Limits: Properties, Theorems, and Standard Results

Algebra of Limits (Sum, Difference, Product, Quotient Rules)

When dealing with limits involving combinations of functions, such as sums, differences, products, or quotients, we can often evaluate the limit of the combined function by evaluating the limits of the individual functions and then combining the results according to specific rules. These rules, collectively known as the Algebra of Limits or Limit Properties, are applicable when the limits of the individual functions exist and are finite real numbers.

Let $f(x)$ and $g(x)$ be two functions, and let $a$ be a real number. Suppose that the limits of $f(x)$ and $g(x)$ as $x$ approaches $a$ exist and are finite:

Then, the following limit properties hold:

1. Sum Rule

The limit of the sum of two functions is equal to the sum of their individual limits.

$\lim\limits_{x \to a} [f(x) + g(x)] = \lim\limits_{x \to a} f(x) + \lim\limits_{x \to a} g(x) = L + M$

2. Difference Rule

The limit of the difference of two functions is equal to the difference of their individual limits.

$\lim\limits_{x \to a} [f(x) - g(x)] = \lim\limits_{x \to a} f(x) - \lim\limits_{x \to a} g(x) = L - M$

3. Constant Multiple Rule

The limit of a constant times a function is equal to the constant times the limit of the function.

For any constant $k \in \mathbb{R}$:

$\lim\limits_{x \to a} [k \cdot f(x)] = k \cdot \lim\limits_{x \to a} f(x) = kL$

4. Product Rule

The limit of the product of two functions is equal to the product of their individual limits.

$\lim\limits_{x \to a} [f(x) \cdot g(x)] = \left(\lim\limits_{x \to a} f(x)\right) \cdot \left(\lim\limits_{x \to a} g(x)\right) = L \cdot M$

5. Quotient Rule

The limit of the quotient of two functions is equal to the quotient of their individual limits, provided that the limit of the denominator is not zero.

If $\lim\limits_{x \to a} g(x) = M \neq 0$, then:

$\lim\limits_{x \to a} \left[\frac{f(x)}{g(x)}\right] = \frac{\lim\limits_{x \to a} f(x)}{\lim\limits_{x \to a} g(x)} = \frac{L}{M}$

If $\lim\limits_{x \to a} g(x) = 0$, the Quotient Rule cannot be directly applied, and the limit may be unbounded or indeterminate ($\frac{L}{0}$ forms or $\frac{0}{0}$ forms). Special techniques are required for such cases.

6. Power Rule

The limit of a function raised to a rational power is equal to the limit of the function raised to that power, provided the result is a defined real number.

For any rational number $n$ (i.e., $n = p/q$ where $p, q$ are integers and $q \neq 0$), provided $L^n$ is a real number:

$\lim\limits_{x \to a} [f(x)]^n = \left[\lim\limits_{x \to a} f(x)\right]^n = L^n$

This rule applies, for example, to taking roots, as $\sqrt[q]{f(x)} = [f(x)]^{1/q}$.

7. Limit of a Constant Function

The limit of a constant function $f(x) = k$ as $x$ approaches any value $a$ is always equal to the constant itself.

$\lim\limits_{x \to a} k = k$ (where $k$ is any constant)

These rules are fundamental and allow us to evaluate limits of complex functions by breaking them down into simpler parts whose limits are easier to find (often using direct substitution). They are based on the rigorous epsilon-delta definition of limits, although their proofs are typically covered in more advanced texts.

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Some Important Theorems on Limits (Squeeze Play Theorem)

Besides the algebraic properties, there are other theorems that are highly useful for evaluating limits, particularly when direct substitution or algebraic manipulation is not straightforward. One such powerful theorem is the Squeeze Theorem.

Squeeze Theorem (Also known as the Sandwich Theorem or Pinching Theorem)

The Squeeze Theorem is used to confirm the limit of a function by comparing it to two other functions whose limits are known and equal. It is especially helpful when the function in question is bounded between two simpler functions.

Statement of the Squeeze Theorem:

Let $f(x)$, $g(x)$, and $h(x)$ be three functions. Suppose that for all $x$ in some open interval containing $a$ (but not necessarily at $a$ itself), the following inequality holds:

$g(x) \le f(x) \le h(x)$

If, additionally, the limits of the bounding functions $g(x)$ and $h(x)$ as $x$ approaches $a$ both exist and are equal to the same finite value $L$, that is:

Then, the limit of the function $f(x)$ as $x$ approaches $a$ must also exist and be equal to the same value $L$:

$\lim\limits_{x \to a} f(x) = L$

Intuition: Imagine the graphs of the three functions. The function $f(x)$ is "squeezed" between the graphs of $g(x)$ and $h(x)$ near $x=a$. If both the upper function $h(x)$ and the lower function $g(x)$ converge to the same height $L$ as $x$ approaches $a$, then the function $f(x)$, being trapped between them, has no choice but to also converge to that same height $L$.

