Chapter 13 Limits and Derivatives (Concepts)

Limits

The concept of a limit is the fundamental idea upon which all of calculus is built. Before we can understand concepts like the instantaneous speed of a car or the exact slope of a curved line, we must first understand limits. A limit describes the behavior of a function as its input gets closer and closer to a particular value, without actually reaching it.

The Intuitive Idea of a Limit: The Foundation of Calculus

The Core Concept: A Question of Approach, Not Arrival

Imagine you are walking along a winding path drawn on a graph, where your height at any position $x$ is given by a function, $f(x)$. You want to know the height of a specific landmark located at position $x=a$.

In many simple cases, you can just walk to the landmark, stand on it, and measure your height. This would be finding the value of $f(a)$. However, the idea of a limit asks a more subtle and powerful question: What height are you approaching as you get infinitely close to the landmark, without ever actually standing on it?

Think of it as a prediction. Two friends are walking towards the landmark from opposite directions.

• One friend walks from the left side, getting closer and closer: $x = a - 0.1$, then $x = a - 0.01$, then $x = a - 0.001$, and so on.
• The other friend walks from the right side, getting closer and closer: $x = a + 0.1$, then $x = a + 0.01$, then $x = a + 0.001$, and so on.

Why Is This "Approaching" Idea So Important?

The power of the limit comes from situations where you *cannot* stand on the landmark itself. Consider these three scenarios for the landmark at $x=a$:

1. A Smooth Path: The landmark is on solid ground. The height you are approaching is the same as the height of the ground at that exact spot. Here, the limit equals the function's value.
2. A Broken Bridge (A "Hole" in the Graph): The landmark is a bottomless pit. The path exists right up to the edge on both sides, but at the exact spot $x=a$, there is no ground to stand on. The function is undefined at $x=a$. However, your two friends can still look across the tiny gap and agree on the height of the edge they are approaching. The limit exists, even though the function value does not!
3. A Sudden Cliff (A "Jump" in the Graph): The path is broken, with the left side ending at a different height than the right side. Your friend approaching from the left predicts one height, while the friend approaching from the right predicts a completely different height. Since they do not agree, there is no single, well-defined height that is being approached. In this case, the limit does not exist.

Calculus is deeply concerned with the second scenario—analyzing functions at points where they might be undefined. This is precisely what happens when we try to find the slope of a curve at a single point, which initially leads to an expression like $\frac{0}{0}$.

Decoding the Mathematical Notation

Let's break down the formal notation based on this intuition:

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

• lim: This is the limit operator. It signals that we are not just plugging a number in; we are performing this special "approaching" process.
• $x \to a$: This is read as "x approaches a". It means we are considering values of $x$ that get arbitrarily close to $a$, from both sides, but importantly, $x$ is never equal to $a$.
• $f(x)$: This is the function we are investigating—the "path" whose height we want to predict.
• $= L$: This states that the predicted value, or the number that $f(x)$ gets arbitrarily close to during this process, is the number $L$.

So, the entire statement reads: "The value that the function $f(x)$ is expected to have as its input $x$ gets infinitely close to (but not equal to) $a$, is the number $L$."

Left-Hand and Right-Hand Limits

When approaching the landmark at $x=a$, you could be coming from the left side (where $x < a$) or from the right side (where $x > a$). For the limit to be well-defined, the height you approach must be the same regardless of the direction from which you approach.

Left-Hand Limit (LHL)

This is the value that $f(x)$ approaches as $x$ gets closer to $a$ from the left side (i.e., from values smaller than $a$). We denote this with a small minus sign superscript:

$\lim\limits_{x \to a^-} f(x)$

Right-Hand Limit (RHL)

This is the value that $f(x)$ approaches as $x$ gets closer to $a$ from the right side (i.e., from values larger than $a$). We denote this with a small plus sign superscript:

$\lim\limits_{x \to a^+} f(x)$

The Existence of a Limit

A limit at a point $x=a$ is said to exist only if you approach the exact same value from both the left and the right. If you approach a cliff edge from the left at a height of 100 meters, but from the right you approach a different cliff edge at a height of 50 meters, there is no single, well-defined height at that point. In this case, the limit does not exist.

