression is $\frac{0}{\cos 2βx - 1} \cdot x$. For $x$ near 0 ($x \neq 0$), the denominator $\cos 2βx - 1 \neq 0$. Thus, the expression is $0 \cdot x = 0$. The limit is $\lim\limits_{x \to 0} 0 = 0$. Our formula $\frac{2α}{α^2 - β^2} = \frac{2(0)}{0 - β^2} = 0$ matches this case when $β \neq 0$.

If $α = 0$ and $β = 0$, both the numerator and the denominator are identically zero. The expression is $\frac{0}{0} \cdot x$, which is an indeterminate form for all $x$. The limit cannot be uniquely determined in this specific case without further context or redefinition of the function.

Therefore, the limit evaluates to $\frac{2α}{α^2 - β^2}$ when $α^2 \neq β^2$, and the limit does not exist when $α = \pm β \neq 0$. The case $α=β=0$ is indeterminate.

The question implies that the limit exists, so we provide the result for the general case where it is well-defined.

The final answer is $\frac{2α}{α^2 - β^2}$, assuming $α^2 \neq β^2$.

Given:

The limit to be evaluated is:

$L = \lim\limits_{x \to \frac{π}{4}} \frac{\tan^3 x − \tan x}{\cos x + \frac{π}{4}}$

To Evaluate:

We need to find the value of the given limit.

Solution (Method 1: Using the Definition of the Derivative):

We observe that as $x \to \frac{π}{4}$, the numerator $\tan^3 x - \tan x$ approaches $\tan^3(\frac{π}{4}) - \tan(\frac{π}{4}) = 1^3 - 1 = 0$.

The denominator is given as $\cos x + \frac{π}{4}$. Assuming the denominator is intended to be $\cos\left(x + \frac{π}{4}\right)$ for the limit to be an indeterminate form $\frac{0}{0}$, let's proceed with this assumption which is standard in such calculus problems. If the denominator were $(\cos x) + \frac{π}{4}$, the limit would be trivial.

Assuming the denominator is $\cos\left(x + \frac{π}{4}\right)$, as $x \to \frac{π}{4}$, the denominator approaches $\cos\left(\frac{π}{4} + \frac{π}{4}\right) = \cos\left(\frac{π}{2}\right) = 0$.

The limit is of the indeterminate form $\frac{0}{0}$.

Let $f(x) = \tan^3 x - \tan x$ and $g(x) = \cos\left(x + \frac{π}{4}\right)$. The limit is $\lim\limits_{x \to \frac{π}{4}} \frac{f(x)}{g(x)}$.

We note that $f(\frac{π}{4}) = 0$ and $g(\frac{π}{4}) = 0$. So we can rewrite the limit as:

$L = \lim\limits_{x \to \frac{π}{4}} \frac{f(x) - f(\frac{π}{4})}{g(x) - g(\frac{π}{4})}$

This can be expressed using the definition of the derivative:

$L = \lim\limits_{x \to \frac{π}{4}} \frac{\frac{f(x) - f(\frac{π}{4})}{x - \frac{π}{4}}}{\frac{g(x) - g(\frac{π}{4})}{x - \frac{π}{4}}}$

If $f$ and $g$ are differentiable at $x = \frac{π}{4}$ and $g'(\frac{π}{4}) \neq 0$, this limit is equal to $\frac{f'(\frac{π}{4})}{g'(\frac{π}{4})}$.

First, find the derivative of $f(x) = \tan^3 x - \tan x$:

$f'(x) = \frac{d}{dx}(\tan^3 x) - \frac{d}{dx}(\tan x)$

$f'(x) = 3\tan^2 x \cdot \frac{d}{dx}(\tan x) - \sec^2 x$

$f'(x) = 3\tan^2 x \cdot \sec^2 x - \sec^2 x$

$f'(x) = \sec^2 x (3\tan^2 x - 1)$

Evaluate $f'(\frac{π}{4})$:

$f'(\frac{π}{4}) = \sec^2(\frac{π}{4}) (3\tan^2(\frac{π}{4}) - 1)$

Since $\sec(\frac{π}{4}) = \sqrt{2}$ and $\tan(\frac{π}{4}) = 1$:

$f'(\frac{π}{4}) = (\sqrt{2})^2 (3(1)^2 - 1) = 2 (3 - 1) = 2(2) = 4$

Next, find the derivative of $g(x) = \cos\left(x + \frac{π}{4}\right)$ using the chain rule:

$g'(x) = \frac{d}{dx}\left(\cos\left(x + \frac{π}{4}\right)\right)$

$g'(x) = -\sin\left(x + \frac{π}{4}\right) \cdot \frac{d}{dx}\left(x + \frac{π}{4}\right)$

$g'(x) = -\sin\left(x + \frac{π}{4}\right) \cdot 1$

$g'(x) = -\sin\left(x + \frac{π}{4}\right)$

Evaluate $g'(\frac{π}{4})$:

$g'(\frac{π}{4}) = -\sin\left(\frac{π}{4} + \frac{π}{4}\right) = -\sin\left(\frac{π}{2}\right)$

Since $\sin(\frac{π}{2}) = 1$:

$g'(\frac{π}{4}) = -1$

Now, substitute the values of $f'(\frac{π}{4})$ and $g'(\frac{π}{4})$ into the limit expression:

$L = \frac{f'(\frac{π}{4})}{g'(\frac{π}{4})} = \frac{4}{-1} = -4$

Thus, the value of the limit is -4.

$\lim\limits_{x \to \frac{π}{4}} \frac{\tan^3 x − \tan x}{\cos\left(x + \frac{π}{4}\right)} = -4$

Alternate Solution (Method 2: Using L'Hopital's Rule):

As evaluated earlier, the limit is of the indeterminate form $\frac{0}{0}$. We can apply L'Hopital's Rule, which states that if $\lim\limits_{x \to c} \frac{f(x)}{g(x)}$ is of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim\limits_{x \to c} \frac{f(x)}{g(x)} = \lim\limits_{x \to c} \frac{f'(x)}{g'(x)}$, provided the latter limit exists.

Let $f(x) = \tan^3 x - \tan x$ and $g(x) = \cos\left(x + \frac{π}{4}\right)$.

We found the derivatives in Method 1:

$f'(x) = \sec^2 x (3\tan^2 x - 1)$

$g'(x) = -\sin\left(x + \frac{π}{4}\right)$

Now, evaluate the limit of the ratio of the derivatives as $x \to \frac{π}{4}$:

$L = \lim\limits_{x \to \frac{π}{4}} \frac{\sec^2 x (3\tan^2 x - 1)}{-\sin\left(x + \frac{π}{4}\right)}$

Substitute $x = \frac{π}{4}$ into the expression:

$L = \frac{\sec^2(\frac{π}{4}) (3\tan^2(\frac{π}{4}) - 1)}{-\sin(\frac{π}{4} + \frac{π}{4})}$

$L = \frac{(\sqrt{2})^2 (3(1)^2 - 1)}{-\sin(\frac{π}{2})}$

$L = \frac{2 (3 - 1)}{-1}$

$L = \frac{2(2)}{-1} = \frac{4}{-1} = -4$

Both methods yield the same result.

$\lim\limits_{x \to \frac{π}{4}} \frac{\tan^3 x − \tan x}{\cos\left(x + \frac{π}{4}\right)} = -4$

Given:

The limit to be evaluated is:

$L = \lim\limits_{x \to π} \frac{1 − \sin \frac{x}{2}}{\cos \frac{x}{2} \cos \frac{x}{4} − \sin \frac{x}{4}}$

To Evaluate:

We need to find the value of the given limit.

Solution:

We evaluate the numerator and the denominator separately as $x$ approaches $π$.

Consider the numerator as $x \to π$:

Numerator $= 1 - \sin\left(\frac{π}{2}\right)$

Numerator $= 1 - 1$

Numerator $= 0$

So, as $x \to π$, the numerator approaches 0.

Consider the denominator as $x \to π$:

Denominator $= \cos\left(\frac{π}{2}\right) \cos\left(\frac{π}{4}\right) − \sin\left(\frac{π}{4}\right)$

We know that $\cos\left(\frac{π}{2}\right) = 0$, $\cos\left(\frac{π}{4}\right) = \frac{\sqrt{2}}{2}$, and $\sin\left(\frac{π}{4}\right) = \frac{\sqrt{2}}{2}$.

Denominator $= (0) \left(\frac{\sqrt{2}}{2}\right) − \left(\frac{\sqrt{2}}{2}\right)$

Denominator $= 0 − \frac{\sqrt{2}}{2}$

Denominator $= -\frac{\sqrt{2}}{2}$

So, as $x \to π$, the denominator approaches $-\frac{\sqrt{2}}{2}$.

The limit is therefore of the form $\frac{0}{-\frac{\sqrt{2}}{2}}$.

When the numerator of a fraction approaches 0 and the denominator approaches a non-zero value, the limit of the fraction is 0.

$L = \frac{\lim\limits_{x \to π} (1 − \sin \frac{x}{2})}{\lim\limits_{x \to π} (\cos \frac{x}{2} \cos \frac{x}{4} − \sin \frac{x}{4})}$

$L = \frac{0}{-\frac{\sqrt{2}}{2}}$

$L = 0$

Thus, the value of the limit is 0.

$\lim\limits_{x \to π} \frac{1 − \sin \frac{x}{2}}{\cos \frac{x}{2} \cos \frac{x}{4} − \sin \frac{x}{4}} = 0$

Given:

The limit to be evaluated is $\lim\limits_{x \to 4} \frac{|x − 4|}{x − 4}$.

To Show:

We need to show that the given limit does not exist.

Solution:

For a limit $\lim\limits_{x \to c} f(x)$ to exist, the left-hand limit ($\lim\limits_{x \to c^-} f(x)$) and the right-hand limit ($\lim\limits_{x \to c^+} f(x)$) must exist and be equal.

The given function is $f(x) = \frac{|x − 4|}{x − 4}$.

The definition of the absolute value function $|u|$ is:

$|u| = u$, if $u \geq 0$

$|u| = -u$, if $u < 0$

In this case, $u = x - 4$.

We consider the cases when $x > 4$ and when $x < 4$. Note that $x \neq 4$ in the limit definition.

Case 1: Right-hand limit (as $x \to 4^+$)

As $x \to 4^+$ means that $x$ approaches 4 from values greater than 4. This implies $x > 4$, so $x - 4 > 0$.

For $x > 4$, $|x - 4| = x - 4$.

So, for $x > 4$, the function is $f(x) = \frac{x - 4}{x - 4} = 1$.

The right-hand limit is:

$\lim\limits_{x \to 4^+} \frac{|x − 4|}{x − 4} = \lim\limits_{x \to 4^+} 1$

The limit of a constant is the constant itself.

$\lim\limits_{x \to 4^+} \frac{|x − 4|}{x − 4} = 1$

Case 2: Left-hand limit (as $x \to 4^-$)

As $x \to 4^-$ means that $x$ approaches 4 from values less than 4. This implies $x < 4$, so $x - 4 < 0$.

For $x < 4$, $|x - 4| = -(x - 4)$.

So, for $x < 4$, the function is $f(x) = \frac{-(x - 4)}{x - 4} = -1$.

The left-hand limit is:

$\lim\limits_{x \to 4^-} \frac{|x − 4|}{x − 4} = \lim\limits_{x \to 4^-} (-1)$

The limit of a constant is the constant itself.

$\lim\limits_{x \to 4^-} \frac{|x − 4|}{x − 4} = -1$

Comparing the left-hand limit and the right-hand limit, we have:

Left-hand limit $= -1$

Right-hand limit $= 1$

Since $\lim\limits_{x \to 4^-} \frac{|x − 4|}{x − 4} \neq \lim\limits_{x \to 4^+} \frac{|x − 4|}{x − 4}$ (because $-1 \neq 1$), the limit $\lim\limits_{x \to 4} \frac{|x − 4|}{x − 4}$ does not exist.

Hence, shown that $\lim\limits_{x \to 4} \frac{|x − 4|}{x − 4}$ does not exists.

Given:

The function $f(x)$ is defined as:

$f(x) = \begin{cases} \frac{k \cos x}{π − 2x} & when\; x ≠ \frac{π}{2} \\ 3 & when\; x = \frac{π}{2} \end{cases}$

It is also given that $\lim\limits_{x \to \frac{π}{2}} f(x) = f \left( \frac{π}{2} \right)$.

To Find:

We need to find the value of the constant $k$.

Solution:

The given condition $\lim\limits_{x \to \frac{π}{2}} f(x) = f \left( \frac{π}{2} \right)$ means that the function $f(x)$ is continuous at $x = \frac{π}{2}$.

From the definition of the function, we are given that $f \left( \frac{π}{2} \right) = 3$.

So, the condition becomes $\lim\limits_{x \to \frac{π}{2}} f(x) = 3$.

For $x \neq \frac{π}{2}$, the function is given by $f(x) = \frac{k \cos x}{π − 2x}$.

We need to evaluate the limit $\lim\limits_{x \to \frac{π}{2}} \frac{k \cos x}{π − 2x}$.

As $x \to \frac{π}{2}$, the numerator $k \cos x \to k \cos(\frac{π}{2}) = k \cdot 0 = 0$.

As $x \to \frac{π}{2}$, the denominator $π − 2x \to π − 2(\frac{π}{2}) = π - π = 0$.

The limit is of the indeterminate form $\frac{0}{0}$. We can use L'Hopital's Rule or a substitution to evaluate this limit.

Method 1: Using Substitution

Let $y = x - \frac{π}{2}$. As $x \to \frac{π}{2}$, $y \to 0$. Then $x = y + \frac{π}{2}$.