Utility: The Squeeze Theorem is particularly powerful for evaluating limits of functions involving terms like $\sin\left(\frac{1}{x}\right)$ or $\cos\left(\frac{1}{x}\right)$ as $x \to 0$, where the argument of the trigonometric function approaches infinity, causing rapid oscillation and making direct substitution or simple algebraic manipulation impossible. By bounding the oscillating term (e.g., $-1 \le \sin(\text{anything}) \le 1$), we can create bounding functions $g(x)$ and $h(x)$ whose limits can be easily found.

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Some Standard Algebraic Results on Limits

In addition to the general algebra of limits, certain specific functional forms appear frequently, and their limits can be determined using standard results derived from basic principles or other theorems. Mastering these standard results allows for quicker evaluation of many limit problems.

Standard Limit: $\lim\limits_{x \to a} \frac{x^n - a^n}{x - a}$

One of the most fundamental standard limits in calculus, particularly useful for deriving differentiation rules, is the limit of the difference of powers divided by the difference of the variables. The result is:

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

This result is valid for any rational number $n$, provided that $a^{n-1}$ is defined and finite. If $n$ is negative or fractional, care must be taken regarding the domain and potential $a=0$ case.

• If $n$ is a positive integer, the result holds for all real $a$.
• If $n=0$, the limit becomes $\lim\limits_{x \to a} \frac{x^0 - a^0}{x - a} = \lim\limits_{x \to a} \frac{1 - 1}{x - a} = \lim\limits_{x \to a} \frac{0}{x - a} = 0$. The formula gives $0 \cdot a^{0-1} = 0 \cdot a^{-1}$ which is 0 (if $a \neq 0$). If $a=0$, the original expression is undefined.
• If $n$ is a negative integer, say $n = -m$ where $m > 0$ is an integer, the result holds for $a \neq 0$.
• If $n$ is a rational non-integer, say $n=p/q$, the result holds for $a > 0$. If $a=0$, the limit might exist or not depending on $n$.

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Derivation for Positive Integer $n$

We can derive this result for the case where $n$ is a positive integer using algebraic factorization.

Recall the algebraic identity for the difference of $n$-th powers:

Now, consider the expression $\frac{x^n - a^n}{x - a}$. For $x \neq a$, the term $(x - a)$ is non-zero, so we can divide both sides of identity (i) by $(x - a)$:

Cancelling the common factor $(x - a)$ (which is valid as we are considering the limit as $x \to a$, meaning $x \neq a$):

The expression on the right side of equation (ii) is a polynomial in $x$. It consists of $n$ terms. Let's look at the general form of these terms:

• Term 1: $x^{n-1} = x^{n-1} a^0$ (Here, the power of $x$ is $n-1$, power of $a$ is 0)
• Term 2: $x^{n-2}a = x^{n-2} a^1$ (Power of $x$ is $n-2$, power of $a$ is 1)
• Term 3: $x^{n-3}a^2$ (Power of $x$ is $n-3$, power of $a$ is 2)
• ...
• Term $k$: $x^{n-k} a^{k-1}$ (The sum of the powers is always $(n-k) + (k-1) = n-1$)
• ...
• Term $n$: $a^{n-1} = x^{n-n} a^{n-1} = x^0 a^{n-1}$ (Power of $x$ is 0, power of $a$ is $n-1$)

There are exactly $n$ such terms.

Now, we take the limit of the expression as $x \to a$. Since the right side of equation (ii) is a polynomial, we can evaluate the limit by direct substitution:

$\lim\limits_{x \to a} \frac{x^n - a^n}{x - a} = \lim\limits_{x \to a} (x^{n-1} + x^{n-2}a + x^{n-3}a^2 + \dots + xa^{n-2} + a^{n-1})$

Substitute $x = a$ into each term:

$= (a)^{n-1} + (a)^{n-2}a + (a)^{n-3}a^2 + \dots + (a)a^{n-2} + a^{n-1}$

Simplify each term using the rule $a^m \cdot a^k = a^{m+k}$:

$= a^{n-1} + a^{(n-2)+1} + a^{(n-3)+2} + \dots + a^{1+(n-2)} + a^{n-1}$

$= a^{n-1} + a^{n-1} + a^{n-1} + \dots + a^{n-1} + a^{n-1}$

We have a sum of $n$ identical terms, each equal to $a^{n-1}$.

$= n \cdot a^{n-1}$

Thus, we have derived the formula for positive integer $n$. The proof for rational $n$ is more advanced, often using the properties of derivatives or a substitution like $x = y^q$ if $n=p/q$.