The formal condition is:

A limit $\lim\limits_{x \to a} f(x)$ exists and equals $L$ if and only if:

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

Answer:

We want to find $\lim\limits_{x \to 3} (x + 2)$.

Tabular Approach:

We can create a table of values for $x$ approaching 3 from both sides.

$x$ (approaching from left)	$f(x) = x+2$
2.9	4.9
2.99	4.99
2.999	4.999

From the table, as $x \to 3^-$, $f(x) \to 5$. So, LHL = 5.

$x$ (approaching from right)	$f(x) = x+2$
3.1	5.1
3.01	5.01
3.001	5.001

From the table, as $x \to 3^+$, $f(x) \to 5$. So, RHL = 5.

Conclusion: Since LHL = RHL = 5, the limit exists and is 5. $\lim\limits_{x \to 3} (x + 2) = 5$.

Graphical Approach:

The graph of $f(x) = x+2$ is a straight line. As we trace the line from the left towards $x=3$ and from the right towards $x=3$, both paths converge to the same point, which has a y-value of 5.

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The Indeterminate Form $\frac{0}{0}$

Sometimes, if we try to find a limit by simply plugging in the value, we get a nonsensical answer like $\frac{0}{0}$. This is called an indeterminate form. It doesn't mean the limit is zero or that it doesn't exist. It's a signal that we need to do more work to find the answer. The value of the function at that exact point might be undefined (like a hole in the graph), but the limit—the value the function is *approaching*—can still exist.

The most common technique to solve this is to use algebra to simplify the function first.

Answer:

Step 1: Try direct substitution.

If we plug in $x=1$, we get $\frac{1^2 - 1}{1 - 1} = \frac{0}{0}$. This is the indeterminate form.

Step 2: Tabular Approach.

Let's see what happens for values of $x$ close to 1.

$x$	$f(x) = \frac{x^2 - 1}{x - 1}$
0.9	$\frac{(0.9)^2 - 1}{0.9 - 1} = \frac{-0.19}{-0.1} = 1.9$
0.99	$\frac{(0.99)^2 - 1}{0.99 - 1} = \frac{-0.0199}{-0.01} = 1.99$
1.01	$\frac{(1.01)^2 - 1}{1.01 - 1} = \frac{0.0201}{0.01} = 2.01$
1.1	$\frac{(1.1)^2 - 1}{1.1 - 1} = \frac{0.21}{0.1} = 2.1$

The table strongly suggests that as $x$ approaches 1 from either side, the value of $f(x)$ approaches 2.

Step 3: Algebraic and Graphical Approach.

We simplify the function by factoring the numerator: $f(x) = \frac{(x - 1)(x + 1)}{x - 1}$.

Since the limit definition means $x \neq 1$, we can safely cancel the $(x-1)$ terms:

$f(x) = x + 1$ (for all $x \neq 1$).

This means the graph of our original function is identical to the line $y = x + 1$, but with a single point removed—a "hole"—at $x=1$.

Now we evaluate the limit of the simplified function:

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

Using direct substitution on the simplified form: $1 + 1 = 2$.

The limit is 2.

Answer:

Because the function's definition changes at $x=2$, we must evaluate the left-hand and right-hand limits separately.

Left-Hand Limit (LHL):

To find the LHL, we approach $x=2$ from the left side, so we use the piece of the function where $x < 2$, which is $f(x) = x+1$.

$\lim\limits_{x \to 2^-} f(x) = \lim\limits_{x \to 2^-} (x+1) = 2+1 = 3$.

Right-Hand Limit (RHL):

To find the RHL, we approach $x=2$ from the right side, so we use the piece of the function where $x \ge 2$, which is $f(x) = x^2-1$.

$\lim\limits_{x \to 2^+} f(x) = \lim\limits_{x \to 2^+} (x^2-1) = (2)^2 - 1 = 4 - 1 = 3$.

Conclusion:

Since the Left-Hand Limit (3) is equal to the Right-Hand Limit (3), the overall limit exists.

$\lim\limits_{x \to 2} f(x) = 3$.

Note that in this case, the limit is equal to the function's value, $f(2) = 2^2-1=3$. This means the two pieces of the graph meet perfectly at $x=2$.