Substitute this into the expression:

Numerator $= k \cos x = k \cos\left(y + \frac{π}{2}\right)$

Using the identity $\cos\left(\theta + \frac{π}{2}\right) = -\sin \theta$, we have:

Numerator $= k (-\sin y) = -k \sin y$

Denominator $= π - 2x = π - 2\left(y + \frac{π}{2}\right) = π - 2y - π = -2y$

Now, the limit becomes:

$\lim\limits_{y \to 0} \frac{-k \sin y}{-2y} = \lim\limits_{y \to 0} \frac{k \sin y}{2y}$

We can factor out the constant $\frac{k}{2}$:

$= \frac{k}{2} \lim\limits_{y \to 0} \frac{\sin y}{y}$

Using the standard limit $\lim\limits_{y \to 0} \frac{\sin y}{y} = 1$:

$= \frac{k}{2} \cdot 1 = \frac{k}{2}$

So, $\lim\limits_{x \to \frac{π}{2}} f(x) = \frac{k}{2}$.

Given that $\lim\limits_{x \to \frac{π}{2}} f(x) = f \left( \frac{π}{2} \right) = 3$, we have:

$\frac{k}{2} = 3$

Multiply both sides by 2:

$k = 6$

Method 2: Using L'Hopital's Rule

The limit is of the form $\frac{0}{0}$ as $x \to \frac{π}{2}$. We can apply L'Hopital's Rule.

Let $N(x) = k \cos x$ and $D(x) = π - 2x$.

Find the derivative of the numerator $N'(x)$:

$N'(x) = \frac{d}{dx}(k \cos x) = k (-\sin x) = -k \sin x$

Find the derivative of the denominator $D'(x)$:

$D'(x) = \frac{d}{dx}(π - 2x) = 0 - 2 = -2$

Apply L'Hopital's Rule:

$\lim\limits_{x \to \frac{π}{2}} \frac{k \cos x}{π − 2x} = \lim\limits_{x \to \frac{π}{2}} \frac{-k \sin x}{-2}$

Simplify the expression:

$= \lim\limits_{x \to \frac{π}{2}} \frac{k \sin x}{2}$

Substitute $x = \frac{π}{2}$ into the expression:

$= \frac{k \sin(\frac{π}{2})}{2}$

Since $\sin(\frac{π}{2}) = 1$:

$= \frac{k \cdot 1}{2} = \frac{k}{2}$

So, $\lim\limits_{x \to \frac{π}{2}} f(x) = \frac{k}{2}$.

Given that $\lim\limits_{x \to \frac{π}{2}} f(x) = f \left( \frac{π}{2} \right) = 3$, we have:

$\frac{k}{2} = 3$

$k = 6$

Both methods give the same value for $k$.

The value of $k$ that makes the function continuous at $x = \frac{π}{2}$ is 6.

The value of k is 6.

Given:

The function $f(x)$ is defined as:

$f (x) = \begin{cases} x + 2 & x ≤ -1 \\ cx^2 & x > -1 \end{cases}$

It is given that $\lim\limits_{x \to −1} f(x)$ exists.

To Find:

We need to find the value of the constant $c$.

Solution:

For the limit $\lim\limits_{x \to −1} f(x)$ to exist, the left-hand limit (LHL) and the right-hand limit (RHL) of the function $f(x)$ at $x = -1$ must exist and be equal.

First, we calculate the left-hand limit as $x$ approaches -1 from the left ($x < -1$). For $x \leq -1$, $f(x) = x + 2$.

LHL $= \lim\limits_{x \to -1^-} f(x) = \lim\limits_{x \to -1^-} (x + 2)$

Since $x + 2$ is a polynomial, the limit can be evaluated by direct substitution:

LHL $= (-1) + 2 = 1$

Next, we calculate the right-hand limit as $x$ approaches -1 from the right ($x > -1$). For $x > -1$, $f(x) = cx^2$.

RHL $= \lim\limits_{x \to -1^+} f(x) = \lim\limits_{x \to -1^+} (cx^2)$

Since $cx^2$ is a polynomial, the limit can be evaluated by direct substitution:

RHL $= c(-1)^2 = c(1) = c$

For the limit $\lim\limits_{x \to −1} f(x)$ to exist, we must have LHL = RHL.

$1 = c$

Thus, the value of $c$ is 1.

The value of c is 1.

The given limit is $\lim\limits_{x \to \pi} \frac{\sin x}{x - \pi}$.

As $x \to \pi$, the numerator $\sin x \to \sin \pi = 0$ and the denominator $x - \pi \to \pi - \pi = 0$.

This is an indeterminate form of type $\frac{0}{0}$.

Method 1: Using Substitution

Let $y = x - \pi$.

As $x \to \pi$, $y \to 0$.

From $y = x - \pi$, we get $x = y + \pi$.

So, $\sin x = \sin(y + \pi) = -\sin y$.

Substituting these into the limit, we get:

$\lim\limits_{x \to \pi} \frac{\sin x}{x - \pi} = \lim\limits_{y \to 0} \frac{-\sin y}{y}$

$= - \lim\limits_{y \to 0} \frac{\sin y}{y}$

Using the standard limit $\lim\limits_{y \to 0} \frac{\sin y}{y} = 1$, we have:

$= -(1) = -1$

Method 2: Using L'Hopital's Rule

Since the limit is of the form $\frac{0}{0}$, we can apply L'Hopital's Rule.

L'Hopital's Rule states that if $\lim\limits_{x \to c} \frac{f(x)}{g(x)}$ is of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim\limits_{x \to c} \frac{f(x)}{g(x)} = \lim\limits_{x \to c} \frac{f'(x)}{g'(x)}$, provided the latter limit exists.

Here, $f(x) = \sin x$ and $g(x) = x - \pi$.

$f'(x) = \frac{d}{dx}(\sin x) = \cos x$

$g'(x) = \frac{d}{dx}(x - \pi) = 1$

Applying L'Hopital's Rule:

$\lim\limits_{x \to \pi} \frac{\sin x}{x - \pi} = \lim\limits_{x \to \pi} \frac{\cos x}{1}$

Substitute $x = \pi$:

$= \frac{\cos \pi}{1}$

Since $\cos \pi = -1$, we get:

$= \frac{-1}{1} = -1$

Both methods yield the same result.

The value of the limit is -1.

The correct option is (C) –1.

The given limit is $\lim\limits_{x \to 0} \frac{x^2 \cos x}{1 − \cos x}$.

Let's evaluate the numerator and denominator as $x \to 0$:

Numerator: $x^2 \cos x \to (0)^2 \cos(0) = 0 \times 1 = 0$.

Denominator: $1 - \cos x \to 1 - \cos(0) = 1 - 1 = 0$.

The limit is of the indeterminate form $\frac{0}{0}$.

We can rewrite the expression to use standard limits. We know the standard limit $\lim\limits_{\theta \to 0} \frac{1 - \cos \theta}{\theta^2} = \frac{1}{2}$.

We can manipulate the given expression as follows:

$\frac{x^2 \cos x}{1 - \cos x} = \frac{\cos x}{\frac{1 - \cos x}{x^2}}$

Now, we can take the limit as $x \to 0$:

$\lim\limits_{x \to 0} \frac{x^2 \cos x}{1 - \cos x} = \lim\limits_{x \to 0} \frac{\cos x}{\frac{1 - \cos x}{x^2}}$

Using the properties of limits, the limit of a quotient is the quotient of the limits (provided the denominator limit is not zero):

$= \frac{\lim\limits_{x \to 0} \cos x}{\lim\limits_{x \to 0} \frac{1 - \cos x}{x^2}}$

Evaluate the limits separately:

$\lim\limits_{x \to 0} \cos x = \cos(0) = 1$.

$\lim\limits_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$ (Standard limit).

Substitute these values back into the expression:

$= \frac{1}{\frac{1}{2}} = 1 \times 2 = 2$.

Alternatively, using L'Hopital's Rule (first application):

$\lim\limits_{x \to 0} \frac{\frac{d}{dx}(x^2 \cos x)}{\frac{d}{dx}(1 - \cos x)} = \lim\limits_{x \to 0} \frac{2x \cos x - x^2 \sin x}{\sin x}$

This is still of the form $\frac{0}{0}$. Apply L'Hopital's Rule again:

$\lim\limits_{x \to 0} \frac{\frac{d}{dx}(2x \cos x - x^2 \sin x)}{\frac{d}{dx}(\sin x)} = \lim\limits_{x \to 0} \frac{(2 \cos x - 2x \sin x) - (2x \sin x + x^2 \cos x)}{\cos x}$

$= \lim\limits_{x \to 0} \frac{2 \cos x - 4x \sin x - x^2 \cos x}{\cos x}$

Now substitute $x=0$:

$= \frac{2 \cos(0) - 4(0) \sin(0) - (0)^2 \cos(0)}{\cos(0)}$

$= \frac{2(1) - 0 - 0}{1} = \frac{2}{1} = 2$.

Both methods confirm the result.

The value of the limit is 2.

The correct option is (A) 2.

The given limit is $\lim\limits_{x \to 0} \frac{(1 + x)^n − 1}{x}$.

Let's evaluate the numerator and denominator as $x \to 0$:

Numerator: $(1 + 0)^n - 1 = 1^n - 1 = 1 - 1 = 0$.

Denominator: $0$.

The limit is of the indeterminate form $\frac{0}{0}$.

Method 1: Using the definition of the derivative

Recall the definition of the derivative of a function $f(x)$ at a point $a$:

$f'(a) = \lim\limits_{x \to a} \frac{f(x) - f(a)}{x - a}$

We can rewrite the given limit to match this form.

Let $f(x) = (1+x)^n$. We are interested in the limit as $x \to 0$.

Evaluate $f(x)$ at $x=0$: $f(0) = (1+0)^n = 1^n = 1$.

The given limit is $\lim\limits_{x \to 0} \frac{(1+x)^n - 1}{x}$. This can be written as $\lim\limits_{x \to 0} \frac{(1+x)^n - (1+0)^n}{x - 0}$.

This is exactly the definition of the derivative of the function $f(x) = (1+x)^n$ evaluated at $x = 0$.

Let's find the derivative of $f(x) = (1+x)^n$ with respect to $x$:

$f'(x) = \frac{d}{dx} (1+x)^n = n(1+x)^{n-1} \cdot \frac{d}{dx}(1+x) = n(1+x)^{n-1} \cdot 1 = n(1+x)^{n-1}$.

Now, evaluate the derivative at $x=0$:

$f'(0) = n(1+0)^{n-1} = n(1)^{n-1} = n \times 1 = n$.

Therefore, the limit is equal to $f'(0) = n$.

Method 2: Using L'Hopital's Rule

Since the limit is of the form $\frac{0}{0}$, we can apply L'Hopital's Rule.

According to L'Hopital's Rule, if $\lim\limits_{x \to c} \frac{f(x)}{g(x)}$ is of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim\limits_{x \to c} \frac{f(x)}{g(x)} = \lim\limits_{x \to c} \frac{f'(x)}{g'(x)}$, provided the latter limit exists.

Here, let $f(x) = (1+x)^n - 1$ and $g(x) = x$.

Find the derivative of $f(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx}((1+x)^n - 1) = n(1+x)^{n-1} \cdot \frac{d}{dx}(1+x) - 0 = n(1+x)^{n-1}$.

Find the derivative of $g(x)$ with respect to $x$:

$g'(x) = \frac{d}{dx}(x) = 1$.

Apply L'Hopital's Rule:

$\lim\limits_{x \to 0} \frac{(1 + x)^n − 1}{x} = \lim\limits_{x \to 0} \frac{n(1+x)^{n-1}}{1}$.

Now substitute $x=0$ into the expression:

$= \frac{n(1+0)^{n-1}}{1} = \frac{n(1)^{n-1}}{1} = \frac{n \times 1}{1} = n$.

Thus, the limit is $n$.

Method 3: Using Binomial Expansion (for integer n)

If $n$ is a positive integer, we can use the binomial expansion:

$(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots + x^n$.

Then, $(1+x)^n - 1 = nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots + x^n$.

Divide by $x$ (for $x \neq 0$):

$\frac{(1+x)^n - 1}{x} = \frac{nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots + x^n}{x}$

$= n + \frac{n(n-1)}{2!}x + \frac{n(n-1)(n-2)}{3!}x^2 + \dots + x^{n-1}$.

Now, take the limit as $x \to 0$:

$\lim\limits_{x \to 0} \left(n + \frac{n(n-1)}{2!}x + \frac{n(n-1)(n-2)}{3!}x^2 + \dots + x^{n-1}\right)$

As $x \to 0$, all terms containing $x$ raised to a positive power tend to 0.

$= n + 0 + 0 + \dots + 0 = n$.

This method confirms the result for positive integer $n$. The result is valid for any real number $n$ using the generalized binomial theorem or the first two methods.

All methods lead to the same result.

The value of the limit is n.

The correct option is (A) n.

The given limit is $\lim\limits_{x \to 1} \frac{x^m − 1}{x^n − 1}$.

Let's evaluate the numerator and denominator as $x \to 1$:

Numerator: $x^m - 1 \to 1^m - 1 = 1 - 1 = 0$.

Denominator: $x^n - 1 \to 1^n - 1 = 1 - 1 = 0$.

The limit is of the indeterminate form $\frac{0}{0}$.

Method 1: Using L'Hopital's Rule

Since the limit is of the form $\frac{0}{0}$, we can apply L'Hopital's Rule.

L'Hopital's Rule states that if $\lim\limits_{x \to c} \frac{f(x)}{g(x)}$ is of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim\limits_{x \to c} \frac{f(x)}{g(x)} = \lim\limits_{x \to c} \frac{f'(x)}{g'(x)}$, provided the latter limit exists.

Here, let $f(x) = x^m - 1$ and $g(x) = x^n - 1$.

Find the derivative of $f(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx}(x^m - 1) = mx^{m-1}$.