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Some Important Theorems on Limits of Trigonometric Functions

There are two pivotal limits involving trigonometric functions as the argument approaches zero. These limits form the basis for evaluating many other limits involving trigonometric functions and are essential for the differentiation of trigonometric functions.

Theorem 1: Limit of $\frac{\sin \theta}{\theta}$ as $\theta \to 0$

The first fundamental trigonometric limit is:

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

A crucial condition for this theorem is that the angle $\theta$ must be measured in radians. This result does not hold if the angle is measured in degrees.

Derivation Outline (using Squeeze Theorem):

This limit can be rigorously derived using the Squeeze Theorem and a geometric argument involving the areas of a triangle, a sector, and another triangle within a unit circle.

1. Consider a unit circle centered at the origin O. Let A be the point (1, 0) on the positive x-axis. Let P be a point on the circle in the first quadrant such that the angle $\angle AOP = \theta$ radians, where $0 < \theta < \pi/2$. The coordinates of P are $(\cos \theta, \sin \theta)$.
2. Draw a line segment from P perpendicular to the x-axis, meeting the x-axis at M. $\triangle OMP$ is a right-angled triangle with height PM = $\sin \theta$ and base OM = $\cos \theta$.
3. Draw a line segment from A perpendicular to the x-axis (which is a vertical line at x=1) and extend OP to meet this line at Q. $\triangle OAQ$ is a right-angled triangle with base OA = 1 and height AQ. In $\triangle OAQ$, $\tan \theta = \frac{AQ}{OA} = \frac{AQ}{1} = AQ$. So, AQ = $\tan \theta$.
4. Compare the areas of the triangle $\triangle OMP$, the sector $OAP$, and the triangle $\triangle OAQ$. From the geometry for $0 < \theta < \pi/2$, we can see that the area of $\triangle OMP$ is less than or equal to the area of sector $OAP$, which is less than or equal to the area of $\triangle OAQ$. (Note: In the common derivation, $\triangle OAP$ is used, not $\triangle OMP$, with height $\sin \theta$ and base OA=1. Let's follow the standard path using $\triangle OAP$ where base is OA and height is the y-coordinate of P, $\sin\theta$.)

   Area($\triangle OAP$) $\le$ Area(Sector $OAP$) $\le$ Area($\triangle OAQ$)

   ... (iii)

5. Calculate the areas:

   Area of $\triangle OAP = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times OA \times (\text{y-coordinate of P}) = \frac{1}{2} \times 1 \times \sin \theta = \frac{1}{2} \sin \theta$.

   Area of Sector $OAP = \frac{1}{2} r^2 \theta = \frac{1}{2} (1)^2 \theta = \frac{1}{2} \theta$ (Formula for area of sector in radians).

   Area of $\triangle OAQ = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times OA \times AQ = \frac{1}{2} \times 1 \times \tan \theta = \frac{1}{2} \tan \theta$.

6. Substitute these areas into the inequality (iii):

   $\frac{1}{2} \sin \theta \le \frac{1}{2} \theta \le \frac{1}{2} \tan \theta$

7. Multiply the inequality by 2:

   $\sin \theta \le \theta \le \tan \theta$

8. Since we are considering $0 < \theta < \pi/2$, $\sin \theta > 0$. We can divide the inequality by $\sin \theta$ without changing the direction of the inequalities:

   $\frac{\sin \theta}{\sin \theta} \le \frac{\theta}{\sin \theta} \le \frac{\tan \theta}{\sin \theta}$

   $1 \le \frac{\theta}{\sin \theta} \le \frac{\sin \theta / \cos \theta}{\sin \theta}$

   $1 \le \frac{\theta}{\sin \theta} \le \frac{1}{\cos \theta}$

9. Take the reciprocal of each part of the inequality. This reverses the direction of the inequalities:

   $\frac{1}{1} \ge \frac{\sin \theta}{\theta} \ge \cos \theta$

   Rearranging this gives:

   $\cos \theta \le \frac{\sin \theta}{\theta} \le 1$

   ... (iv)

10. This inequality was derived for $0 < \theta < \pi/2$. It can be shown to hold also for $-\pi/2 < \theta < 0$ using the property that $\frac{\sin (-\theta)}{-\theta} = \frac{-\sin \theta}{-\theta} = \frac{\sin \theta}{\theta}$ and $\cos (-\theta) = \cos \theta$. So, inequality (iv) holds for all $\theta$ in the interval $(-\pi/2, \pi/2)$ except possibly at $\theta=0$.
11. Now apply the Squeeze Theorem to inequality (iv). Let $g(\theta) = \cos \theta$, $f(\theta) = \frac{\sin \theta}{\theta}$, and $h(\theta) = 1$.

