1.5 Special limits

Achievable AP Calculus AB

1. Limits

Our AP Calculus AB course is now in "early access" - get 50% off for a limited time.

Special limits

9 min read

Font

Discuss

Feedback

What you’ll learn

Limits at infinity for exponential and logarithmic functions
Applying the Squeeze theorem to find limits of oscillating functions
Special trigonometric limits involving sine and cosine

Growth of exponential and logarithmic functions

Dominant term analysis can be extended to exponential and logarithmic functions by comparing their growth rates to algebraic functions (such as polynomials or roots).

As $x \to \infty$ , the growth rates of these functions can be ranked in the following order:

$Logarithmic < Algebraic < Exponential$

This comparison is specifically for large positive values of $x$ . In expressions containing a mix of these functions, the faster-growing function dominates the end behavior.

Example 1: $x \to \infty$

Evaluate:

a) $x \to \infty lim \frac{x ^{1000}}{e ^{x}}$

b) $x \to \infty lim \frac{x}{ln x}$

c) $x \to \infty lim \frac{2 ^{x}}{5 ^{x}}$

Solutions

a) $x \to \infty lim \frac{x ^{1000}}{e ^{x}}$

(spoiler)

Direct substitution gives the indeterminate form $\frac{\infty}{\infty}$ .

Even with the massive power of $x$ , the exponential function $e^{x}$ in the denominator eventually dominates the polynomial $x^{1000}$ in the numerator for large values of $x$ .

This behavior fits case 1 from the horizontal asymptotes section, and the limit is $0$ .

b) $x \to \infty lim \frac{x}{l n x}$

(spoiler)

Direct substitution gives $\frac{\infty}{\infty}$ .

Since $x$ , an algebraic function, grows faster than the natural logarithm $ln x$ , the numerator eventually dominates.

This behavior fits case 3 and the limit is $\infty$ .

c) $x \to \infty lim \frac{2 ^{x}}{5 ^{x}}$

(spoiler)

Both are exponential functions, but the denominator dominates because it has a larger base. Since this fits case 1, the limit is $0$ .

Alternatively, we can use exponent rules to rewrite the expression as a single function

$x \to \infty lim (\frac{2}{5})^{x}$

Because $0 < \frac{2}{5} < 1$ , repeated multiplication by $\frac{2}{5}$ makes the expression shrink toward $0$ . Therefore, the limit is $0$ .

Example 2: $x \to - \infty$

As $x$ approaches $- \infty$ , basic exponential functions such as $e^{x}$ approach $0$ instead, which leads to different end behavior.

Evaluate:

$x \to - \infty lim \frac{x + 1}{e ^{x}}$

(spoiler)

As $x \to - \infty$ , the pieces behave like this:

Numerator: $(x + 1) \to - \infty$
Denominator: $e^{x} \to 0^{+}$ (a very small positive number)

When a very large negative number is divided by a very small positive number, the expression becomes unbounded in the negative direction. You can observe this by calculating with concrete values such as $\frac{- 1000}{0.001}$ .

Therefore,

$x \to - \infty lim \frac{x + 1}{e ^{x}} = - \infty$

This fits case 3: the numerator dominates because the denominator shrinks toward $0$ .

Let

$f (x) = \frac{2 ^{x} + 6 ^{x} + 1}{5 ^{x} - 2}$

Evaluate:

a) $x \to \infty lim f (x)$

b) $x \to - \infty lim f (x)$

Answers

(spoiler)

a) $\infty$
b) $- \frac{1}{2}$

Solutions

a) $x \to \infty lim f (x)$

(spoiler)

As $x \to \infty$ :

In the numerator, $6^{x}$ dominates $2^{x}$ and the constant $1$ .
In the denominator, $5^{x}$ dominates the constant $- 2$ .

So for large $x$ , the expression can be reduced to

$x \to \infty lim f (x) \approx x \to \infty lim \frac{6 ^{x}}{5 ^{x}} = x \to \infty lim (\frac{6}{5})^{x}$

This fits case 3. You may also consider that because the base ( $\frac{6}{5}$ ) is greater than $1$ , repeated multiplication causes the expression to grow without bound, and the limit is $\infty$ .

b) $x \to - \infty lim f (x)$

(spoiler)

As $x \to - \infty$ , the exponential functions $2^{x}, 6^{x},$ and $5^{x}$ all approach $0$ . Substituting those end behaviors into the expression gives

$x \to - \infty lim \frac{2 ^{x} + 6 ^{x} + 1}{5 ^{x} - 2} = \frac{0 + 0 + 1}{0 - 2} = - \frac{1}{2}$

Squeeze theorem

Also called the sandwich theorem, the Squeeze theorem helps you evaluate limits by trapping a function between two others whose limits are already known.