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Some Standard Results on Limits

While the intuitive idea of a limit involves checking values from the left and right, this process can be tedious. To evaluate limits more efficiently, mathematicians have established a set of rules and standard results. These results act as powerful shortcuts, especially for common types of functions like polynomials and rational functions.

Limit of Polynomial and Rational Functions

Polynomials are among the "best-behaved" functions in calculus. They are smooth and continuous everywhere, meaning their graphs have no breaks, jumps, or holes. This well-behaved nature leads to a very simple rule for finding their limits.

Limit of a Polynomial Function

A polynomial function is an expression of the form $P(x) = c_n x^n + c_{n-1} x^{n-1} + \dots + c_1 x + c_0$. Because there are no divisions or other operations that could cause issues at any real number, the value a polynomial *approaches* as $x$ gets close to $a$ is simply the value of the polynomial *at* $a$.

Rule: To find the limit of any polynomial function as $x$ approaches a number $a$, you can simply substitute $a$ for $x$ in the function.

Limit of a Rational Function

A rational function is a fraction where both the numerator and the denominator are polynomials, like $f(x) = \frac{P(x)}{Q(x)}$. These functions are also well-behaved and continuous, except at the points where the denominator becomes zero (which correspond to holes or vertical asymptotes in the graph).

Rule: To find the limit of a rational function, you can substitute $a$ for $x$, as long as this substitution does not make the denominator zero.

If the denominator $Q(a)$ is zero, we encounter an indeterminate form (like $\frac{0}{0}$) or a form indicating an infinite limit. These cases require more advanced techniques like factoring, rationalizing, or using the special formulas discussed below.

A Key Formula for the Indeterminate Form $\frac{0}{0}$

One of the most important and frequently encountered limits in calculus is of the following form, which always results in $\frac{0}{0}$ upon direct substitution:

This powerful formula works for any real number $n$ (positive integers, negative integers, fractions, etc.) and for any $a>0$ when $n$ is not an integer. It provides a direct way to find the limit without needing to factor or use other complex methods. Geometrically, this limit represents the slope of the tangent line to the curve $y=x^n$ at the point $x=a$.

Proof of the Formula (for positive integer n)

Let's prove the result for the general case where $n$ is any positive integer: $\lim\limits_{x \to a} \frac{x^n - a^n}{x - a}$.

Direct substitution gives $\frac{a^n - a^n}{a - a} = \frac{0}{0}$, so we must simplify.

We use the polynomial factorization formula for a difference of powers:

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

The second factor is a sum of $n$ terms. So, our limit becomes:

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

Since the limit process involves $x$ getting close to $a$ but not being equal to $a$, the term $(x-a)$ is not zero, so we can cancel it from the numerator and denominator.

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

We are now left with a simple polynomial. We can find its limit by direct substitution (plugging in $x=a$).

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

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

Since there are $n$ terms in the sum, and each term simplifies to $a^{n-1}$, the total sum is:

$= \underbrace{a^{n-1} + a^{n-1} + \dots + a^{n-1}}_{n \text{ times}} = n a^{n-1}$

This proves the formula for any positive integer $n$. The proof for rational and real numbers is more advanced but yields the same result.

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Some Important Theorems on Limits

While the concept of a limit is defined by the idea of "approaching" a value, calculating every limit this way would be incredibly time-consuming. To work with limits efficiently, we use a set of rules, often called theorems or properties, that allow us to break down complex functions into simpler parts. These rules are collectively known as the algebra of limits.

The Algebra of Limits

Let's assume we have two well-behaved functions, $f(x)$ and $g(x)$. We know that as $x$ approaches some number $a$, the limit of $f(x)$ is $L$ and the limit of $g(x)$ is $M$. In mathematical terms:

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

With these assumptions, the following rules hold true:

1. The Sum and Difference Rule

The limit of the sum (or difference) of two functions is simply the sum (or difference) of their individual limits. This allows us to handle functions that are added or subtracted term by term, like polynomials.

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

In simpler terms: If you know what two functions are approaching, you can find what their sum is approaching by just adding those two values together.

2. The Product Rule

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

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

3. The Constant Multiple Rule

A very useful special case of the product rule is the limit of a constant times a function. The limit of a constant function, like $f(x)=c$, is always just $c$. Therefore, the limit of a constant times a function is the constant times the limit of the function.