    Evaluate the limits of the bounding functions as $\theta \to 0$:

    $\lim\limits_{\theta \to 0} g(\theta) = \lim\limits_{\theta \to 0} \cos \theta = \cos 0 = 1$ (by direct substitution, as $\cos \theta$ is continuous at $\theta=0$).

    $\lim\limits_{\theta \to 0} h(\theta) = \lim\limits_{\theta \to 0} 1 = 1$ (Limit of a constant is the constant).

    $\lim\limits_{\theta \to 0} \cos \theta = 1$

    ... (v)

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

    ... (vi)

12. Since $\cos \theta \le \frac{\sin \theta}{\theta} \le 1$ for $\theta$ near 0 (but $\theta \neq 0$), and $\lim\limits_{\theta \to 0} \cos \theta = 1 = \lim\limits_{\theta \to 0} 1$, by the Squeeze Theorem, the limit of the middle function must also be 1.

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

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Theorem 2: Limit of $\frac{1 - \cos \theta}{\theta}$ as $\theta \to 0$

The second fundamental trigonometric limit is:

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

Again, the angle $\theta$ must be in radians.

Derivation:

We can derive this limit using algebraic manipulation and the first fundamental trigonometric limit.

1. Attempt direct substitution at $\theta=0$: $\frac{1 - \cos 0}{0} = \frac{1 - 1}{0} = \frac{0}{0}$. This is an indeterminate form, so we need another method.
2. Multiply the numerator and the denominator by the conjugate of the numerator, which is $(1 + \cos \theta)$:

   $\lim\limits_{\theta \to 0} \frac{1 - \cos \theta}{\theta} = \lim\limits_{\theta \to 0} \left( \frac{1 - \cos \theta}{\theta} \times \frac{1 + \cos \theta}{1 + \cos \theta} \right)$

3. Simplify the numerator using the difference of squares formula $(A-B)(A+B) = A^2 - B^2$:

   $1 - \cos^2 \theta = \sin^2 \theta$ (using the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$).

   The expression becomes:

   $= \lim\limits_{\theta \to 0} \frac{1^2 - \cos^2 \theta}{\theta (1 + \cos \theta)} = \lim\limits_{\theta \to 0} \frac{\sin^2 \theta}{\theta (1 + \cos \theta)}$

   [Using $1-\cos^2 \theta = \sin^2 \theta$]

4. Rewrite the expression by splitting the terms to utilize the first fundamental limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:

   $= \lim\limits_{\theta \to 0} \left( \frac{\sin \theta}{\theta} \cdot \frac{\sin \theta}{1 + \cos \theta} \right)$

5. Apply the product rule for limits, assuming the individual limits exist:

   $= \left( \lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} \right) \cdot \left( \lim\limits_{\theta \to 0} \frac{\sin \theta}{1 + \cos \theta} \right)$

   [Using Product Rule for Limits]

6. Evaluate the two limits:

   The first limit is the standard result from Theorem 1:

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

   [From Theorem 1]

   The second limit can be evaluated by direct substitution, as the denominator $(1 + \cos \theta)$ is non-zero at $\theta=0$ ($1 + \cos 0 = 1 + 1 = 2$):

   $\lim\limits_{\theta \to 0} \frac{\sin \theta}{1 + \cos \theta} = \frac{\sin 0}{1 + \cos 0} = \frac{0}{1 + 1} = \frac{0}{2} = 0$

   [Direct Substitution]

7. Combine the results using the product:

   $= (1) \cdot (0) = 0$.

Thus, $\lim\limits_{\theta \to 0} \frac{1 - \cos \theta}{\theta} = 0$.

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Related Standard Trigonometric Limits

From the first fundamental limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$, we can derive other related limits:

$\lim\limits_{\theta \to 0} \frac{\theta}{\sin \theta} = \lim\limits_{\theta \to 0} \frac{1}{(\sin \theta)/\theta} = \frac{1}{\lim\limits_{\theta \to 0} (\sin \theta)/\theta} = \frac{1}{1} = 1$ (using Quotient Rule for limits)

$\lim\limits_{\theta \to 0} \frac{\tan \theta}{\theta} = \lim\limits_{\theta \to 0} \frac{\sin \theta/\cos \theta}{\theta} = \lim\limits_{\theta \to 0} \left( \frac{\sin \theta}{\theta} \cdot \frac{1}{\cos \theta} \right) = \left(\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta}\right) \cdot \left(\lim\limits_{\theta \to 0} \frac{1}{\cos \theta}\right) = 1 \cdot \frac{1}{1} = 1$ (using Product and Direct Substitution)

$\lim\limits_{\theta \to 0} \frac{\theta}{\tan \theta} = \lim\limits_{\theta \to 0} \frac{1}{(\tan \theta)/\theta} = \frac{1}{\lim\limits_{\theta \to 0} (\tan \theta)/\theta} = \frac{1}{1} = 1$ (using Quotient Rule)

These results are also valid provided $\theta$ is in radians.