Squeeze theorem:

If $g (x) \leq f (x) \leq h (x)$ for all $x$ near $a$ , and

$x \to a lim g (x) = x \to a lim h (x) = L,$

then

$x \to a lim f (x) = L$

As shown in the graph, if the lower bounding function $g (x)$ (in green) and the upper bounding function $h (x)$ (in red) converge to the same limit $L$ at a point, the middle function $f (x)$ (in blue) is forced to that limit.

Example

Let $g$ and $h$ be the functions defined by $g (x) = 2 x^{2} + 4 x + 1$ and $h (x) = x^{3} - 3 x + 8$ .

If $f$ is a function such that $g (x) \leq f (x) \leq h (x)$ for all $x > 0$ , find $x \to 1 lim f (x)$ .

(spoiler)

Since $f (x)$ is squeezed between the bounding functions $g (x)$ and $h (x)$ as $x \to 1$ , evaluate the limits of the bounds:

$x \to 1 lim g (x) = x \to 1 lim (2 x^{2} + 4 x + 1) = 7$

and

$x \to 1 lim h (x) = x \to 1 lim (x^{3} - 3 x + 8) = 7$

Because the two limits are equal, $x \to 1 lim f (x) = 7$ by the Squeeze theorem.

Bounded functions at infinity

Consider the limit

$x \to \infty lim \frac{sin x}{x}$

Trigonometric functions like $sin x$ and $cos x$ oscillate indefinitely between $- 1$ and $1$ , so direct substitution does not result in an indeterminate form.

Since the denominator grows without bound while the numerator remains bounded, dominant term analysis suggests that the fraction approaches $0$ .

This can be proven using the Squeeze theorem. Since $sin x$ is bounded for all $x$ ,

$- 1 \leq sin x \leq 1$

Dividing each part of the inequality by $x$ gives:

$- \frac{1}{x} \leq \frac{sin x}{x} \leq \frac{1}{x}$

The limits of both bounding functions equal $0$ as $x$ approaches $\infty$ . Therefore, by the Squeeze theorem,

$x \to \infty lim \frac{sin x}{x} = 0$

Special trigonometric limits

If $x$ approaches $0$ instead of $\infty$ , the previous approach no longer applies. Although the following limits can be proven using the Squeeze theorem, you are only expected to memorize them:

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

Their reciprocals are the same:

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

AP tip:

Know these standard trig limits well.

Examples

A common variant changes the argument of the sine function.

Evaluate

$x \to 0 lim \frac{sin ( 3 x )}{x}$

(spoiler)

In order to use the special trig limit, the denominator must match the argument of sine, or $3 x$ . The trick is to multiply by $1$ in the form of $\frac{3}{3}$ :

$= = = x \to 0 lim \frac{sin ( 3 x )}{x} \cdot \frac{3}{3} x \to 0 lim \frac{sin ( 3 x )}{3 x} \cdot 3 1 \cdot 3 3$

Evaluate

$x \to 0 lim \frac{sin ( 3 x )}{sin ( 4 x )}$

(spoiler)

Here, there is no $x$ outside the sine functions, so we create the special limit forms by multiplying by $\frac{3 x}{3 x}$ and $\frac{4 x}{4 x}$ .

Start by rewriting the expression as a product:

$x \to 0 lim \frac{sin ( 3 x )}{1} \times \frac{1}{sin ( 4 x )}$

Now create the special limit form:

$x \to 0 lim \frac{sin ( 3 x )}{1} \times \frac{3 x}{3 x} \times \frac{1}{sin ( 4 x )} \times \frac{4 x}{4 x}$

Rearrange the expression to form the special limits:

$x \to 0 lim \frac{sin ( 3 x )}{3 x} \times 3 x \times \frac{4 x}{sin ( 4 x )} \times \frac{1}{4 x} = x \to 0 lim \frac{sin ( 3 x )}{3 x} \times \frac{4 x}{sin ( 4 x )} \times \frac{3 x}{4 x} = 1 \times 1 \times \frac{3}{4} = \frac{3}{4}$

As a general shortcut,

$x \to 0 lim \frac{sin ( a x )}{b x} = \frac{a}{b}$

and

$x \to 0 lim \frac{a x}{sin ( b x )} = \frac{a}{b}$

Challenge problem

$x \to \infty lim x sin (\frac{2}{x}) =$

(spoiler)

Direct substitution gives the indeterminate form $\infty \times 0$ .