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

4. The Quotient Rule

The limit of the quotient (division) of two functions is the quotient of their individual limits, with one very important condition: the limit of the function in the denominator cannot be zero.

$\lim\limits_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim\limits_{x \to a} f(x)}{\lim\limits_{x \to a} g(x)} = \frac{L}{M}$, provided that $M \neq 0$

If the limit of the denominator is zero, the limit of the quotient might be infinite or might be an indeterminate form requiring more analysis.

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The Squeeze Theorem (or Sandwich Theorem)

The Squeeze Theorem is a beautiful and intuitive idea for finding the limit of a complicated function by "trapping" or "squeezing" it between two simpler functions whose limits are known.

Imagine you have three functions, $f(x)$, $g(x)$, and $h(x)$. Let's say you know that for all the x-values near a certain point $a$, the function $g(x)$ is always "sandwiched" between $f(x)$ and $h(x)$. That is, $f(x) \le g(x) \le h(x)$.

Now, suppose you also know that the two "outer" functions, $f(x)$ and $h(x)$ (the two slices of bread), are both approaching the exact same limit, $L$, as $x$ approaches $a$.

Since $g(x)$ is trapped in the middle, it has no choice but to be "squeezed" towards that same limit $L$.

Formal Statement: If $f(x) \le g(x) \le h(x)$ for all $x$ near $a$ (except possibly at $a$ itself), and if $\lim\limits_{x \to a} f(x) = \lim\limits_{x \to a} h(x) = L$, then $\lim\limits_{x \to a} g(x) = L$.

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Problem Analysis:

We cannot use the product rule directly, because while $\lim\limits_{x \to 0} x^2 = 0$, the limit $\lim\limits_{x \to 0} \sin(\frac{1}{x})$ does not exist. The function $\sin(\frac{1}{x})$ oscillates infinitely fast between -1 and 1 as $x$ approaches 0. This is a classic case for the Squeeze Theorem.

Solution:

Step 1: Find two functions to "sandwich" our function.

We know that the sine function is always bounded between -1 and 1, no matter what its input is.

$-1 \le \sin(\frac{1}{x}) \le 1$ for all $x \neq 0$.

Now, we can multiply all parts of this inequality by $x^2$. Since $x^2$ is always non-negative, the inequality signs will not change.

$-1 \cdot x^2 \le x^2 \sin(\frac{1}{x}) \le 1 \cdot x^2$

$-x^2 \le x^2 \sin(\frac{1}{x}) \le x^2$

We have successfully squeezed our difficult function, $g(x) = x^2 \sin(\frac{1}{x})$, between two simpler functions: $f(x) = -x^2$ and $h(x) = x^2$.

Step 2: Find the limits of the outer functions.

Now, let's find the limits of our "sandwich bread" as $x$ approaches 0.

$\lim\limits_{x \to 0} f(x) = \lim\limits_{x \to 0} (-x^2) = -(0)^2 = 0$.

$\lim\limits_{x \to 0} h(x) = \lim\limits_{x \to 0} (x^2) = (0)^2 = 0$.

Step 3: Apply the Squeeze Theorem.

Since our function $x^2 \sin(\frac{1}{x})$ is squeezed between two functions that both approach a limit of 0, it must also approach 0.

Therefore, by the Squeeze Theorem, $\lim\limits_{x \to 0} x^2 \sin(\frac{1}{x}) = 0$.

Some Important Limits of Trigonometric Functions

When working with calculus, we often encounter limits involving trigonometric functions like sine, cosine, and tangent. Direct substitution often fails for these limits, especially as the variable approaches zero, leading to the indeterminate form $\frac{0}{0}$. To solve these, mathematicians have established a few fundamental trigonometric limits that serve as essential tools for evaluating more complex expressions.

The Most Important Trigonometric Limit

The cornerstone for all trigonometric limits is the following result. It describes the profound relationship between the sine of a very small angle and the angle itself.

Important Note: This formula is only valid when the angle $x$ is measured in radians. This is because the geometric proof relies on the property that the area of a circular sector is $\frac{1}{2}r^2\theta$, which is true only for radians.

If you try to plug in $x=0$, you get $\frac{\sin 0}{0} = \frac{0}{0}$, which is undefined. However, the limit tells us that for angles very close to zero, the value of $\sin x$ is almost identical to the value of $x$ itself. The ratio $\frac{\sin x}{x}$ gets closer and closer to 1 as $x$ gets closer to 0.