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Limits of Exponential Functions

Exponential functions are a class of functions where the independent variable appears in the exponent. They are of the form $f(x) = a^x$, where the base $a$ is a positive constant and $a \neq 1$. The domain of $a^x$ is all real numbers $(-\infty, \infty)$. A particularly important exponential function in calculus is the natural exponential function, $f(x) = e^x$, where $e$ is Euler's number, approximately 2.71828.

Understanding the limits of exponential functions is crucial, especially as $x$ approaches finite values or infinity.

Continuity and Limits at a Finite Point

Exponential functions, both $a^x$ (for $a>0, a\neq 1$) and $e^x$, are continuous for all real numbers. This means that there are no breaks, jumps, or holes in their graphs. Due to this continuity property, the limit of an exponential function as $x$ approaches any finite real number $c$ can be evaluated by simple direct substitution.

For any real number $c$:

$\lim\limits_{x \to c} a^x = a^c$ (where $a > 0, a \neq 1$)

And specifically for the natural exponential function:

$\lim\limits_{x \to c} e^x = e^c$

This is a direct application of the definition of continuity at a point: $\lim\limits_{x \to c} f(x) = f(c)$.

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Limits at Infinity ($x \to \infty$)

The behavior of the exponential function $a^x$ as $x$ approaches positive infinity depends significantly on the value of the base $a$.

• Case 1: $a > 1$ (e.g., $a = 2, 10, e$)

  If the base $a$ is greater than 1, the function $a^x$ grows without bound as $x$ increases. The graph rises steeply as $x$ moves to the right.

  $\lim\limits_{x \to \infty} a^x = \infty \quad \text{(for } a > 1)$

  This means for any large positive number $M$, we can find an $x$ such that $a^x > M$.

  For the natural exponential function, since $e \approx 2.718 > 1$:

  $\lim\limits_{x \to \infty} e^x = \infty$

• Case 2: $0 < a < 1$ (e.g., $a = 1/2, 0.5, 0.1$)

  If the base $a$ is between 0 and 1, the function $a^x$ decays towards zero as $x$ increases. The graph approaches the x-axis as $x$ moves to the right.

  $\lim\limits_{x \to \infty} a^x = 0 \quad \text{(for } 0 < a < 1)$

  This means for any small positive number $\epsilon$, we can find an $x$ such that $a^x < \epsilon$.

  Example: $\lim\limits_{x \to \infty} \left(\frac{1}{2}\right)^x = \lim\limits_{x \to \infty} \frac{1}{2^x}$. As $x \to \infty$, $2^x \to \infty$, so $\frac{1}{2^x} \to \frac{1}{\infty} \to 0$.

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Limits at Negative Infinity ($x \to -\infty$)

The behavior of the exponential function $a^x$ as $x$ approaches negative infinity also depends on the base $a$. We can analyze this by considering the substitution $y = -x$. As $x \to -\infty$, $y \to \infty$. Then $\lim\limits_{x \to -\infty} a^x = \lim\limits_{y \to \infty} a^{-y} = \lim\limits_{y \to \infty} \frac{1}{a^y} = \lim\limits_{y \to \infty} \left(\frac{1}{a}\right)^y$.

• Case 1: $a > 1$ (e.g., $a = 2, 10, e$)

  If the base $a$ is greater than 1, then $\frac{1}{a}$ is between 0 and 1 ($0 < \frac{1}{a} < 1$). As $y \to \infty$, $(\frac{1}{a})^y$ approaches 0 (from the previous section, Case 2 for $y \to \infty$).

  $\lim\limits_{x \to -\infty} a^x = 0 \quad \text{(for } a > 1)$

  For the natural exponential function:

  $\lim\limits_{x \to -\infty} e^x = 0$

  Example: $\lim\limits_{x \to -\infty} 2^x$. Let $y = -x$. As $x \to -\infty$, $y \to \infty$. $\lim\limits_{y \to \infty} 2^{-y} = \lim\limits_{y \to \infty} \frac{1}{2^y}$. As $y \to \infty$, $2^y \to \infty$, so $\frac{1}{2^y} \to 0$.