Find the derivative of $g(x)$ with respect to $x$:

$g'(x) = \frac{d}{dx}(x^n - 1) = nx^{n-1}$.

Apply L'Hopital's Rule:

$\lim\limits_{x \to 1} \frac{x^m − 1}{x^n − 1} = \lim\limits_{x \to 1} \frac{mx^{m-1}}{nx^{n-1}}$

Now substitute $x=1$ into the expression:

$= \frac{m(1)^{m-1}}{n(1)^{n-1}}$

Since $1$ raised to any power is $1$, we have:

$= \frac{m \times 1}{n \times 1} = \frac{m}{n}$.

Method 2: Using Standard Limit Formula

We can use the standard limit formula: $\lim\limits_{x \to a} \frac{x^p - a^p}{x - a} = pa^{p-1}$.

We can rewrite the given expression by dividing both the numerator and the denominator by $(x-1)$:

$\frac{x^m − 1}{x^n − 1} = \frac{\frac{x^m - 1}{x - 1}}{\frac{x^n - 1}{x - 1}}$

Now, take the limit as $x \to 1$:

$\lim\limits_{x \to 1} \frac{x^m − 1}{x^n − 1} = \lim\limits_{x \to 1} \frac{\frac{x^m - 1^m}{x - 1}}{\frac{x^n - 1^n}{x - 1}}$

Using the property that the limit of a quotient is the quotient of the limits (provided the denominator limit is non-zero):

$= \frac{\lim\limits_{x \to 1} \frac{x^m - 1^m}{x - 1}}{\lim\limits_{x \to 1} \frac{x^n - 1^n}{x - 1}}$

Applying the standard limit formula $\lim\limits_{x \to a} \frac{x^p - a^p}{x - a} = pa^{p-1}$ to both the numerator and denominator limits (with $a=1$):

Numerator limit: $\lim\limits_{x \to 1} \frac{x^m - 1^m}{x - 1} = m(1)^{m-1} = m \times 1 = m$.

Denominator limit: $\lim\limits_{x \to 1} \frac{x^n - 1^n}{x - 1} = n(1)^{n-1} = n \times 1 = n$.

Substitute these values back into the expression:

$= \frac{m}{n}$.

Both methods yield the same result.

The value of the limit is $\frac{m}{n}$.

The correct option is (B) $\frac{m}{n}$.

The given limit is $\lim\limits_{\theta \to 0} \frac{1 − \cos 4θ}{1 − \cos 6θ}$. Note that the variable in the limit is given as $x \to 0$, but the expression uses $\theta$. We will assume the limit is with respect to $\theta$, so $\lim\limits_{\theta \to 0}$.

Let's evaluate the numerator and denominator as $\theta \to 0$:

Numerator: $1 - \cos(4 \times 0) = 1 - \cos(0) = 1 - 1 = 0$.

Denominator: $1 - \cos(6 \times 0) = 1 - \cos(0) = 1 - 1 = 0$.

The limit is of the indeterminate form $\frac{0}{0}$.

Method 1: Using Standard Limit Formula

We use the standard limit $\lim\limits_{y \to 0} \frac{1 - \cos y}{y^2} = \frac{1}{2}$.

We can rewrite the expression by dividing the numerator by $(4\theta)^2$ and the denominator by $(6\theta)^2$, adjusting with multiplication:

$\frac{1 − \cos 4θ}{1 − \cos 6θ} = \frac{\frac{1 − \cos 4θ}{(4θ)^2} \cdot (4θ)^2}{\frac{1 − \cos 6θ}{(6θ)^2} \cdot (6θ)^2}$

$= \frac{\frac{1 − \cos 4θ}{16θ^2} \cdot 16θ^2}{\frac{1 − \cos 6θ}{36θ^2} \cdot 36θ^2}$

$= \frac{\frac{1 − \cos 4θ}{(4θ)^2}}{\frac{1 − \cos 6θ}{(6θ)^2}} \cdot \frac{16θ^2}{36θ^2}$

For $\theta \neq 0$, we can cancel $\theta^2$:

$= \frac{\frac{1 − \cos 4θ}{(4θ)^2}}{\frac{1 − \cos 6θ}{(6θ)^2}} \cdot \frac{16}{36}$

$= \frac{\frac{1 − \cos 4θ}{(4θ)^2}}{\frac{1 − \cos 6θ}{(6θ)^2}} \cdot \frac{4}{9}$

Now, take the limit as $\theta \to 0$:

$\lim\limits_{\theta \to 0} \frac{1 − \cos 4θ}{1 − \cos 6θ} = \lim\limits_{\theta \to 0} \frac{\frac{1 − \cos 4θ}{(4θ)^2}}{\frac{1 − \cos 6θ}{(6θ)^2}} \cdot \frac{4}{9}$

Let $y = 4\theta$. As $\theta \to 0$, $y \to 0$. The numerator limit becomes $\lim\limits_{y \to 0} \frac{1 - \cos y}{y^2} = \frac{1}{2}$.

Let $z = 6\theta$. As $\theta \to 0$, $z \to 0$. The denominator limit becomes $\lim\limits_{z \to 0} \frac{1 - \cos z}{z^2} = \frac{1}{2}$.

So, the limit is:

$= \frac{\lim\limits_{\theta \to 0} \frac{1 − \cos 4θ}{(4θ)^2}}{\lim\limits_{\theta \to 0} \frac{1 − \cos 6θ}{(6θ)^2}} \cdot \frac{4}{9} = \frac{\frac{1}{2}}{\frac{1}{2}} \cdot \frac{4}{9} = 1 \cdot \frac{4}{9} = \frac{4}{9}$.

Method 2: Using L'Hopital's Rule

Since the limit is of the form $\frac{0}{0}$, we can apply L'Hopital's Rule.

Let $f(\theta) = 1 - \cos 4\theta$ and $g(\theta) = 1 - \cos 6\theta$.

Find the derivatives:

$f'(\theta) = \frac{d}{d\theta}(1 - \cos 4\theta) = 0 - (-\sin 4\theta) \cdot \frac{d}{d\theta}(4\theta) = 4 \sin 4\theta$.

$g'(\theta) = \frac{d}{d\theta}(1 - \cos 6\theta) = 0 - (-\sin 6\theta) \cdot \frac{d}{d\theta}(6\theta) = 6 \sin 6\theta$.

Apply L'Hopital's Rule once:

$\lim\limits_{\theta \to 0} \frac{1 − \cos 4θ}{1 − \cos 6θ} = \lim\limits_{\theta \to 0} \frac{4 \sin 4θ}{6 \sin 6θ}$.

As $\theta \to 0$, $4 \sin 4\theta \to 4 \sin(0) = 0$ and $6 \sin 6\theta \to 6 \sin(0) = 0$. This is still of the form $\frac{0}{0}$. Apply L'Hopital's Rule again.

Find the second derivatives:

$f''(\theta) = \frac{d}{d\theta}(4 \sin 4\theta) = 4 (\cos 4\theta) \cdot \frac{d}{d\theta}(4\theta) = 16 \cos 4\theta$.

$g''(\theta) = \frac{d}{d\theta}(6 \sin 6\theta) = 6 (\cos 6\theta) \cdot \frac{d}{d\theta}(6\theta) = 36 \cos 6\theta$.

Apply L'Hopital's Rule a second time:

$\lim\limits_{\theta \to 0} \frac{4 \sin 4θ}{6 \sin 6θ} = \lim\limits_{\theta \to 0} \frac{16 \cos 4θ}{36 \cos 6θ}$.

Now, substitute $\theta = 0$:

$= \frac{16 \cos(4 \times 0)}{36 \cos(6 \times 0)} = \frac{16 \cos(0)}{36 \cos(0)} = \frac{16 \times 1}{36 \times 1} = \frac{16}{36}$.

Simplify the fraction:

$= \frac{16}{36} = \frac{4 \times 4}{9 \times 4} = \frac{4}{9}$.

Both methods give the same result.

The value of the limit is $\frac{4}{9}$.

The correct option is (A) $\frac{4}{9}$.

The given limit is $\lim\limits_{x \to 0} \frac{\text{cosec}\; x − \cot x}{x}$.

Let's evaluate the numerator and denominator as $x \to 0$.

Numerator: $\text{cosec}\; x − \cot x$. As $x \to 0^+$, $\text{cosec}\; x \to \infty$ and $\cot x \to \infty$. This is an indeterminate form of type $\infty - \infty$.

Denominator: $x \to 0$.

We first rewrite the numerator in terms of $\sin x$ and $\cos x$:

$\text{cosec}\; x = \frac{1}{\sin x}$

$\cot x = \frac{\cos x}{\sin x}$

So, $\text{cosec}\; x − \cot x = \frac{1}{\sin x} - \frac{\cos x}{\sin x} = \frac{1 - \cos x}{\sin x}$.

The given limit becomes:

$\lim\limits_{x \to 0} \frac{\frac{1 - \cos x}{\sin x}}{x} = \lim\limits_{x \to 0} \frac{1 - \cos x}{x \sin x}$.

Now, let's evaluate this new form as $x \to 0$:

Numerator: $1 - \cos x \to 1 - \cos(0) = 1 - 1 = 0$.

Denominator: $x \sin x \to 0 \cdot \sin(0) = 0 \cdot 0 = 0$.

The limit is of the indeterminate form $\frac{0}{0}$.

Method 1: Using Standard Limits

We know the standard limits: $\lim\limits_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$ and $\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$.

We can rewrite the expression $\frac{1 - \cos x}{x \sin x}$ by dividing the numerator by $x^2$ and the denominator by $x^2$ (or by cleverly grouping terms):

$\frac{1 - \cos x}{x \sin x} = \frac{1 - \cos x}{x \cdot (\sin x)} = \frac{1 - \cos x}{x \cdot x \cdot (\frac{\sin x}{x})} = \frac{1 - \cos x}{x^2 \cdot \frac{\sin x}{x}} = \frac{\frac{1 - \cos x}{x^2}}{\frac{\sin x}{x}}$.

Now, take the limit as $x \to 0$:

$\lim\limits_{x \to 0} \frac{1 - \cos x}{x \sin x} = \lim\limits_{x \to 0} \frac{\frac{1 - \cos x}{x^2}}{\frac{\sin x}{x}}$

Using the property that the limit of a quotient is the quotient of the limits (provided the denominator limit is non-zero):

$= \frac{\lim\limits_{x \to 0} \frac{1 - \cos x}{x^2}}{\lim\limits_{x \to 0} \frac{\sin x}{x}}$

Substitute the values of the standard limits:

$= \frac{\frac{1}{2}}{1} = \frac{1}{2}$.

Method 2: Using L'Hopital's Rule

The limit $\lim\limits_{x \to 0} \frac{1 - \cos x}{x \sin x}$ is of the form $\frac{0}{0}$, so we can apply L'Hopital's Rule.

Let $f(x) = 1 - \cos x$ and $g(x) = x \sin x$.

Find the derivatives:

$f'(x) = \frac{d}{dx}(1 - \cos x) = \sin x$.

$g'(x) = \frac{d}{dx}(x \sin x) = 1 \cdot \sin x + x \cdot \cos x = \sin x + x \cos x$.

Apply L'Hopital's Rule:

$\lim\limits_{x \to 0} \frac{1 - \cos x}{x \sin x} = \lim\limits_{x \to 0} \frac{\sin x}{\sin x + x \cos x}$.

Evaluating this new limit as $x \to 0$: Numerator $\to 0$, Denominator $\to 0 + 0 \cdot 1 = 0$. This is still of the form $\frac{0}{0}$. Apply L'Hopital's Rule again.

Find the second derivatives:

$f''(x) = \frac{d}{dx}(\sin x) = \cos x$.

$g''(x) = \frac{d}{dx}(\sin x + x \cos x) = \cos x + (1 \cdot \cos x + x \cdot (-\sin x)) = \cos x + \cos x - x \sin x = 2 \cos x - x \sin x$.

Apply L'Hopital's Rule a second time:

$\lim\limits_{x \to 0} \frac{\sin x}{\sin x + x \cos x} = \lim\limits_{x \to 0} \frac{\cos x}{2 \cos x - x \sin x}$.

Now, substitute $x=0$:

$= \frac{\cos(0)}{2 \cos(0) - 0 \cdot \sin(0)} = \frac{1}{2 \cdot 1 - 0} = \frac{1}{2}$.

Both methods confirm the result.

The value of the limit is $\frac{1}{2}$.

The correct option is (C) $\frac{1}{2}$.

The given limit is $\lim\limits_{x \to 0} \frac{\sin x}{\sqrt{x + 1} − \sqrt{1 − x}}$.

Let's evaluate the numerator and denominator as $x \to 0$:

Numerator: $\sin x \to \sin 0 = 0$.

Denominator: $\sqrt{x + 1} − \sqrt{1 − x} \to \sqrt{0 + 1} − \sqrt{1 − 0} = \sqrt{1} − \sqrt{1} = 1 - 1 = 0$.

The limit is of the indeterminate form $\frac{0}{0}$.

Method 1: Rationalizing the Denominator

Multiply the numerator and denominator by the conjugate of the denominator, which is $\sqrt{x+1} + \sqrt{1-x}$.