First, rewrite the expression as

$x \to \infty lim \frac{sin ( \frac{2}{x} )}{\frac{1}{x}}$

Next, make a substitution by letting $\frac{2}{x} = u$ . Then $\frac{1}{x} = \frac{1}{2} u$ .

As $x$ approaches $\infty, u$ approaches $0$ , and the limit rewritten in terms of $u$ is

$u \to 0 lim \frac{sin ( u )}{\frac{1}{2} u} = u \to 0 lim 2 \cdot \frac{sin ( u )}{u} = 2 \cdot 1 = 2$

Note that substitution doesn’t change the function. We’ve only changed the form it’s written in so that the special limit can be applied.

Growth of exponential and logarithmic functions

Exponential functions ( $e^{x}$ ) outgrow any polynomial as $x \to \infty$
Logarithmic functions ( $ln (x)$ ) grow slower than any power of $x$ as $x \to \infty$
For limits at infinity:
- If denominator grows faster (e.g., $e^{x}$ ), limit $\to 0$
- If numerator grows faster as denominator $\to 0$ , limit $\to \pm \infty$
- Compare dominant terms for large $x$

Squeeze theorem

Used to evaluate limits by bounding $f (x)$ between two functions with known limits
If $g (x) \leq f (x) \leq h (x)$ and $lim_{x \to a} g (x) = lim_{x \to a} h (x) = L$ , then $lim_{x \to a} f (x) = L$
Especially useful for oscillating functions and trigonometric limits

Special trigonometric limits

$lim_{x \to 0} \frac{s i n ( x )}{x} = 1$
$lim_{x \to 0} \frac{1 - c o s ( x )}{x} = 0$
For $lim_{x \to 0} \frac{s i n ( k x )}{x} = k$
- Use substitution or multiply/divide to match standard forms

Example strategies and challenge problems

Use substitution to rewrite limits in standard forms (e.g., $u = k x$ )
Factor denominators to isolate special limit forms
For limits like $\frac{s i n ( a x )}{s i n ( b x )}$ as $x \to 0$ , answer is $\frac{a}{b}$ by matching to $\frac{s i n ( a x )}{a x} \cdot \frac{b x}{s i n ( b x )}$

Special limits

What you’ll learn

Limits at infinity for exponential and logarithmic functions
Applying the Squeeze theorem to find limits of oscillating functions
Special trigonometric limits involving sine and cosine

Growth of exponential and logarithmic functions

Dominant term analysis can be extended to exponential and logarithmic functions by comparing their growth rates to algebraic functions (such as polynomials or roots).

As $x \to \infty$ , the growth rates of these functions can be ranked in the following order:

$Logarithmic < Algebraic < Exponential$

This comparison is specifically for large positive values of $x$ . In expressions containing a mix of these functions, the faster-growing function dominates the end behavior.

Example 1: $x \to \infty$

Evaluate:

a) $x \to \infty lim \frac{x ^{1000}}{e ^{x}}$

b) $x \to \infty lim \frac{x}{ln x}$

c) $x \to \infty lim \frac{2 ^{x}}{5 ^{x}}$

Solutions

a) $x \to \infty lim \frac{x ^{1000}}{e ^{x}}$

(spoiler)

Direct substitution gives the indeterminate form $\frac{\infty}{\infty}$ .

Even with the massive power of $x$ , the exponential function $e^{x}$ in the denominator eventually dominates the polynomial $x^{1000}$ in the numerator for large values of $x$ .

This behavior fits case 1 from the horizontal asymptotes section, and the limit is $0$ .

b) $x \to \infty lim \frac{x}{l n x}$

(spoiler)

Direct substitution gives $\frac{\infty}{\infty}$ .

Since $x$ , an algebraic function, grows faster than the natural logarithm $ln x$ , the numerator eventually dominates.

This behavior fits case 3 and the limit is $\infty$ .

c) $x \to \infty lim \frac{2 ^{x}}{5 ^{x}}$

(spoiler)

Both are exponential functions, but the denominator dominates because it has a larger base. Since this fits case 1, the limit is $0$ .

Alternatively, we can use exponent rules to rewrite the expression as a single function

$x \to \infty lim (\frac{2}{5})^{x}$

Because $0 < \frac{2}{5} < 1$ , repeated multiplication by $\frac{2}{5}$ makes the expression shrink toward $0$ . Therefore, the limit is $0$ .