• Case 2: $0 < a < 1$ (e.g., $a = 1/2, 0.5, 0.1$)

  If the base $a$ is between 0 and 1, then $\frac{1}{a}$ is greater than 1 ($\frac{1}{a} > 1$). As $y \to \infty$, $(\frac{1}{a})^y$ grows without bound (from the previous section, Case 1 for $y \to \infty$).

  $\lim\limits_{x \to -\infty} a^x = \infty \quad \text{(for } 0 < a < 1)$

  Example: $\lim\limits_{x \to -\infty} \left(\frac{1}{2}\right)^x$. Let $y = -x$. As $x \to -\infty$, $y \to \infty$. $\lim\limits_{y \to \infty} \left(\frac{1}{2}\right)^{-y} = \lim\limits_{y \to \infty} \left(\frac{1}{1/2}\right)^y = \lim\limits_{y \to \infty} 2^y = \infty$.

These limit behaviors are clearly observable in the graphs of exponential functions.

Limits of Logarithmic Functions

Logarithmic functions are the inverse functions of exponential functions. They are of the form $f(x) = \log_a x$, where the base $a$ is a positive constant ($a > 0, a \neq 1$). The domain of $\log_a x$ is restricted to positive real numbers, $x > 0$, because the range of the exponential function $a^x$ is $y > 0$. The most common base is $e$, giving the natural logarithm $f(x) = \ln x = \log_e x$.

Understanding the limits of logarithmic functions requires considering their behavior at finite positive values, as $x$ approaches positive infinity, and as $x$ approaches 0 from the right side (since the domain is $x>0$).

Continuity and Limits at a Finite Point

Logarithmic functions, both $\log_a x$ (for $a>0, a\neq 1$) and $\ln x$, are continuous throughout their domain, which is $(0, \infty)$. For any finite positive real number $c$ ($c > 0$), the limit of a logarithmic function as $x$ approaches $c$ can be evaluated by direct substitution.

For any $c > 0$:

$\lim\limits_{x \to c} \log_a x = \log_a c$ (where $a > 0, a \neq 1$)

And specifically for the natural logarithmic function:

$\lim\limits_{x \to c} \ln x = \ln c$ (for $c > 0$)

This is a direct consequence of the definition of continuity at a point within the domain of the function.

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Limit as $x \to \infty$

The behavior of the logarithmic function $\log_a x$ as $x$ approaches positive infinity depends on the value of the base $a$.

• Case 1: $a > 1$ (e.g., $a = 2, 10, e$)

  If the base $a$ is greater than 1, the function $\log_a x$ increases without bound as $x$ increases. The graph rises as $x$ moves to the right, albeit very slowly compared to exponential functions.

  $\lim\limits_{x \to \infty} \log_a x = \infty \quad \text{(for } a > 1)$

  For the natural logarithmic function, since $e > 1$:

  $\lim\limits_{x \to \infty} \ln x = \infty$

  This corresponds to the fact that $\lim\limits_{y \to \infty} a^y = \infty$ for $a>1$. Since $y = \log_a x$ is the inverse, as $x \to \infty$, $y$ must also go to $\infty$.

• Case 2: $0 < a < 1$ (e.g., $a = 1/2, 0.5, 0.1$)

  If the base $a$ is between 0 and 1, the function $\log_a x$ decreases without bound towards negative infinity as $x$ increases. The graph falls as $x$ moves to the right.

  $\lim\limits_{x \to \infty} \log_a x = -\infty \quad \text{(for } 0 < a < 1)$

  This corresponds to the fact that $\lim\limits_{y \to -\infty} a^y = \infty$ for $0 < a < 1$. Since $y = \log_a x$ is the inverse, as $x \to \infty$, $y$ must go to $-\infty$.

  Example: $\lim\limits_{x \to \infty} \log_{1/2} x$. Using the change of base formula, $\log_{1/2} x = \frac{\ln x}{\ln(1/2)} = \frac{\ln x}{-\ln 2}$. As $x \to \infty$, $\ln x \to \infty$ and $-\ln 2$ is a negative constant. So, $\lim\limits_{x \to \infty} \frac{\ln x}{-\ln 2} = \frac{\infty}{-\ln 2} = -\infty$.

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Limit as $x \to 0^+$

Since the domain of logarithmic functions is $x > 0$, we can only consider the limit as $x$ approaches 0 from the right side ($x \to 0^+$).

• Case 1: $a > 1$ (e.g., $a = 2, 10, e$)

  If the base $a$ is greater than 1, the function $\log_a x$ decreases without bound towards negative infinity as $x$ approaches 0 from the right. The graph approaches the negative y-axis.

  $\lim\limits_{x \to 0^+} \log_a x = -\infty \quad \text{(for } a > 1)$

  For the natural logarithmic function:

  $\lim\limits_{x \to 0^+} \ln x = -\infty$

  The y-axis (the line $x=0$) is a vertical asymptote.