$\lim\limits_{x \to 0} \frac{\sin x}{\sqrt{x + 1} − \sqrt{1 − x}} = \lim\limits_{x \to 0} \frac{\sin x}{(\sqrt{x + 1} − \sqrt{1 − x})} \times \frac{(\sqrt{x + 1} + \sqrt{1 − x})}{(\sqrt{x + 1} + \sqrt{1 − x})}$

$= \lim\limits_{x \to 0} \frac{\sin x (\sqrt{x + 1} + \sqrt{1 − x})}{(\sqrt{x + 1})^2 − (\sqrt{1 − x})^2}$

$= \lim\limits_{x \to 0} \frac{\sin x (\sqrt{x + 1} + \sqrt{1 − x})}{(x + 1) − (1 − x)}$

$= \lim\limits_{x \to 0} \frac{\sin x (\sqrt{x + 1} + \sqrt{1 − x})}{x + 1 − 1 + x}$

$= \lim\limits_{x \to 0} \frac{\sin x (\sqrt{x + 1} + \sqrt{1 − x})}{2x}$

We can separate this into a product of limits:

$= \lim\limits_{x \to 0} \left(\frac{\sin x}{x}\right) \cdot \left(\frac{\sqrt{x + 1} + \sqrt{1 − x}}{2}\right)$

Using the standard limit $\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$ and evaluating the second part by direct substitution:

$= (1) \cdot \left(\frac{\sqrt{0 + 1} + \sqrt{1 − 0}}{2}\right)$

$= 1 \cdot \left(\frac{\sqrt{1} + \sqrt{1}}{2}\right)$

$= 1 \cdot \left(\frac{1 + 1}{2}\right)$

$= 1 \cdot \left(\frac{2}{2}\right)$

$= 1 \cdot 1 = 1$.

Method 2: Using L'Hopital's Rule

Since the limit is of the form $\frac{0}{0}$, we can apply L'Hopital's Rule.

Let $f(x) = \sin x$ and $g(x) = \sqrt{x + 1} − \sqrt{1 − x}$.

The derivatives are:

$f'(x) = \frac{d}{dx}(\sin x) = \cos x$

$g'(x) = \frac{d}{dx}(\sqrt{x + 1} − \sqrt{1 − x}) = \frac{d}{dx}((x+1)^{1/2}) − \frac{d}{dx}((1−x)^{1/2})$

$= \frac{1}{2}(x+1)^{-1/2} \cdot \frac{d}{dx}(x+1) − \frac{1}{2}(1−x)^{-1/2} \cdot \frac{d}{dx}(1−x)$

$= \frac{1}{2\sqrt{x+1}} \cdot 1 − \frac{1}{2\sqrt{1−x}} \cdot (-1)$

$= \frac{1}{2\sqrt{x+1}} + \frac{1}{2\sqrt{1−x}}$

Apply L'Hopital's Rule:

$\lim\limits_{x \to 0} \frac{\sin x}{\sqrt{x + 1} − \sqrt{1 − x}} = \lim\limits_{x \to 0} \frac{\cos x}{\frac{1}{2\sqrt{x+1}} + \frac{1}{2\sqrt{1−x}}}$

Substitute $x=0$:

$= \frac{\cos(0)}{\frac{1}{2\sqrt{0+1}} + \frac{1}{2\sqrt{1−0}}}

$= \frac{1}{\frac{1}{2\sqrt{1}} + \frac{1}{2\sqrt{1}}}

$= \frac{1}{\frac{1}{2} + \frac{1}{2}} = \frac{1}{1} = 1$.

Both methods give the same result.

The value of the limit is 1.

The correct option is (C) 1.

The given limit is $\lim\limits_{x \to \frac{\pi}{4}} \frac{\sec^2 x − 2}{\tan x − 1}$.

Let's evaluate the numerator and denominator as $x \to \frac{\pi}{4}$:

Numerator: $\sec^2 \left(\frac{\pi}{4}\right) - 2 = (\sqrt{2})^2 - 2 = 2 - 2 = 0$.

Denominator: $\tan \left(\frac{\pi}{4}\right) - 1 = 1 - 1 = 0$.

The limit is of the indeterminate form $\frac{0}{0}$.

Method 1: Using Trigonometric Identities

We can use the identity $\sec^2 x = 1 + \tan^2 x$ to rewrite the numerator.

Numerator = $\sec^2 x - 2 = (1 + \tan^2 x) - 2 = \tan^2 x - 1$.

The numerator $\tan^2 x - 1$ is a difference of squares, which can be factored as $(\tan x - 1)(\tan x + 1)$.

So, the expression becomes:

$\frac{\sec^2 x − 2}{\tan x − 1} = \frac{\tan^2 x - 1}{\tan x − 1} = \frac{(\tan x - 1)(\tan x + 1)}{\tan x − 1}$.

For $x$ approaching $\frac{\pi}{4}$ but not equal to $\frac{\pi}{4}$, $\tan x \neq 1$, so $\tan x - 1 \neq 0$. We can cancel the common factor $(\tan x - 1)$ from the numerator and denominator.

The simplified expression is $\tan x + 1$.

Now, evaluate the limit of the simplified expression:

$\lim\limits_{x \to \frac{\pi}{4}} (\tan x + 1)$

Substitute $x = \frac{\pi}{4}$:

$= \tan\left(\frac{\pi}{4}\right) + 1 = 1 + 1 = 2$.

Method 2: Using L'Hopital's Rule

Since the limit is of the form $\frac{0}{0}$, we can apply L'Hopital's Rule.

L'Hopital's Rule states that if $\lim\limits_{x \to c} \frac{f(x)}{g(x)}$ is of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim\limits_{x \to c} \frac{f(x)}{g(x)} = \lim\limits_{x \to c} \frac{f'(x)}{g'(x)}$, provided the latter limit exists.

Here, let $f(x) = \sec^2 x - 2$ and $g(x) = \tan x - 1$.

Find the derivative of $f(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx}(\sec^2 x - 2) = 2 \sec x \cdot \frac{d}{dx}(\sec x) = 2 \sec x (\sec x \tan x) = 2 \sec^2 x \tan x$.

Find the derivative of $g(x)$ with respect to $x$:

$g'(x) = \frac{d}{dx}(\tan x - 1) = \sec^2 x$.

Apply L'Hopital's Rule:

$\lim\limits_{x \to \frac{\pi}{4}} \frac{\sec^2 x − 2}{\tan x − 1} = \lim\limits_{x \to \frac{\pi}{4}} \frac{2 \sec^2 x \tan x}{\sec^2 x}$.

For $x$ approaching $\frac{\pi}{4}$ but not equal to $\frac{\pi}{4}$, $\sec^2 x \neq 0$. We can cancel $\sec^2 x$ from the numerator and denominator.

The simplified expression is $2 \tan x$.

Now substitute $x = \frac{\pi}{4}$ into the expression:

$= 2 \tan\left(\frac{\pi}{4}\right) = 2 \times 1 = 2$.

Both methods yield the value of the limit as 2.

Upon checking the provided options, the calculated limit value 2 is not present among options (A) 3, (B) 1, (C) 0, or (D) $\sqrt{2}$.

Based on the question as stated, the limit evaluates to 2.

The given limit is $\lim\limits_{x \to 1} \frac{(\sqrt{x} − 1) (2x − 3)}{2x^2 + x − 3}$.

Let's evaluate the numerator and denominator as $x \to 1$:

Numerator: $(\sqrt{1} - 1)(2(1) - 3) = (1 - 1)(2 - 3) = 0 \times (-1) = 0$.

Denominator: $2(1)^2 + (1) - 3 = 2 + 1 - 3 = 0$.

The limit is of the indeterminate form $\frac{0}{0}$.

Method 1: Algebraic Manipulation

We can rationalize the term $(\sqrt{x} - 1)$ in the numerator by multiplying by its conjugate $(\sqrt{x} + 1)$.

We also factor the quadratic denominator $2x^2 + x - 3$. We can see that when $x=1$, the denominator is $2(1)^2 + 1 - 3 = 2 + 1 - 3 = 0$, so $(x-1)$ is a factor. Using polynomial factorization or inspection, we find $2x^2 + x - 3 = (x-1)(2x+3)$.

Rewrite the expression:

$\frac{(\sqrt{x} − 1) (2x − 3)}{2x^2 + x − 3} = \frac{(\sqrt{x} − 1) (2x − 3)}{(x-1)(2x+3)}$

Multiply the numerator and denominator by $(\sqrt{x} + 1)$:

$= \frac{(\sqrt{x} − 1)(\sqrt{x} + 1) (2x − 3)}{(x-1)(2x+3)(\sqrt{x} + 1)}$

Using the difference of squares formula $a^2 - b^2 = (a-b)(a+b)$ on the numerator $(\sqrt{x} − 1)(\sqrt{x} + 1) = (\sqrt{x})^2 - 1^2 = x - 1$:

$= \frac{(x - 1) (2x − 3)}{(x-1)(2x+3)(\sqrt{x} + 1)}$

For $x \to 1$, $x \neq 1$, so $(x-1) \neq 0$. We can cancel the common factor $(x-1)$ from the numerator and the denominator:

$= \frac{2x − 3}{(2x+3)(\sqrt{x} + 1)}$

Now, evaluate the limit of the simplified expression by substituting $x=1$:

$\lim\limits_{x \to 1} \frac{2x − 3}{(2x+3)(\sqrt{x} + 1)} = \frac{2(1) − 3}{(2(1)+3)(\sqrt{1} + 1)}$

$= \frac{2 − 3}{(2+3)(1 + 1)}$

$= \frac{-1}{(5)(2)}$

$= \frac{-1}{10}$

Method 2: Using L'Hopital's Rule

Since the limit is of the form $\frac{0}{0}$, we can apply L'Hopital's Rule.

Let $f(x) = (\sqrt{x} − 1) (2x − 3) = (x^{1/2} - 1)(2x - 3) = 2x^{3/2} - 3x^{1/2} - 2x + 3$.

Let $g(x) = 2x^2 + x - 3$.

Find the derivatives of $f(x)$ and $g(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx}(2x^{3/2} - 3x^{1/2} - 2x + 3) = 2 \cdot \frac{3}{2}x^{3/2 - 1} - 3 \cdot \frac{1}{2}x^{1/2 - 1} - 2 + 0 = 3x^{1/2} - \frac{3}{2}x^{-1/2} - 2 = 3\sqrt{x} - \frac{3}{2\sqrt{x}} - 2$.

$g'(x) = \frac{d}{dx}(2x^2 + x - 3) = 4x + 1 - 0 = 4x + 1$.

Apply L'Hopital's Rule:

$\lim\limits_{x \to 1} \frac{(\sqrt{x} − 1) (2x − 3)}{2x^2 + x − 3} = \lim\limits_{x \to 1} \frac{f'(x)}{g'(x)} = \lim\limits_{x \to 1} \frac{3\sqrt{x} - \frac{3}{2\sqrt{x}} - 2}{4x + 1}$.

Substitute $x=1$ into the expression:

$= \frac{3\sqrt{1} - \frac{3}{2\sqrt{1}} - 2}{4(1) + 1}$

$= \frac{3(1) - \frac{3}{2(1)} - 2}{4 + 1}$

$= \frac{3 - \frac{3}{2} - 2}{5}$

Calculate the numerator: $3 - 2 - \frac{3}{2} = 1 - \frac{3}{2} = \frac{2}{2} - \frac{3}{2} = -\frac{1}{2}$.

So, the limit is $\frac{-\frac{1}{2}}{5} = -\frac{1}{2} \times \frac{1}{5} = -\frac{1}{10}$.

Both methods yield the same result.

The value of the limit is $-\frac{1}{10}$.

The correct option is (B) $\frac{−1}{10}$.

Given the function:

$f(x) = \begin{cases} \frac{\sin [x]}{[x]}, &[x] ≠ 0 \\ 0, & [x] = 0 \end{cases}$

where $[x]$ denotes the greatest integer function.

We need to find $\lim\limits_{x \to 0} f(x)$.

For the limit $\lim\limits_{x \to 0} f(x)$ to exist, the left-hand limit ($\lim\limits_{x \to 0^-} f(x)$) and the right-hand limit ($\lim\limits_{x \to 0^+} f(x)$) must both exist and be equal.

Right-hand limit ($\lim\limits_{x \to 0^+} f(x)$):

As $x$ approaches 0 from the right ($x \to 0^+$), $x$ takes values slightly greater than 0 (e.g., 0.1, 0.001, ...). For any $x$ in the interval $(0, 1)$, the greatest integer $[x]$ is 0.

According to the definition of $f(x)$, when $[x] = 0$, $f(x) = 0$.

Thus, for all $x$ in the interval $(0, 1)$, $f(x) = 0$.

Therefore, the right-hand limit is:

$\lim\limits_{x \to 0^+} f(x) = \lim\limits_{x \to 0^+} 0 = 0$.

Left-hand limit ($\lim\limits_{x \to 0^-} f(x)$):

As $x$ approaches 0 from the left ($x \to 0^-$), $x$ takes values slightly less than 0 (e.g., -0.1, -0.001, ...). For any $x$ in the interval $(-1, 0)$, the greatest integer $[x]$ is -1.

According to the definition of $f(x)$, when $[x] \neq 0$ (which is the case when $[x] = -1$), $f(x) = \frac{\sin [x]}{[x]}$.

Thus, for all $x$ in the interval $(-1, 0)$, $f(x) = \frac{\sin (-1)}{-1}$.

Therefore, the left-hand limit is:

$\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^-} \frac{\sin [x]}{[x]}$

Since $[x] = -1$ for $x \in (-1, 0)$, we substitute this value:

$= \frac{\sin (-1)}{-1}$.

Using the property $\sin(-\theta) = -\sin(\theta)$, we get:

$= \frac{-\sin 1}{-1} = \sin 1$.

Note that $\sin 1$ is the sine of 1 radian, which is a constant value approximately equal to 0.841. It is not equal to 0.

Comparing the one-sided limits:

$\lim\limits_{x \to 0^+} f(x) = 0$

$\lim\limits_{x \to 0^-} f(x) = \sin 1$

Since the left-hand limit ($\sin 1$) is not equal to the right-hand limit (0), the limit of the function as $x \to 0$ does not exist.

The limit $\lim\limits_{x \to 0} f(x)$ does not exist.

The correct option is (D) None of these.

The given limit is $\lim\limits_{x \to 0} \frac{|\sin x|}{x}$.