Proof using the Squeeze Theorem

This limit is proven using a geometric argument involving a unit circle and the Squeeze Theorem.

Consider a sector of a unit circle (radius = 1) with a small positive angle $x$ (in radians). We can compare the area of a small triangle inside the sector, the area of the sector itself, and the area of a large triangle outside the sector.

From the geometry, we can establish the inequality:

Area(Triangle OAP) $\le$ Area(Sector OAP) $\le$ Area(Triangle OAT)

Calculating each area:

• Area(Triangle OAP) = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}(1)(\sin x) = \frac{1}{2}\sin x$
• Area(Sector OAP) = $\frac{1}{2}r^2 x = \frac{1}{2}(1)^2 x = \frac{1}{2}x$
• Area(Triangle OAT) = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}(1)(\tan x) = \frac{1}{2}\tan x$

Substituting these into our inequality gives:

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

Assuming $x$ is a small positive angle, $\sin x > 0$. We can divide the entire inequality by $\frac{1}{2}\sin x$:

$1 \le \frac{x}{\sin x} \le \frac{\tan x}{\sin x} \implies 1 \le \frac{x}{\sin x} \le \frac{1}{\cos x}$

Taking the reciprocal of all parts reverses the inequality signs:

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

We have now "squeezed" our function, $\frac{\sin x}{x}$, between $\cos x$ and $1$. As $x$ approaches 0, we know that $\lim\limits_{x \to 0} \cos x = 1$ and $\lim\limits_{x \to 0} 1 = 1$. Since both the lower and upper bounds are approaching 1, the Squeeze Theorem guarantees that the function in the middle must also approach 1.

Other Key Trigonometric Limits

Using the fundamental limit and the rules of algebra for limits, we can derive several other important results:

1. The Tangent Limit

Proof: We can rewrite $\tan x$ as $\frac{\sin x}{\cos x}$ and use the product rule for limits.

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

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

2. The Cosine Limit

Proof: We multiply the numerator and denominator by the conjugate $(1 + \cos x)$.

$\lim\limits_{x \to 0} \frac{1 - \cos x}{x} = \lim\limits_{x \to 0} \frac{1 - \cos x}{x} \cdot \frac{1 + \cos x}{1 + \cos x}$

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

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

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

A General Rule

A very practical extension of these rules is that if the argument of the sine or tangent function is not just $x$, but some other function that also goes to zero, the limit still holds. For any constant $k \neq 0$:

$\lim\limits_{x \to 0} \frac{\sin (kx)}{kx} = 1$ and $\lim\limits_{x \to 0} \frac{\tan (kx)}{kx} = 1$

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Limits of Exponential and Logarithmic Functions

Exponential and logarithmic functions are fundamental in describing phenomena involving growth, decay, and scaling. In calculus, their limits are essential for understanding their rates of change. Direct substitution into these functions often leads to unique indeterminate forms like $1^\infty$ and $\frac{0}{0}$, which are resolved by a set of powerful standard limit formulas.

Throughout this section, '$\log$' refers to the natural logarithm (base $e$), also written as '$\ln$'.

Standard Exponential Limits

The most important base for exponential and logarithmic functions in calculus is the natural number $e \approx 2.718$. This number is itself defined by a limit, which gives rise to a family of related limit problems.

1. The Definition of $e$

The number $e$ can be defined as the value that the expression $(1 + \frac{1}{n})^n$ approaches as $n$ becomes infinitely large. A more general form used for limits as a variable approaches zero is:

If you try to substitute $x=0$, you get $(1+0)^{\frac{1}{0}}$, which suggests the form $1^\infty$. This is an indeterminate form, and this limit formula tells us that this specific form resolves to the value of $e$.

2. The Fundamental Exponential Limit

This is arguably the most crucial exponential limit for finding derivatives. It deals with the indeterminate form $\frac{0}{0}$.

This limit tells us that for very small values of $x$, the value of $e^x - 1$ is approximately equal to $x$. Geometrically, it means the slope of the tangent line to the curve $y=e^x$ at the point $(0,1)$ is exactly 1.

3. Exponential Limit with a General Base

We can generalize the above limit for any valid base $a > 0$.