  This corresponds to the fact that $\lim\limits_{y \to -\infty} a^y = 0$ for $a>1$. Since $y = \log_a x$ is the inverse, as $x \to 0^+$, $y$ must go to $-\infty$.

• Case 2: $0 < a < 1$ (e.g., $a = 1/2, 0.5, 0.1$)

  If the base $a$ is between 0 and 1, the function $\log_a x$ increases without bound towards positive infinity as $x$ approaches 0 from the right. The graph approaches the positive y-axis.

  $\lim\limits_{x \to 0^+} \log_a x = \infty \quad \text{(for } 0 < a < 1)$

  The y-axis (the line $x=0$) is a vertical asymptote.

  This corresponds to the fact that $\lim\limits_{y \to \infty} a^y = 0$ for $0 < a < 1$. Since $y = \log_a x$ is the inverse, as $x \to 0^+$, $y$ must go to $\infty$.

These limit behaviors at the boundaries of the domain ($x \to \infty$ and $x \to 0^+$) are key characteristics of logarithmic functions.

Standard Results for Limits of Exponential and Logarithmic Functions

Beyond evaluating limits using direct substitution or analyzing behavior at infinity, there are specific standard limits involving exponential and logarithmic functions, particularly with the base $e$, that are frequently encountered and are fundamental in differential calculus. These limits are often derived using the definition of the derivative or series expansions.

Theorem 1: Limit involving the Natural Exponential Function

One of the most important standard limits is related to the derivative of $e^x$ at $x=0$.

$\lim\limits_{x \to 0} \frac{e^x - 1}{x} = 1$

When $x=0$, direct substitution yields the indeterminate form $\frac{e^0 - 1}{0} = \frac{1 - 1}{0} = \frac{0}{0}$. The theorem states that despite this, the limit exists and is equal to 1.

Derivation using the Definition of the Derivative:

Recall the definition of the derivative of a function $f(t)$ at a point $t=c$:

$f'(c) = \lim\limits_{h \to 0} \frac{f(c+h) - f(c)}{h}$

Consider the natural exponential function $f(t) = e^t$. We know (or will learn later) that the derivative of $e^t$ is $e^t$ itself, i.e., $f'(t) = e^t$.

Let's evaluate the derivative of $f(t) = e^t$ at the specific point $t=0$. The derivative at $t=0$ is $f'(0) = e^0 = 1$.

Now, let's write out the definition of the derivative of $f(t) = e^t$ at $t=0$ using the limit definition:

Simplify the expression:

We know that $f'(0) = 1$. Therefore, equating the derivative and its limit definition:

Replacing the variable $h$ with $x$ (the variable name does not affect the limit value) gives the standard result:

$\lim\limits_{x \to 0} \frac{e^x - 1}{x} = 1$.

Geometrically, this limit represents the slope of the tangent line to the curve $y=e^x$ at the point where $x=0$, which is $(0, e^0) = (0, 1)$. The slope at this point is precisely 1.

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Theorem 2: Limit involving the Natural Logarithmic Function

Another important standard limit involves the natural logarithmic function, related to the derivative of $\ln x$ at $x=1$.

$\lim\limits_{x \to 0} \frac{\ln(1+x)}{x} = 1$

When $x=0$, direct substitution yields the indeterminate form $\frac{\ln(1+0)}{0} = \frac{\ln(1)}{0} = \frac{0}{0}$. The theorem states that this limit is also equal to 1.

Derivation using the Definition of the Derivative:

Consider the natural logarithmic function $f(t) = \ln t$. The domain is $t > 0$. We know (or will learn) that the derivative of $\ln t$ is $1/t$, i.e., $f'(t) = 1/t$.

Let's evaluate the derivative of $f(t) = \ln t$ at the specific point $t=1$ (a point in the domain where we expect a simple derivative value). The derivative at $t=1$ is $f'(1) = 1/1 = 1$.

Now, write out the definition of the derivative of $f(t) = \ln t$ at $t=1$ using the limit definition:

Simplify the expression:

We know that $f'(1) = 1$. Therefore, equating the derivative and its limit definition:

Replacing the variable $h$ with $x$ gives the standard result:

$\lim\limits_{x \to 0} \frac{\ln(1+x)}{x} = 1$.

Alternate Derivation using Theorem 1:

We can also derive this limit using the previously established limit $\lim\limits_{y \to 0} \frac{e^y - 1}{y} = 1$.