The behavior of $|\sin x|$ depends on the sign of $\sin x$. Near $x=0$:

If $x > 0$ and close to 0 (e.g., in $(0, \pi)$), $\sin x > 0$, so $|\sin x| = \sin x$.

If $x < 0$ and close to 0 (e.g., in $(-\pi, 0)$), $\sin x < 0$, so $|\sin x| = -\sin x$.

To evaluate the limit as $x \to 0$, we need to check the left-hand limit and the right-hand limit.

Right-hand limit ($\lim\limits_{x \to 0^+}$):

As $x \to 0$ from the right, $x > 0$. For such values of $x$ near 0, $|\sin x| = \sin x$.

So, the right-hand limit is:

$\lim\limits_{x \to 0^+} \frac{|\sin x|}{x} = \lim\limits_{x \to 0^+} \frac{\sin x}{x}$.

This is a standard limit.

$\lim\limits_{x \to 0^+} \frac{\sin x}{x} = 1$.

Left-hand limit ($\lim\limits_{x \to 0^-}$):

As $x \to 0$ from the left, $x < 0$. For such values of $x$ near 0, $|\sin x| = -\sin x$.

So, the left-hand limit is:

$\lim\limits_{x \to 0^-} \frac{|\sin x|}{x} = \lim\limits_{x \to 0^-} \frac{-\sin x}{x}$.

$= - \lim\limits_{x \to 0^-} \frac{\sin x}{x}$.

The limit $\lim\limits_{x \to 0^-} \frac{\sin x}{x}$ is also a standard limit, and its value is 1.

$= -(1) = -1$.

For the limit $\lim\limits_{x \to 0} \frac{|\sin x|}{x}$ to exist, the left-hand limit must be equal to the right-hand limit.

Right-hand limit = 1

Left-hand limit = -1

Since $1 \neq -1$, the left-hand limit is not equal to the right-hand limit.

Therefore, the limit $\lim\limits_{x \to 0} \frac{|\sin x|}{x}$ does not exist.

The correct option is (C) does not exist.

Given:

The piecewise function $f(x) = \begin{cases} x^2 − 1, & 0 < x < 2 \\ 2x + 3, & 2 ≤ x < 3 \end{cases}$.

To Find:

The quadratic equation whose roots are $\lim\limits_{x \to 2^−} f(x)$ and $\lim\limits_{x \to 2^+} f(x)$.

Solution:

We need to find the left-hand limit and the right-hand limit of $f(x)$ as $x$ approaches 2.

For the left-hand limit, as $x \to 2^−$, $x$ is slightly less than 2. In this case, the definition of $f(x)$ for $0 < x < 2$ applies.

$f(x) = x^2 - 1$ for $0 < x < 2$.

The left-hand limit is:

$\lim\limits_{x \to 2^−} f(x) = \lim\limits_{x \to 2^−} (x^2 - 1)$.

Since $x^2 - 1$ is a polynomial, the limit can be evaluated by direct substitution:

$\lim\limits_{x \to 2^−} (x^2 - 1) = (2)^2 - 1 = 4 - 1 = 3$.

Let the first root be $\alpha = 3$.

For the right-hand limit, as $x \to 2^+$, $x$ is slightly greater than 2. In this case, the definition of $f(x)$ for $2 \leq x < 3$ applies.

$f(x) = 2x + 3$ for $2 \leq x < 3$.

The right-hand limit is:

$\lim\limits_{x \to 2^+} f(x) = \lim\limits_{x \to 2^+} (2x + 3)$.

Since $2x + 3$ is a polynomial, the limit can be evaluated by direct substitution:

$\lim\limits_{x \to 2^+} (2x + 3) = 2(2) + 3 = 4 + 3 = 7$.

Let the second root be $\beta = 7$.

The roots of the quadratic equation are $\alpha = 3$ and $\beta = 7$.

A quadratic equation with roots $\alpha$ and $\beta$ is given by $x^2 - (\alpha + \beta)x + \alpha\beta = 0$.

Calculate the sum of the roots:

Sum of roots $= \alpha + \beta = 3 + 7 = 10$.

Calculate the product of the roots:

Product of roots $= \alpha\beta = 3 \times 7 = 21$.

Substitute the sum and product into the quadratic equation formula:

$x^2 - (10)x + (21) = 0$.

The quadratic equation is $x^2 - 10x + 21 = 0$.

Compare this equation with the given options.

Option (A): $x^2 – 6x + 9 = 0$

Option (B): $x^2 – 7x + 8 = 0$

Option (C): $x^2 – 14x + 49 = 0$

Option (D): $x^2 – 10x + 21 = 0$

The derived equation matches Option (D).

The correct option is (D) $x^2 – 10x + 21 = 0$.

The given limit is $\lim\limits_{x \to 0} \frac{\tan 2x − x}{3x − \sin x}$.

Let's evaluate the numerator and denominator as $x \to 0$:

Numerator: $\tan(2 \times 0) - 0 = \tan(0) - 0 = 0 - 0 = 0$.

Denominator: $3(0) - \sin(0) = 0 - 0 = 0$.

The limit is of the indeterminate form $\frac{0}{0}$.

Method 1: Dividing by $x$ and using standard limits

We can divide both the numerator and the denominator by $x$ (since $x \to 0$, $x \neq 0$).

$\lim\limits_{x \to 0} \frac{\frac{\tan 2x − x}{x}}{\frac{3x − \sin x}{x}}$

$= \lim\limits_{x \to 0} \frac{\frac{\tan 2x}{x} − \frac{x}{x}}{\frac{3x}{x} − \frac{\sin x}{x}}$

$= \lim\limits_{x \to 0} \frac{\frac{\tan 2x}{x} − 1}{3 − \frac{\sin x}{x}}$.

We know the standard limits $\lim\limits_{x \to 0} \frac{\tan kx}{x} = k$ and $\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$.

Using the property that the limit of a quotient is the quotient of the limits (provided the denominator limit is non-zero) and the limit of a sum/difference is the sum/difference of the limits:

$= \frac{\lim\limits_{x \to 0} \left(\frac{\tan 2x}{x} − 1\right)}{\lim\limits_{x \to 0} \left(3 − \frac{\sin x}{x}\right)}$

$= \frac{\lim\limits_{x \to 0} \frac{\tan 2x}{x} − \lim\limits_{x \to 0} 1}{\lim\limits_{x \to 0} 3 − \lim\limits_{x \to 0} \frac{\sin x}{x}}$

Evaluate the standard limits:

$\lim\limits_{x \to 0} \frac{\tan 2x}{x} = 2$ (Using $\lim\limits_{x \to 0} \frac{\tan y}{y} = 1$, let $y=2x$, then $\lim\limits_{x \to 0} \frac{\tan 2x}{x} = \lim\limits_{x \to 0} \frac{\tan 2x}{2x} \cdot 2 = 1 \cdot 2 = 2$).

$\lim\limits_{x \to 0} 1 = 1$.

$\lim\limits_{x \to 0} 3 = 3$.

$\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$.

Substitute these values:

$= \frac{2 − 1}{3 − 1} = \frac{1}{2}$.

Method 2: Using L'Hopital's Rule

Since the limit is of the form $\frac{0}{0}$, we can apply L'Hopital's Rule.

Let $f(x) = \tan 2x - x$ and $g(x) = 3x - \sin x$.

Find the derivatives:

$f'(x) = \frac{d}{dx}(\tan 2x - x) = \sec^2(2x) \cdot \frac{d}{dx}(2x) - 1 = 2 \sec^2(2x) - 1$.

$g'(x) = \frac{d}{dx}(3x - \sin x) = 3 - \cos x$.

Apply L'Hopital's Rule:

$\lim\limits_{x \to 0} \frac{\tan 2x − x}{3x − \sin x} = \lim\limits_{x \to 0} \frac{2 \sec^2(2x) - 1}{3 - \cos x}$.

Now substitute $x=0$:

$= \frac{2 \sec^2(2 \times 0) - 1}{3 - \cos(0)}$

$= \frac{2 \sec^2(0) - 1}{3 - 1}$.

Since $\sec(0) = \frac{1}{\cos(0)} = \frac{1}{1} = 1$, $\sec^2(0) = 1^2 = 1$.

$= \frac{2(1) - 1}{2} = \frac{2 - 1}{2} = \frac{1}{2}$.

Both methods yield the same result.

The value of the limit is $\frac{1}{2}$.

The correct option is (B) $\frac{1}{2}$.

Given:

The function $f(x) = x - [x]$, where $[x]$ denotes the greatest integer function.

The function $f(x) = x - [x]$ is also known as the fractional part of $x$, denoted by $\{x\}$. So, $f(x) = \{x\}$.

To Find:

The derivative of $f(x)$ at $x = \frac{1}{2}$, i.e., $f'\left(\frac{1}{2}\right)$.

Solution:

We need to find the derivative of $f(x) = x - [x]$. The greatest integer function $[x]$ is a step function that is constant between integers.

For $x = \frac{1}{2}$, we need to consider the behavior of $f(x)$ in an open interval around $x = \frac{1}{2}$ that does not contain any integer.

Consider an open interval around $x = \frac{1}{2}$, such as $\left(0, 1\right)$. For any $x$ in the interval $\left(0, 1\right)$, the greatest integer $[x]$ is 0.

For $x \in \left(0, 1\right)$, the function $f(x)$ is defined as:

$f(x) = x - [x] = x - 0 = x$.

So, in the interval $\left(0, 1\right)$, the function is simply $f(x) = x$.

The point $x = \frac{1}{2}$ is within the interval $\left(0, 1\right)$.

The derivative of $f(x) = x$ with respect to $x$ is:

$f'(x) = \frac{d}{dx}(x) = 1$.

Since $f'(x) = 1$ for all $x$ in the interval $\left(0, 1\right)$, the derivative at $x = \frac{1}{2}$ is:

$f'\left(\frac{1}{2}\right) = 1$.

The function $[x]$ is differentiable at any point $c$ that is not an integer, and its derivative at such a point is 0.

The derivative of $x$ is 1.

For any $x$ that is not an integer, $f'(x) = \frac{d}{dx}(x - [x]) = \frac{d}{dx}(x) - \frac{d}{dx}([x])$.

If $x$ is not an integer, $\frac{d}{dx}([x]) = 0$.

So, for any $x$ not an integer, $f'(x) = 1 - 0 = 1$.

Since $\frac{1}{2}$ is not an integer, the derivative of $f(x)$ at $x = \frac{1}{2}$ is 1.

The value of $f'\left(\frac{1}{2}\right)$ is 1.

The correct option is (B) 1.

Given:

The function $y = \sqrt{x} + \frac{1}{\sqrt{x}}$.

We can rewrite the function using exponents: $y = x^{1/2} + x^{-1/2}$.

To Find:

The derivative $\frac{dy}{dx}$ at $x = 1$.

Solution:

We need to find the derivative of $y$ with respect to $x$. We use the power rule for differentiation, which states that $\frac{d}{dx}(x^n) = nx^{n-1}$.

$\frac{dy}{dx} = \frac{d}{dx}\left(x^{1/2} + x^{-1/2}\right)$.

Using the linearity of differentiation, we can differentiate each term separately:

$\frac{dy}{dx} = \frac{d}{dx}(x^{1/2}) + \frac{d}{dx}(x^{-1/2})$.

Apply the power rule to the first term (with $n = \frac{1}{2}$):

$\frac{d}{dx}(x^{1/2}) = \frac{1}{2} x^{1/2 - 1} = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}}$.

Apply the power rule to the second term (with $n = -\frac{1}{2}$):

$\frac{d}{dx}(x^{-1/2}) = -\frac{1}{2} x^{-1/2 - 1} = -\frac{1}{2} x^{-3/2}$.

So, the derivative $\frac{dy}{dx}$ is:

$\frac{dy}{dx} = \frac{1}{2\sqrt{x}} - \frac{1}{2} x^{-3/2}$.

We can rewrite $x^{-3/2}$ as $\frac{1}{x^{3/2}} = \frac{1}{x\sqrt{x}}$.

$\frac{dy}{dx} = \frac{1}{2\sqrt{x}} - \frac{1}{2x\sqrt{x}}$.

Now, we need to evaluate $\frac{dy}{dx}$ at $x = 1$. Substitute $x = 1$ into the derivative expression:

$\left.\frac{dy}{dx}\right|_{x=1} = \frac{1}{2\sqrt{1}} - \frac{1}{2(1)\sqrt{1}}$.

$= \frac{1}{2(1)} - \frac{1}{2(1)(1)}$.

$= \frac{1}{2} - \frac{1}{2}$.

$= 0$.

Alternatively, we could combine the terms in the derivative before substituting $x=1$:

$\frac{dy}{dx} = \frac{1}{2\sqrt{x}} - \frac{1}{2x\sqrt{x}} = \frac{x}{2x\sqrt{x}} - \frac{1}{2x\sqrt{x}} = \frac{x-1}{2x\sqrt{x}}$.

Now substitute $x=1$:

$\left.\frac{dy}{dx}\right|_{x=1} = \frac{1-1}{2(1)\sqrt{1}} = \frac{0}{2(1)} = \frac{0}{2} = 0$.

The value of $\frac{dy}{dx}$ at $x = 1$ is 0.

The correct option is (D) 0.

Given:

The function $f(x) = \frac{x − 4}{2\sqrt{x}}$.

We can rewrite the function as $f(x) = \frac{1}{2} \left(\frac{x}{\sqrt{x}} - \frac{4}{\sqrt{x}}\right) = \frac{1}{2} \left(\sqrt{x} - 4x^{-1/2}\right) = \frac{1}{2} \left(x^{1/2} - 4x^{-1/2}\right)$.

To Find:

The derivative of $f(x)$ at $x = 1$, i.e., $f'(1)$.

Solution:

We need to find the derivative of $f(x)$ with respect to $x$. We use the power rule for differentiation, $\frac{d}{dx}(x^n) = nx^{n-1}$, and the constant multiple rule, $\frac{d}{dx}(cf(x)) = c\frac{d}{dx}(f(x))$.