Proof: We can rewrite $a^x$ using base $e$ as $a^x = e^{\log(a^x)} = e^{x \log a}$.

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

To match the form $\frac{e^u-1}{u}$, we let $u = x \log a$. We multiply and divide by $\log a$:

$\lim\limits_{x \to 0} \frac{e^{x \log a} - 1}{x \log a} \cdot \log a = \left(\lim\limits_{x \to 0} \frac{e^{x \log a} - 1}{x \log a}\right) \cdot \log a$

As $x \to 0$, $u = x \log a \to 0$. The first limit becomes $\lim\limits_{u \to 0} \frac{e^u-1}{u}$, which is 1. So the result is $1 \cdot \log a = \log a$.

Notice that if we set $a=e$ in this formula, we get $\log e = 1$, which gives us back the previous standard limit (Formula ii).

Standard Logarithmic Limits

The corresponding limits for logarithmic functions are closely related to the exponential ones and can be derived from them.

1. The Fundamental Logarithmic Limit

This limit, which also results in the indeterminate form $\frac{0}{0}$, is the counterpart to the fundamental exponential limit.

Proof: Let $y = \log(1+x)$. This means $e^y = 1+x$, or $x = e^y - 1$. As $x \to 0$, $y \to \log(1)=0$. So we can substitute:

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

This result shows that for very small values of $x$, the value of $\log(1+x)$ is approximately equal to $x$.

2. Logarithmic Limit with a General Base

We can generalize this for any valid logarithmic base $a > 0, a \neq 1$.

Proof: Using the change of base formula, $\log_a M = \frac{\log M}{\log a}$.

$\lim\limits_{x \to 0} \frac{\log_a(1 + x)}{x} = \lim\limits_{x \to 0} \frac{\log(1+x)}{x \log a} = \frac{1}{\log a} \cdot \left( \lim\limits_{x \to 0} \frac{\log(1+x)}{x} \right) = \frac{1}{\log a} \cdot 1 = \frac{1}{\log a}$.

A General Rule for Application

A practical way to use these formulas is to recognize that if the argument inside the function is not just $x$, but some other expression (like $kx$) that also goes to zero, the pattern still holds. For any constant $k \neq 0$:

$\lim\limits_{x \to 0} \frac{e^{kx} - 1}{x} = k$ and $\lim\limits_{x \to 0} \frac{\log(1 + kx)}{x} = k$

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Derivatives

After establishing the concept of a limit, we can now introduce the most central concept in differential calculus: the derivative. A derivative is a measure of how a function's output value changes with respect to a change in its input value. In essence, it tells us the instantaneous rate of change of the function at any given point.

Definition of the Derivative (First Principles)

The derivative is defined using a limit. We start by looking at the average rate of change between two points on a function's graph and then see what happens as those two points get infinitely close to each other.

Let $f(x)$ be a function. The average rate of change between a point $x$ and a nearby point $x+h$ is the change in the function's value divided by the change in the input value:

Average Rate of Change = $\frac{\text{Change in } f(x)}{\text{Change in } x} = \frac{f(x+h) - f(x)}{(x+h) - x} = \frac{f(x+h) - f(x)}{h}$

To find the instantaneous rate of change at the point $x$, we take the limit of this average rate of change as the distance between the points, $h$, shrinks to zero.

The derivative of a function $f(x)$, denoted by $f'(x)$ (read as "f prime of x") or $\frac{dy}{dx}$, is defined as:

This limit, if it exists, gives us a new function, $f'(x)$, that tells us the instantaneous rate of change of the original function $f(x)$ at any point $x$. This method of finding a derivative using the limit definition is called finding the derivative from first principles.

The Geometric Interpretation: Slope of the Tangent Line

The derivative has a powerful and intuitive geometric meaning. The expression for the average rate of change, $\frac{f(x+h) - f(x)}{h}$, is precisely the formula for the slope of a straight line (the secant line) that connects the two points $(x, f(x))$ and $(x+h, f(x+h))$ on the curve.

As we take the limit where $h \to 0$, the second point slides along the curve towards the first point. The secant line connecting them pivots until, at the limit, it becomes the tangent line to the curve at the point $(x, f(x))$.