1. Let the substitution be $y = \ln(1+x)$.
2. Find the inverse relationship: If $y = \ln(1+x)$, then by the definition of logarithm, $e^y = 1+x$. Solving for $x$, we get $x = e^y - 1$.
3. Determine the behavior of the new variable $y$ as the original variable $x$ approaches 0: As $x \to 0$, $y = \ln(1+x) \to \ln(1+0) = \ln(1) = 0$. So, as $x \to 0$, $y \to 0$.
4. Substitute $y$ and $x$ in terms of $y$ into the original limit expression:

   $\lim\limits_{x \to 0} \frac{\ln(1+x)}{x} = \lim\limits_{y \to 0} \frac{y}{e^y - 1}$

5. Rewrite the expression to match the form of Theorem 1:

   $= \lim\limits_{y \to 0} \frac{1}{\frac{e^y - 1}{y}}$

   [Taking reciprocal]

6. Apply the Quotient Rule for limits (since the limit of the denominator is 1, which is non-zero):

   $= \frac{\lim\limits_{y \to 0} 1}{\lim\limits_{y \to 0} \frac{e^y - 1}{y}}$

   [Using Quotient Rule]

7. Evaluate the limits: The limit of a constant is the constant itself, and the limit in the denominator is 1 (from Theorem 1).

   $= \frac{1}{1} = 1$.

Thus, $\lim\limits_{x \to 0} \frac{\ln(1+x)}{x} = 1$.

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Generalization of Limits for Arbitrary Base $a$

The standard limits for base $e$ can be generalized to any valid base $a$ ($a>0, a \neq 1$) using the change of base formulas for exponentials and logarithms.

• Limit involving $a^x$:

  We want to evaluate $\lim\limits_{x \to 0} \frac{a^x - 1}{x}$.

  Use the identity $a^x = e^{x \ln a}$.

  $\lim\limits_{x \to 0} \frac{a^x - 1}{x} = \lim\limits_{x \to 0} \frac{e^{x \ln a} - 1}{x}$

  Let $u = x \ln a$. As $x \to 0$, $u \to 0 \cdot \ln a = 0$. Note that $\ln a$ is a constant.

  Also, $x = \frac{u}{\ln a}$.

  Substitute $u$ and $x$ into the limit (assuming $a \neq 1$, so $\ln a \neq 0$):

  $\lim\limits_{x \to 0} \frac{e^{x \ln a} - 1}{x} = \lim\limits_{u \to 0} \frac{e^u - 1}{u/\ln a} = \lim\limits_{u \to 0} (\ln a) \cdot \frac{e^u - 1}{u}$

  Using the Constant Multiple Rule for limits:

  $= (\ln a) \cdot \lim\limits_{u \to 0} \frac{e^u - 1}{u}$

  From Theorem 1, $\lim\limits_{u \to 0} \frac{e^u - 1}{u} = 1$.

  So, the limit is: $(\ln a) \cdot 1 = \ln a$.

  $\lim\limits_{x \to 0} \frac{a^x - 1}{x} = \ln a$ (for $a > 0, a \neq 1$)

• Limit involving $\log_a(1+x)$:

  We want to evaluate $\lim\limits_{x \to 0} \frac{\log_a(1+x)}{x}$.

  Use the change of base formula for logarithms: $\log_a(1+x) = \frac{\ln(1+x)}{\ln a}$.

  $\lim\limits_{x \to 0} \frac{\log_a(1+x)}{x} = \lim\limits_{x \to 0} \frac{\frac{\ln(1+x)}{\ln a}}{x} = \lim\limits_{x \to 0} \frac{\ln(1+x)}{x \ln a}$

  Rewrite the expression (assuming $a \neq 1$, so $\ln a \neq 0$):

  $= \lim\limits_{x \to 0} \frac{1}{\ln a} \cdot \frac{\ln(1+x)}{x}$

  Using the Constant Multiple Rule for limits:

  $= \frac{1}{\ln a} \cdot \lim\limits_{x \to 0} \frac{\ln(1+x)}{x}$

  From Theorem 2, $\lim\limits_{x \to 0} \frac{\ln(1+x)}{x} = 1$.

  So, the limit is: $\frac{1}{\ln a} \cdot 1 = \frac{1}{\ln a}$.

  $\lim\limits_{x \to 0} \frac{\log_a(1+x)}{x} = \frac{1}{\ln a}$ (for $a > 0, a \neq 1$)

These four standard limits ($\lim\limits_{x \to 0} \frac{e^x - 1}{x}$, $\lim\limits_{x \to 0} \frac{\ln(1+x)}{x}$, $\lim\limits_{x \to 0} \frac{a^x - 1}{x}$, $\lim\limits_{x \to 0} \frac{\log_a(1+x)}{x}$) are extremely useful for evaluating more complex limits involving exponential and logarithmic functions, often by transforming the expression into one of these forms using algebraic manipulation or substitution.

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