$f'(x) = \frac{d}{dx} \left[\frac{1}{2} \left(x^{1/2} - 4x^{-1/2}\right)\right]$.

Using the constant multiple rule:

$f'(x) = \frac{1}{2} \frac{d}{dx} \left(x^{1/2} - 4x^{-1/2}\right)$.

Using the difference rule:

$f'(x) = \frac{1}{2} \left[\frac{d}{dx}(x^{1/2}) - \frac{d}{dx}(4x^{-1/2})\right]$.

Apply the power rule to the first term (with $n = \frac{1}{2}$):

$\frac{d}{dx}(x^{1/2}) = \frac{1}{2} x^{1/2 - 1} = \frac{1}{2} x^{-1/2}$.

Apply the constant multiple rule and power rule to the second term (with $c=4$, $n = -\frac{1}{2}$):

$\frac{d}{dx}(4x^{-1/2}) = 4 \cdot \left(-\frac{1}{2}\right) x^{-1/2 - 1} = -2 x^{-3/2}$.

Substitute these derivatives back into the expression for $f'(x)$:

$f'(x) = \frac{1}{2} \left[\frac{1}{2} x^{-1/2} - (-2 x^{-3/2})\right]$.

$f'(x) = \frac{1}{2} \left[\frac{1}{2} x^{-1/2} + 2 x^{-3/2}\right]$.

$f'(x) = \frac{1}{4} x^{-1/2} + x^{-3/2}$.

Now, we need to evaluate $f'(x)$ at $x = 1$. Substitute $x = 1$ into the derivative expression:

$f'(1) = \frac{1}{4} (1)^{-1/2} + (1)^{-3/2}$.

Since $1$ raised to any power is $1$, $(1)^{-1/2} = 1$ and $(1)^{-3/2} = 1$.

$f'(1) = \frac{1}{4} (1) + (1)$.

$f'(1) = \frac{1}{4} + 1$.

Combine the terms:

$f'(1) = \frac{1}{4} + \frac{4}{4} = \frac{1+4}{4} = \frac{5}{4}$.

Alternatively, using the Quotient Rule: If $f(x) = \frac{u(x)}{v(x)}$, then $f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$.

Let $u(x) = x - 4$ and $v(x) = 2\sqrt{x} = 2x^{1/2}$.

Find the derivatives of $u(x)$ and $v(x)$:

$u'(x) = \frac{d}{dx}(x - 4) = 1 - 0 = 1$.

$v'(x) = \frac{d}{dx}(2x^{1/2}) = 2 \cdot \frac{1}{2} x^{1/2 - 1} = x^{-1/2} = \frac{1}{\sqrt{x}}$.

Apply the Quotient Rule:

$f'(x) = \frac{(1)(2\sqrt{x}) - (x - 4)(\frac{1}{\sqrt{x}})}{(2\sqrt{x})^2}$.

$f'(x) = \frac{2\sqrt{x} - \left(\frac{x}{\sqrt{x}} - \frac{4}{\sqrt{x}}\right)}{4x}$.

$f'(x) = \frac{2\sqrt{x} - \left(\sqrt{x} - \frac{4}{\sqrt{x}}\right)}{4x}$.

$f'(x) = \frac{2\sqrt{x} - \sqrt{x} + \frac{4}{\sqrt{x}}}{4x}$.

$f'(x) = \frac{\sqrt{x} + \frac{4}{\sqrt{x}}}{4x}$.

To simplify the numerator, find a common denominator:

$\sqrt{x} + \frac{4}{\sqrt{x}} = \frac{(\sqrt{x})(\sqrt{x})}{\sqrt{x}} + \frac{4}{\sqrt{x}} = \frac{x + 4}{\sqrt{x}}$.

Substitute this back into the expression for $f'(x)$:

$f'(x) = \frac{\frac{x + 4}{\sqrt{x}}}{4x} = \frac{x + 4}{\sqrt{x}} \cdot \frac{1}{4x} = \frac{x + 4}{4x\sqrt{x}}$.

Now, evaluate $f'(x)$ at $x = 1$. Substitute $x = 1$:

$f'(1) = \frac{1 + 4}{4(1)\sqrt{1}} = \frac{5}{4(1)(1)} = \frac{5}{4}$.

Both methods yield the same result.

The value of $f'(1)$ is $\frac{5}{4}$.

The correct option is (A) $\frac{5}{4}$.

Given:

The function $y = \frac{1 + \frac{1}{x^2}}{1 − \frac{1}{x^2}}$.

To Find:

The derivative $\frac{dy}{dx}$.

Solution:

First, let's simplify the expression for $y$. Multiply the numerator and the denominator by $x^2$ to eliminate the fractions within the numerator and denominator:

$y = \frac{\left(1 + \frac{1}{x^2}\right) \cdot x^2}{\left(1 − \frac{1}{x^2}\right) \cdot x^2}$

$y = \frac{1 \cdot x^2 + \frac{1}{x^2} \cdot x^2}{1 \cdot x^2 − \frac{1}{x^2} \cdot x^2}$

$y = \frac{x^2 + 1}{x^2 - 1}$.

Now we differentiate $y = \frac{x^2 + 1}{x^2 - 1}$ with respect to $x$. We can use the Quotient Rule. The Quotient Rule states that if $y = \frac{u(x)}{v(x)}$, then $\frac{dy}{dx} = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$.

Let $u(x) = x^2 + 1$ and $v(x) = x^2 - 1$.

Find the derivative of $u(x)$:

$u'(x) = \frac{d}{dx}(x^2 + 1) = 2x + 0 = 2x$.

Find the derivative of $v(x)$:

$v'(x) = \frac{d}{dx}(x^2 - 1) = 2x - 0 = 2x$.

Apply the Quotient Rule:

$\frac{dy}{dx} = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$

$\frac{dy}{dx} = \frac{(2x)(x^2 - 1) - (x^2 + 1)(2x)}{(x^2 - 1)^2}$.

Simplify the numerator:

Numerator $= 2x(x^2 - 1) - 2x(x^2 + 1)$

Numerator $= 2x^3 - 2x - (2x^3 + 2x)$

Numerator $= 2x^3 - 2x - 2x^3 - 2x$

Numerator $= (2x^3 - 2x^3) + (-2x - 2x)$

Numerator $= 0 - 4x = -4x$.

The denominator is $(x^2 - 1)^2$.

So, $\frac{dy}{dx} = \frac{-4x}{(x^2 - 1)^2}$.

The derivative $\frac{dy}{dx}$ is $\frac{-4x}{(x^2 - 1)^2}$.

Compare this result with the given options.

Option (A): $\frac{−4x}{(x^2 − 1)^2}$

Option (B): $\frac{−4x}{x^2 − 1}$

Option (C): $\frac{1 − x^2}{4x}$

Option (D): $\frac{4x}{x^2 − 1}$

The derived derivative matches Option (A).

The correct option is (A) $\frac{−4x}{(x^2 − 1)^2}$.

Given:

The function $y = \frac{\sin x + \cos x}{\sin x − \cos x}$.

To Find:

The derivative $\frac{dy}{dx}$ at $x = 0$.

Solution:

We need to find the derivative of $y$ with respect to $x$. We use the Quotient Rule. The Quotient Rule states that if $y = \frac{u(x)}{v(x)}$, then $\frac{dy}{dx} = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$.

Let $u(x) = \sin x + \cos x$ and $v(x) = \sin x − \cos x$.

Find the derivative of $u(x)$:

$u'(x) = \frac{d}{dx}(\sin x + \cos x) = \cos x - \sin x$.

Find the derivative of $v(x)$:

$v'(x) = \frac{d}{dx}(\sin x − \cos x) = \cos x - (-\sin x) = \cos x + \sin x$.

Apply the Quotient Rule:

$\frac{dy}{dx} = \frac{(\cos x - \sin x)(\sin x − \cos x) - (\sin x + \cos x)(\cos x + \sin x)}{(\sin x − \cos x)^2}$.

Simplify the numerator:

Numerator $= -(\sin x - \cos x)(\sin x - \cos x) - (\sin x + \cos x)(\sin x + \cos x)$

Numerator $= -(\sin x - \cos x)^2 - (\sin x + \cos x)^2$.

Recall the identities: $(a-b)^2 = a^2 - 2ab + b^2$ and $(a+b)^2 = a^2 + 2ab + b^2$.

Numerator $= -(\sin^2 x - 2 \sin x \cos x + \cos^2 x) - (\sin^2 x + 2 \sin x \cos x + \cos^2 x)$.

Using the identity $\sin^2 x + \cos^2 x = 1$:

Numerator $= -(1 - 2 \sin x \cos x) - (1 + 2 \sin x \cos x)$.

Numerator $= -1 + 2 \sin x \cos x - 1 - 2 \sin x \cos x$.

Numerator $= (-1 - 1) + (2 \sin x \cos x - 2 \sin x \cos x)$.

Numerator $= -2 + 0 = -2$.

The denominator is $(\sin x − \cos x)^2$.

So, $\frac{dy}{dx} = \frac{-2}{(\sin x − \cos x)^2}$.

Now, we need to evaluate $\frac{dy}{dx}$ at $x = 0$. Substitute $x = 0$ into the derivative expression:

$\left.\frac{dy}{dx}\right|_{x=0} = \frac{-2}{(\sin(0) − \cos(0))^2}$.

Evaluate $\sin(0)$ and $\cos(0)$:

$\sin(0) = 0$

$\cos(0) = 1$

Substitute these values:

$= \frac{-2}{(0 − 1)^2}$.

$= \frac{-2}{(-1)^2}$.

$= \frac{-2}{1} = -2$.

Alternatively, simplify $y$ first by dividing the numerator and denominator by $\cos x$ (assuming $\cos x \neq 0$):

$y = \frac{\frac{\sin x}{\cos x} + \frac{\cos x}{\cos x}}{\frac{\sin x}{\cos x} − \frac{\cos x}{\cos x}} = \frac{\tan x + 1}{\tan x − 1}$.

Now differentiate $y = \frac{\tan x + 1}{\tan x − 1}$ using the Quotient Rule.

Let $u(x) = \tan x + 1$ and $v(x) = \tan x - 1$.

$u'(x) = \frac{d}{dx}(\tan x + 1) = \sec^2 x$.

$v'(x) = \frac{d}{dx}(\tan x - 1) = \sec^2 x$.

Apply the Quotient Rule:

$\frac{dy}{dx} = \frac{(\sec^2 x)(\tan x - 1) - (\tan x + 1)(\sec^2 x)}{(\tan x − 1)^2}$.

Factor out $\sec^2 x$ from the numerator:

Numerator $= \sec^2 x [(\tan x - 1) - (\tan x + 1)]$

Numerator $= \sec^2 x [\tan x - 1 - \tan x - 1]$

Numerator $= \sec^2 x [-2] = -2 \sec^2 x$.

So, $\frac{dy}{dx} = \frac{-2 \sec^2 x}{(\tan x − 1)^2}$.

Now, evaluate $\frac{dy}{dx}$ at $x = 0$. Substitute $x = 0$:

$\left.\frac{dy}{dx}\right|_{x=0} = \frac{-2 \sec^2(0)}{(\tan(0) − 1)^2}$.

Evaluate $\sec(0)$ and $\tan(0)$:

$\sec(0) = 1$

$\tan(0) = 0$

Substitute these values:

$= \frac{-2 (1)^2}{(0 − 1)^2}$.

$= \frac{-2}{(-1)^2}$.

$= \frac{-2}{1} = -2$.

Both methods yield the same result.

The value of $\frac{dy}{dx}$ at $x = 0$ is –2.

The correct option is (A) –2.

Given:

The function $y = \frac{\sin (x + 9)}{\cos x}$.

To Find:

The derivative $\frac{dy}{dx}$ at $x = 0$.

Solution:

We need to find the derivative of $y$ with respect to $x$. We use the Quotient Rule. The Quotient Rule states that if $y = \frac{u(x)}{v(x)}$, then $\frac{dy}{dx} = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$.

Let $u(x) = \sin (x + 9)$ and $v(x) = \cos x$.

Find the derivative of $u(x)$ using the chain rule:

$u'(x) = \frac{d}{dx}(\sin (x + 9)) = \cos (x + 9) \cdot \frac{d}{dx}(x + 9) = \cos (x + 9) \cdot 1 = \cos (x + 9)$.

Find the derivative of $v(x)$:

$v'(x) = \frac{d}{dx}(\cos x) = -\sin x$.

Apply the Quotient Rule:

$\frac{dy}{dx} = \frac{(\cos (x + 9))(\cos x) - (\sin (x + 9))(-\sin x)}{(\cos x)^2}$.

Simplify the numerator:

Numerator $= \cos (x + 9) \cos x + \sin (x + 9) \sin x$.

Recall the trigonometric identity for the cosine of a difference: $\cos(A - B) = \cos A \cos B + \sin A \sin B$.

Let $A = x + 9$ and $B = x$.

Numerator $= \cos((x + 9) - x) = \cos(9)$.

The denominator is $(\cos x)^2 = \cos^2 x$.

So, the derivative $\frac{dy}{dx}$ is:

$\frac{dy}{dx} = \frac{\cos 9}{\cos^2 x}$.

Now, we need to evaluate $\frac{dy}{dx}$ at $x = 0$. Substitute $x = 0$ into the derivative expression:

$\left.\frac{dy}{dx}\right|_{x=0} = \frac{\cos 9}{\cos^2(0)}$.

Evaluate $\cos(0)$:

$\cos(0) = 1$.

Substitute this value:

$= \frac{\cos 9}{(1)^2}$.

$= \frac{\cos 9}{1} = \cos 9$.

The value of $\frac{dy}{dx}$ at $x = 0$ is $\cos 9$.