Therefore, the derivative $f'(x)$ at a specific point $x$ represents the slope of the tangent line to the graph of $y=f(x)$ at that point. It is the ultimate measure of the steepness or slope of the curve at that single, instantaneous point.

Differentiability and Continuity

If the limit in the definition of the derivative exists at a point $x=c$, we say the function is differentiable at that point. This means the function is "smooth" at that point and doesn't have any sharp corners, cusps, or vertical tangents.

There is a crucial relationship between a function being differentiable and being continuous (unbroken):

Theorem: If a function is differentiable at a point, it must be continuous at that point.

However, the reverse is not always true. A function can be continuous but not differentiable. The classic example is the absolute value function, $f(x)=|x|$, which has a sharp corner at $x=0$. You can draw it without lifting your pen (so it's continuous), but there is no single, well-defined tangent line at that sharp point (so it's not differentiable there).

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Some Standard Results on Derivatives

While finding a derivative from first principles (the limit definition) is fundamental to understanding the concept, it is often a long and complex process. To make differentiation more efficient, mathematicians have used the first principle to establish a set of standard derivative formulas for common functions. By memorizing these basic results and a few rules for combining them, we can differentiate almost any elementary function.

Derivatives of Basic Functions

The following are the foundational building blocks for differentiation. Each one can be proven using the limit definition of a derivative.

1. The Power Rule

This is one of the most frequently used rules. It applies to any function of the form $f(x) = x^n$, where $n$ is any real number.

In words: To find the derivative of $x$ raised to a power, you bring the power down as a coefficient and then subtract one from the original power.

• Derivative of a constant ($f(x)=c=cx^0$): $\frac{d}{dx}(c) = 0$. The slope of a horizontal line is always zero.
• Derivative of $x$: $\frac{d}{dx}(x) = \frac{d}{dx}(x^1) = 1 \cdot x^{1-1} = 1 \cdot x^0 = 1$. The slope of the line $y=x$ is always 1.
• Derivative of $\sqrt{x}$: $\frac{d}{dx}(\sqrt{x}) = \frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{1/2 - 1} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$.
• Derivative of $\frac{1}{x}$: $\frac{d}{dx}(\frac{1}{x}) = \frac{d}{dx}(x^{-1}) = -1 \cdot x^{-1-1} = -1 \cdot x^{-2} = -\frac{1}{x^2}$.

2. Derivatives of Trigonometric Functions

The derivatives of the six standard trigonometric functions (when $x$ is in radians) are:

3. Derivatives of Exponential and Logarithmic Functions

Here, $\log x$ refers to the natural logarithm, $\ln x$.

• The function $e^x$ has the unique property that it is its own derivative: $\frac{d}{dx}(e^x) = e^x$.
• For any other base $a$: $\frac{d}{dx}(a^x) = a^x \log a$.
• The derivative of the natural logarithm is: $\frac{d}{dx}(\log x) = \frac{1}{x}$.
• For any other base $a$: $\frac{d}{dx}(\log_a x) = \frac{1}{x \log a}$.

The Algebra of Derivatives

These rules tell us how to find the derivative of functions that are created by adding, subtracting, multiplying, or dividing the basic functions we just learned.

Let $u(x)$ and $v(x)$ be differentiable functions.

1. Sum, Difference, and Constant Multiple Rules

Differentiation is a "linear" operation, which means it can be distributed over addition and subtraction, and constants can be factored out.

• Sum/Difference Rule: $\frac{d}{dx}[u(x) \pm v(x)] = \frac{d}{dx}u(x) \pm \frac{d}{dx}v(x) = u'(x) \pm v'(x)$
• Constant Multiple Rule: $\frac{d}{dx}[c \cdot u(x)] = c \cdot \frac{d}{dx}u(x) = c \cdot u'(x)$

2. The Product Rule

To differentiate the product of two functions, you do not simply multiply their derivatives. The correct rule is derived from the first principle and is as follows:

In words: "The derivative of the first times the second, plus the first times the derivative of the second."

3. The Quotient Rule

To differentiate a fraction (or quotient) of two functions, the rule is:

Mnemonic: "Low d-High minus High d-Low, over the square of what's below." (Here, "Low" is the denominator $v(x)$, "High" is the numerator $u(x)$, and "d" means the derivative of).

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