The correct option is (A) cos 9.

Given:

The function $f(x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{3} + \dots + \frac{x^{100}}{100}$.

This is a polynomial function.

To Find:

The derivative of $f(x)$ at $x = 1$, i.e., $f'(1)$.

Solution:

We need to find the derivative of $f(x)$ with respect to $x$. We can differentiate each term of the polynomial using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$ and the derivative of a constant $\frac{d}{dx}(c) = 0$.

$f'(x) = \frac{d}{dx}\left(1 + x + \frac{x^2}{2} + \frac{x^3}{3} + \dots + \frac{x^{100}}{100}\right)$.

Differentiate each term:

$\frac{d}{dx}(1) = 0$.

$\frac{d}{dx}(x) = 1$.

$\frac{d}{dx}\left(\frac{x^2}{2}\right) = \frac{1}{2} \frac{d}{dx}(x^2) = \frac{1}{2} (2x) = x$.

$\frac{d}{dx}\left(\frac{x^3}{3}\right) = \frac{1}{3} \frac{d}{dx}(x^3) = \frac{1}{3} (3x^2) = x^2$.

... and so on for the intermediate terms.

The general term is $\frac{x^k}{k}$. Its derivative is $\frac{d}{dx}\left(\frac{x^k}{k}\right) = \frac{1}{k} \frac{d}{dx}(x^k) = \frac{1}{k} (kx^{k-1}) = x^{k-1}$ (for $k \geq 1$).

The derivative of the last term is $\frac{d}{dx}\left(\frac{x^{100}}{100}\right) = \frac{1}{100} \frac{d}{dx}(x^{100}) = \frac{1}{100} (100x^{99}) = x^{99}$.

So, the derivative function $f'(x)$ is:

$f'(x) = 0 + 1 + x + x^2 + x^3 + \dots + x^{99}$.

$f'(x) = 1 + x + x^2 + x^3 + \dots + x^{99}$.

Now, we need to evaluate $f'(x)$ at $x = 1$. Substitute $x = 1$ into the expression for $f'(x)$:

$f'(1) = 1 + (1) + (1)^2 + (1)^3 + \dots + (1)^{99}$.

Each term in this sum is $1$. We need to find the number of terms in the sum $1 + x + x^2 + \dots + x^{99}$.

This is a geometric series starting with $x^0$ and ending with $x^{99}$. The powers range from 0 to 99, inclusive.

The number of terms is $99 - 0 + 1 = 100$.

So, $f'(1)$ is the sum of 100 terms, each equal to 1.

$f'(1) = \underbrace{1 + 1 + 1 + \dots + 1}_{\text{100 terms}}$.

$f'(1) = 100 \times 1 = 100$.

The value of $f'(1)$ is 100.

The correct option is (B) 100.

Given:

The function $f(x) = \frac{x^n − a^n}{x − a}$, for $x \neq a$. The function is not defined at $x=a$.

To Find:

The value of $f'(a)$.

Solution:

The domain of the function $f(x)$ as given is $\{x \in \mathbb{R} \mid x \neq a\}$. Since $a$ is not in the domain of $f(x)$, the derivative $f'(a)$ is strictly speaking undefined at this point based on the standard definition of the derivative at a point $c$, which requires $c$ to be in the domain of the function.

However, in the context of calculus problems involving removable singularities, when asked for the derivative at such a point, it often implies finding the derivative of the continuous extension of the function at that point, if it exists.

Let's consider the limit of $f(x)$ as $x \to a$. We know the standard limit form $\lim\limits_{x \to a} \frac{x^n - a^n}{x - a} = na^{n-1}$.

Let $F(x)$ be the continuous extension of $f(x)$, defined as:

$F(x) = \begin{cases} \frac{x^n - a^n}{x - a}, & x \neq a \\ na^{n-1}, & x = a \end{cases}$

We interpret the question as asking for the derivative of this continuous extension $F(x)$ at $x=a$, i.e., $F'(a)$.

By the definition of the derivative:

$F'(a) = \lim\limits_{x \to a} \frac{F(x) - F(a)}{x - a}$

$F'(a) = \lim\limits_{x \to a} \frac{\frac{x^n - a^n}{x - a} - na^{n-1}}{x - a}$

$F'(a) = \lim\limits_{x \to a} \frac{x^n - a^n - na^{n-1}(x - a)}{(x - a)^2}$.

As $x \to a$, the numerator approaches $a^n - a^n - na^{n-1}(a - a) = 0$, and the denominator approaches $(a - a)^2 = 0$. This is an indeterminate form $\frac{0}{0}$. We can use L'Hopital's Rule twice.

Let $N(x) = x^n - a^n - na^{n-1}(x - a)$ and $D(x) = (x - a)^2$.

$N'(x) = \frac{d}{dx}(x^n - a^n - na^{n-1}x + na^n) = nx^{n-1} - 0 - na^{n-1} + 0 = nx^{n-1} - na^{n-1}$.

$D'(x) = \frac{d}{dx}((x - a)^2) = 2(x - a) \cdot \frac{d}{dx}(x - a) = 2(x - a) \cdot 1 = 2(x - a)$.

So, $F'(a) = \lim\limits_{x \to a} \frac{nx^{n-1} - na^{n-1}}{2(x - a)}$.

This is still of the form $\frac{0}{0}$ as $x \to a$. Apply L'Hopital's Rule again.

$N''(x) = \frac{d}{dx}(nx^{n-1} - na^{n-1}) = n(n-1)x^{n-2}$.

$D''(x) = \frac{d}{dx}(2(x - a)) = 2$.

So, $F'(a) = \lim\limits_{x \to a} \frac{n(n-1)x^{n-2}}{2}$.

Substitute $x=a$:

$F'(a) = \frac{n(n-1)a^{n-2}}{2}$.

The derivative of the continuous extension at $x=a$ is $\frac{n(n-1)}{2}a^{n-2}$.

For this result to match one of the constant options (A), (B), or (D), the expression $\frac{n(n-1)}{2}a^{n-2}$ must be a constant value independent of $n$ and $a$, or correspond to specific values of $n$ that yield one of these options for any valid $a$.

Let's consider specific integer values of $n$, as the form $\frac{x^n - a^n}{x - a}$ is common for integer powers.

If $n=2$, $F'(a) = \frac{2(2-1)}{2}a^{2-2} = \frac{2 \cdot 1}{2} a^0 = 1 \cdot 1 = 1$ (assuming $a^0=1$, which holds for $a \neq 0$). If $a=0$ and $n=2$, $f(x) = \frac{x^2}{x} = x$ for $x \neq 0$. The continuous extension is $F(x) = x$. $F'(x) = 1$. $F'(0) = 1$. So for $n=2$, $F'(a)=1$ for all $a$. This matches Option (A).

If $n=1$, $F'(a) = \frac{1(1-1)}{2}a^{1-2} = \frac{1 \cdot 0}{2} a^{-1} = 0 \cdot a^{-1} = 0$ (assuming $a \neq 0$). If $a=0$ and $n=1$, $f(x) = \frac{x}{x} = 1$ for $x \neq 0$. The continuous extension is $F(x) = 1$. $F'(x) = 0$. $F'(0) = 0$. So for $n=1$, $F'(a)=0$ for all $a$. This matches Option (B).

If $n=0$, $F'(a) = \frac{0(0-1)}{2}a^{0-2} = \frac{0 \cdot (-1)}{2} a^{-2} = 0 \cdot a^{-2} = 0$ (assuming $a \neq 0$). If $a=0$ and $n=0$, $f(x) = \frac{x^0 - a^0}{x - a} = \frac{1-1}{x-a} = 0$ for $x \neq a, a \neq 0$. If $a=0$, $f(x) = \frac{1-1}{x} = 0$ for $x \neq 0$. $F(x)=0$. $F'(x)=0$. $F'(0)=0$. So for $n=0$, $F'(a)=0$ for all $a$ (where the function is defined). This matches Option (B).

Since the question uses a general 'n' but provides specific numerical options, it is likely that the intended question refers to a specific, commonly encountered value of $n$ that yields one of the given constant options regardless of the value of 'a'. The case $n=2$ gives a constant result 1, which is Option (A). The cases $n=0$ and $n=1$ give a constant result 0, which is Option (B).

Without further clarification, the question is ambiguous. However, if a single option must be chosen, the case $n=2$ is a very common and simple example related to this function form.

Assuming the intended question corresponds to the case $n=2$, the derivative $f'(a)$ (interpreted as the derivative of the continuous extension) is 1.

Final Answer based on the most likely intended interpretation ($n=2$):

If $n=2$, $f(x) = \frac{x^2 - a^2}{x - a} = x + a$ for $x \neq a$.

The continuous extension is $F(x) = x + a$.

The derivative $F'(x) = 1$.

Evaluating at $x=a$, $F'(a) = 1$.

The correct option is (A) 1.

Given:

The function $f(x) = x^{100} + x^{99} + \dots + x + 1$.

This is a polynomial function, which can be written in standard form as $f(x) = x^{100} + x^{99} + \dots + x^1 + x^0$.

To Find:

The derivative of $f(x)$ at $x = 1$, i.e., $f'(1)$.

Solution:

We need to find the derivative of $f(x)$ with respect to $x$. We differentiate each term of the polynomial using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$ and the derivative of a constant $\frac{d}{dx}(c) = 0$.

$f'(x) = \frac{d}{dx}(x^{100}) + \frac{d}{dx}(x^{99}) + \dots + \frac{d}{dx}(x^1) + \frac{d}{dx}(x^0)$.

Applying the power rule to each term:

$\frac{d}{dx}(x^{100}) = 100x^{100-1} = 100x^{99}$.

$\frac{d}{dx}(x^{99}) = 99x^{99-1} = 99x^{98}$.

... (The pattern continues for all powers down to $x^1$)

$\frac{d}{dx}(x^2) = 2x^{2-1} = 2x^1 = 2x$.

$\frac{d}{dx}(x^1) = 1x^{1-1} = 1x^0 = 1$.

$\frac{d}{dx}(x^0) = \frac{d}{dx}(1) = 0$.

Summing these derivatives, we get $f'(x)$:

$f'(x) = 100x^{99} + 99x^{98} + \dots + 2x + 1 + 0$.

$f'(x) = 100x^{99} + 99x^{98} + \dots + 2x + 1$.

Now, we need to evaluate $f'(x)$ at $x = 1$. Substitute $x = 1$ into the expression for $f'(x)$:

$f'(1) = 100(1)^{99} + 99(1)^{98} + \dots + 2(1)^1 + 1(1)^0$.

Since $1$ raised to any integer power is $1$, this simplifies to:

$f'(1) = 100 \times 1 + 99 \times 1 + \dots + 2 \times 1 + 1 \times 1$.

$f'(1) = 100 + 99 + \dots + 2 + 1$.

This is the sum of the first 100 positive integers (from 1 to 100).

The sum of the first $N$ positive integers is given by the formula $S_N = \frac{N(N+1)}{2}$.

Here, $N = 100$.

The sum is:

$f'(1) = S_{100} = \frac{100(100+1)}{2}$.

$f'(1) = \frac{100 \times 101}{2}$.

Cancel the 2 with 100:

$f'(1) = \frac{\cancel{100}^{50} \times 101}{\cancel{2}_{1}}$.

$f'(1) = 50 \times 101$.

$50 \times 101 = 50 \times (100 + 1) = 50 \times 100 + 50 \times 1 = 5000 + 50 = 5050$.

The value of $f'(1)$ is 5050.

This matches Option (A).

Given:

The function $f(x) = 1 – x + x^2 – x^3 + \dots – x^{99} + x^{100}$.

This is a polynomial function.

To Find:

The derivative of $f(x)$ at $x = 1$, i.e., $f'(1)$.

Solution:

The function can be written as a sum: $f(x) = \sum_{k=0}^{100} (-1)^k x^k$.

We need to find the derivative of $f(x)$ with respect to $x$. We can differentiate each term of the polynomial using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$ and the derivative of a constant $\frac{d}{dx}(c) = 0$.

$f'(x) = \frac{d}{dx}\left(\sum_{k=0}^{100} (-1)^k x^k\right) = \sum_{k=0}^{100} \frac{d}{dx}((-1)^k x^k)$.

For $k=0$, the term is $(-1)^0 x^0 = 1$, and its derivative is $\frac{d}{dx}(1) = 0$.

For $k \geq 1$, the derivative of $(-1)^k x^k$ is $(-1)^k \frac{d}{dx}(x^k) = (-1)^k k x^{k-1}$.

So, $f'(x) = 0 + \sum_{k=1}^{100} (-1)^k k x^{k-1}$.

$f'(x) = (-1)^1 (1) x^{1-1} + (-1)^2 (2) x^{2-1} + (-1)^3 (3) x^{3-1} + \dots + (-1)^{100} (100) x^{100-1}$.

$f'(x) = -1 x^0 + 2 x^1 - 3 x^2 + 4 x^3 - \dots - 99 x^{98} + 100 x^{99}$.

$f'(x) = -1 + 2x - 3x^2 + 4x^3 - \dots - 99x^{98} + 100x^{99}$.

Now, we need to evaluate $f'(x)$ at $x = 1$. Substitute $x = 1$ into the expression for $f'(x)$:

$f'(1) = -1 + 2(1) - 3(1)^2 + 4(1)^3 - \dots - 99(1)^{98} + 100(1)^{99}$.

Since $1$ raised to any integer power is $1$, this simplifies to:

$f'(1) = -1 + 2 - 3 + 4 - \dots - 99 + 100$.

This is an alternating series. We can group the terms in pairs:

$f'(1) = (-1 + 2) + (-3 + 4) + (-5 + 6) + \dots + (-99 + 100)$.

Each pair sums to 1:

$(-1 + 2) = 1$

$(-3 + 4) = 1$

$(-5 + 6) = 1$

... and so on.

The pairs cover the integers from 1 to 100. There are $100$ numbers, so there are $100/2 = 50$ such pairs.

The sum is the sum of 50 terms, each equal to 1:

$f'(1) = \underbrace{1 + 1 + \dots + 1}_{\text{50 terms}}$.

$f'(1) = 50 \times 1 = 50$.

Alternative Method using Geometric Series:

The function $f(x) = 1 - x + x^2 - x^3 + \dots + x^{100}$ is a finite geometric series with first term $a=1$, common ratio $r = -x$, and number of terms $N=101$ (from $k=0$ to $k=100$).

The sum of this series is $f(x) = a \frac{1 - r^N}{1 - r} = 1 \cdot \frac{1 - (-x)^{101}}{1 - (-x)} = \frac{1 - (-1)^{101} x^{101}}{1 + x}$.

Since 101 is odd, $(-1)^{101} = -1$.

$f(x) = \frac{1 - (-1) x^{101}}{1 + x} = \frac{1 + x^{101}}{1 + x}$, for $x \neq -1$.

Now we find the derivative of $f(x) = \frac{1 + x^{101}}{1 + x}$ using the Quotient Rule $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$.

Let $u(x) = 1 + x^{101}$, so $u'(x) = 101x^{100}$.

Let $v(x) = 1 + x$, so $v'(x) = 1$.

$f'(x) = \frac{(101x^{100})(1 + x) - (1 + x^{101})(1)}{(1 + x)^2}$.

Now, evaluate $f'(1)$ by substituting $x=1$:

$f'(1) = \frac{(101(1)^{100})(1 + 1) - (1 + (1)^{101})(1)}{(1 + 1)^2}$.

$f'(1) = \frac{(101)(2) - (1 + 1)(1)}{(2)^2}$.

$f'(1) = \frac{202 - (2)(1)}{4}$.

$f'(1) = \frac{202 - 2}{4} = \frac{200}{4}$.

$f'(1) = 50$.

Both methods yield the same result.

The value of $f'(1)$ is 50.

The correct option is (D) 50.

The given limit is $\lim\limits_{x \to \pi} \frac{\tan x}{x − π}$.

As $x \to \pi$, the numerator $\tan x \to \tan \pi = 0$.

The denominator $x - \pi \to \pi - \pi = 0$.

This is an indeterminate form of type $\frac{0}{0}$.

Method 1: Using Substitution

Let $y = x - \pi$. As $x \to \pi$, $y \to 0$.

From $y = x - \pi$, we have $x = y + \pi$.

Substituting $x = y + \pi$ into the expression:

$\frac{\tan x}{x - \pi} = \frac{\tan(y + \pi)}{y}$.

Using the trigonometric identity $\tan(\theta + \pi) = \tan \theta$, we have $\tan(y + \pi) = \tan y$.

So the expression becomes $\frac{\tan y}{y}$.

The limit becomes:

$\lim\limits_{y \to 0} \frac{\tan y}{y}$.

This is a standard limit.

$\lim\limits_{y \to 0} \frac{\tan y}{y} = 1$.

Method 2: Using L'Hopital's Rule

Since the limit is of the form $\frac{0}{0}$, we can apply L'Hopital's Rule.

Let $f(x) = \tan x$ and $g(x) = x - \pi$.

Find the derivatives:

$f'(x) = \frac{d}{dx}(\tan x) = \sec^2 x$.

$g'(x) = \frac{d}{dx}(x - \pi) = 1$.

Apply L'Hopital's Rule:

$\lim\limits_{x \to \pi} \frac{\tan x}{x - \pi} = \lim\limits_{x \to \pi} \frac{\sec^2 x}{1}$.

Substitute $x = \pi$:

$= \frac{\sec^2 \pi}{1}$.

Since $\cos \pi = -1$, $\sec \pi = \frac{1}{\cos \pi} = \frac{1}{-1} = -1$.

So, $\sec^2 \pi = (-1)^2 = 1$.

The limit is $\frac{1}{1} = 1$.

Both methods yield the same result.

The value of the limit is 1.

The blank should be filled with 1.

The given limit is $\lim\limits_{x \to 0} \sin mx \cot \frac{x}{\sqrt{3}} = 2$.

We can rewrite the cotangent term in terms of sine and cosine:

$\cot \theta = \frac{\cos \theta}{\sin \theta}$.

So, the limit becomes:

$\lim\limits_{x \to 0} \sin mx \frac{\cos \frac{x}{\sqrt{3}}}{\sin \frac{x}{\sqrt{3}}}$.

We can rearrange the terms:

$\lim\limits_{x \to 0} \frac{\sin mx}{\sin \frac{x}{\sqrt{3}}} \cos \frac{x}{\sqrt{3}}$.

Let's evaluate the limit of each part separately, if possible.

As $x \to 0$, $\cos \frac{x}{\sqrt{3}} \to \cos(0) = 1$.

For the term $\frac{\sin mx}{\sin \frac{x}{\sqrt{3}}}$, as $x \to 0$, both the numerator ($\sin(m \times 0) = \sin 0 = 0$) and the denominator ($\sin(0/\sqrt{3}) = \sin 0 = 0$) approach 0.

This is an indeterminate form $\frac{0}{0}$. We can use standard limits or L'Hopital's Rule.

Method 1: Using Standard Limits

We know the standard limit $\lim\limits_{y \to 0} \frac{\sin y}{y} = 1$.

We can rewrite the fraction $\frac{\sin mx}{\sin \frac{x}{\sqrt{3}}}$ as:

$\frac{\sin mx}{\sin \frac{x}{\sqrt{3}}} = \frac{\frac{\sin mx}{mx} \cdot mx}{\frac{\sin \frac{x}{\sqrt{3}}}{\frac{x}{\sqrt{3}}} \cdot \frac{x}{\sqrt{3}}}$.

For $x \neq 0$, we can cancel $x$ from the numerator and denominator:

$= \frac{\frac{\sin mx}{mx} \cdot m}{\frac{\sin \frac{x}{\sqrt{3}}}{\frac{x}{\sqrt{3}}} \cdot \frac{1}{\sqrt{3}}}$.

Now, take the limit as $x \to 0$. Note that as $x \to 0$, $mx \to 0$ and $\frac{x}{\sqrt{3}} \to 0$.

$\lim\limits_{x \to 0} \frac{\sin mx}{mx} = 1$

$\lim\limits_{x \to 0} \frac{\sin \frac{x}{\sqrt{3}}}{\frac{x}{\sqrt{3}}} = 1$.

So, $\lim\limits_{x \to 0} \frac{\sin mx}{\sin \frac{x}{\sqrt{3}}} = \frac{1 \cdot m}{1 \cdot \frac{1}{\sqrt{3}}} = \frac{m}{1/\sqrt{3}} = m\sqrt{3}$.

The original limit is the product of the limits:

$\lim\limits_{x \to 0} \frac{\sin mx}{\sin \frac{x}{\sqrt{3}}} \cdot \lim\limits_{x \to 0} \cos \frac{x}{\sqrt{3}} = (m\sqrt{3}) \cdot (1) = m\sqrt{3}$.

Method 2: Using L'Hopital's Rule

Consider the form $\lim\limits_{x \to 0} \frac{\sin mx}{\sin \frac{x}{\sqrt{3}}}$. This is $\frac{0}{0}$. Apply L'Hopital's Rule.

$\lim\limits_{x \to 0} \frac{\frac{d}{dx}(\sin mx)}{\frac{d}{dx}(\sin \frac{x}{\sqrt{3}})} = \lim\limits_{x \to 0} \frac{m \cos mx}{\frac{1}{\sqrt{3}} \cos \frac{x}{\sqrt{3}}}$.

Substitute $x=0$:

$= \frac{m \cos(m \cdot 0)}{\frac{1}{\sqrt{3}} \cos(0/\sqrt{3})} = \frac{m \cos 0}{\frac{1}{\sqrt{3}} \cos 0} = \frac{m \cdot 1}{\frac{1}{\sqrt{3}} \cdot 1} = \frac{m}{1/\sqrt{3}} = m\sqrt{3}$.

The original limit is $\lim\limits_{x \to 0} \frac{\sin mx}{\sin \frac{x}{\sqrt{3}}} \cdot \lim\limits_{x \to 0} \cos \frac{x}{\sqrt{3}} = (m\sqrt{3}) \cdot 1 = m\sqrt{3}$.

We are given that this limit is equal to 2.

So, $m\sqrt{3} = 2$.

Solving for $m$:

$m = \frac{2}{\sqrt{3}}$.

The blank should be filled with $\frac{2}{\sqrt{3}}$.

Given:

The function $y = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$.

This is an infinite series.

To Find:

The derivative $\frac{dy}{dx}$.

Solution:

We can find the derivative by differentiating the series term by term. The derivative of a constant is 0, and the derivative of $x^n$ is $nx^{n-1}$.

$y = 1 + \frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots$

$\frac{dy}{dx} = \frac{d}{dx}(1) + \frac{d}{dx}\left(\frac{x^1}{1!}\right) + \frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right) + \dots$

Differentiating each term:

$\frac{d}{dx}(1) = 0$

$\frac{d}{dx}\left(\frac{x^1}{1!}\right) = \frac{1}{1!} \cdot \frac{d}{dx}(x) = \frac{1}{1!} \cdot 1 = \frac{1}{1!} = 1$.

$\frac{d}{dx}\left(\frac{x^2}{2!}\right) = \frac{1}{2!} \cdot \frac{d}{dx}(x^2) = \frac{1}{2!} \cdot 2x = \frac{2}{2!} x = \frac{2}{2 \times 1!} x = \frac{1}{1!} x = x$.

$\frac{d}{dx}\left(\frac{x^3}{3!}\right) = \frac{1}{3!} \cdot \frac{d}{dx}(x^3) = \frac{1}{3!} \cdot 3x^2 = \frac{3}{3!} x^2 = \frac{3}{3 \times 2!} x^2 = \frac{1}{2!} x^2$.

$\frac{d}{dx}\left(\frac{x^4}{4!}\right) = \frac{1}{4!} \cdot \frac{d}{dx}(x^4) = \frac{1}{4!} \cdot 4x^3 = \frac{4}{4!} x^3 = \frac{4}{4 \times 3!} x^3 = \frac{1}{3!} x^3$.

In general, for the term $\frac{x^n}{n!}$ where $n \geq 1$, its derivative is $\frac{d}{dx}\left(\frac{x^n}{n!}\right) = \frac{1}{n!} \cdot n x^{n-1} = \frac{n}{n!} x^{n-1} = \frac{n}{n(n-1)!} x^{n-1} = \frac{1}{(n-1)!} x^{n-1}$.

Summing the derivatives of all terms:

$\frac{dy}{dx} = 0 + 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$.

Rearranging the terms:

$\frac{dy}{dx} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$.

Comparing this resulting series with the original series for $y$, we see that they are identical.

$y = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$

$\frac{dy}{dx} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$

Thus, $\frac{dy}{dx} = y$.

Alternate Method: Using known series expansion

The given series $1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$ is the well-known Maclaurin series expansion for the exponential function $e^x$.

So, $y = e^x$.

The derivative of $y = e^x$ with respect to $x$ is:

$\frac{dy}{dx} = \frac{d}{dx}(e^x) = e^x$.

Since $y = e^x$, we can write $\frac{dy}{dx} = y$.

The derivative $\frac{dy}{dx}$ is equal to $y$, or $e^x$, or the series $1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$. Based on the structure of the question, either $y$ or the series itself is the most direct answer derived from the definition provided.

The blank should be filled with $y$ (or $e^x$ or $1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$).

The given limit is $\lim\limits_{x \to 3^+} \frac{x}{[x]}$, where $[x]$ denotes the greatest integer function.

We are considering the right-hand limit as $x$ approaches 3.

As $x \to 3^+$, $x$ takes values slightly greater than 3. For example, $x$ could be 3.01, 3.001, 3.0001, and so on.

For any number $x$ such that $3 \leq x < 4$, the greatest integer $[x]$ is 3.

Since we are considering $x \to 3^+$, $x$ will be in an interval like $(3, 4)$ for values sufficiently close to 3 from the right.

For all $x$ in the interval $(3, 4)$, $[x] = 3$.

So, for $x$ approaching 3 from the right, the expression $\frac{x}{[x]}$ becomes $\frac{x}{3}$.

Now, we can evaluate the limit:

$\lim\limits_{x \to 3^+} \frac{x}{[x]} = \lim\limits_{x \to 3^+} \frac{x}{3}$.

Since $\frac{x}{3}$ is a continuous function, we can evaluate the limit by direct substitution:

$= \frac{3}{3} = 1$.

For completeness, let's consider the left-hand limit as $x \to 3^-$.

As $x \to 3^-$, $x$ takes values slightly less than 3. For example, $x$ could be 2.9, 2.99, 2.999, and so on.

For any number $x$ such that $2 \leq x < 3$, the greatest integer $[x]$ is 2.

For $x$ approaching 3 from the left, $x$ will be in an interval like $(2, 3)$ for values sufficiently close to 3 from the left.

For all $x$ in the interval $(2, 3)$, $[x] = 2$.

So, for $x$ approaching 3 from the left, the expression $\frac{x}{[x]}$ becomes $\frac{x}{2}$.

The left-hand limit is:

$\lim\limits_{x \to 3^-} \frac{x}{[x]} = \lim\limits_{x \to 3^-} \frac{x}{2}$.

Evaluate by direct substitution:

$= \frac{3}{2}$.

Since the left-hand limit ($\frac{3}{2}$) is not equal to the right-hand limit (1), the overall limit $\lim\limits_{x \to 3} \frac{x}{[x]}$ does not exist. However, the question only asks for the right-hand limit.

The value of the right-hand limit is 1.

The blank should be filled with 1.