Example 2: $x \to - \infty$

As $x$ approaches $- \infty$ , basic exponential functions such as $e^{x}$ approach $0$ instead, which leads to different end behavior.

Evaluate:

$x \to - \infty lim \frac{x + 1}{e ^{x}}$

(spoiler)

As $x \to - \infty$ , the pieces behave like this:

Numerator: $(x + 1) \to - \infty$
Denominator: $e^{x} \to 0^{+}$ (a very small positive number)

Therefore,

$x \to - \infty lim \frac{x + 1}{e ^{x}} = - \infty$

This fits case 3: the numerator dominates because the denominator shrinks toward $0$ .

Let

$f (x) = \frac{2 ^{x} + 6 ^{x} + 1}{5 ^{x} - 2}$

Evaluate:

a) $x \to \infty lim f (x)$

b) $x \to - \infty lim f (x)$

Answers

(spoiler)

a) $\infty$
b) $- \frac{1}{2}$

Solutions

a) $x \to \infty lim f (x)$

(spoiler)

As $x \to \infty$ :

In the numerator, $6^{x}$ dominates $2^{x}$ and the constant $1$ .
In the denominator, $5^{x}$ dominates the constant $- 2$ .

So for large $x$ , the expression can be reduced to

$x \to \infty lim f (x) \approx x \to \infty lim \frac{6 ^{x}}{5 ^{x}} = x \to \infty lim (\frac{6}{5})^{x}$

This fits case 3. You may also consider that because the base ( $\frac{6}{5}$ ) is greater than $1$ , repeated multiplication causes the expression to grow without bound, and the limit is $\infty$ .

b) $x \to - \infty lim f (x)$

(spoiler)

As $x \to - \infty$ , the exponential functions $2^{x}, 6^{x},$ and $5^{x}$ all approach $0$ . Substituting those end behaviors into the expression gives

$x \to - \infty lim \frac{2 ^{x} + 6 ^{x} + 1}{5 ^{x} - 2} = \frac{0 + 0 + 1}{0 - 2} = - \frac{1}{2}$

Squeeze theorem

Also called the sandwich theorem, the Squeeze theorem helps you evaluate limits by trapping a function between two others whose limits are already known.

Squeeze theorem:

If $g (x) \leq f (x) \leq h (x)$ for all $x$ near $a$ , and

$x \to a lim g (x) = x \to a lim h (x) = L,$

then

$x \to a lim f (x) = L$

Example

Let $g$ and $h$ be the functions defined by $g (x) = 2 x^{2} + 4 x + 1$ and $h (x) = x^{3} - 3 x + 8$ .

If $f$ is a function such that $g (x) \leq f (x) \leq h (x)$ for all $x > 0$ , find $x \to 1 lim f (x)$ .

(spoiler)

Since $f (x)$ is squeezed between the bounding functions $g (x)$ and $h (x)$ as $x \to 1$ , evaluate the limits of the bounds:

$x \to 1 lim g (x) = x \to 1 lim (2 x^{2} + 4 x + 1) = 7$

and

$x \to 1 lim h (x) = x \to 1 lim (x^{3} - 3 x + 8) = 7$

Because the two limits are equal, $x \to 1 lim f (x) = 7$ by the Squeeze theorem.

Bounded functions at infinity

Consider the limit

$x \to \infty lim \frac{sin x}{x}$

Trigonometric functions like $sin x$ and $cos x$ oscillate indefinitely between $- 1$ and $1$ , so direct substitution does not result in an indeterminate form.

Since the denominator grows without bound while the numerator remains bounded, dominant term analysis suggests that the fraction approaches $0$ .

This can be proven using the Squeeze theorem. Since $sin x$ is bounded for all $x$ ,

$- 1 \leq sin x \leq 1$

Dividing each part of the inequality by $x$ gives:

$- \frac{1}{x} \leq \frac{sin x}{x} \leq \frac{1}{x}$

The limits of both bounding functions equal $0$ as $x$ approaches $\infty$ . Therefore, by the Squeeze theorem,

$x \to \infty lim \frac{sin x}{x} = 0$

Special trigonometric limits

If $x$ approaches $0$ instead of $\infty$ , the previous approach no longer applies. Although the following limits can be proven using the Squeeze theorem, you are only expected to memorize them:

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

Their reciprocals are the same:

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

AP tip:

Know these standard trig limits well.

Examples

A common variant changes the argument of the sine function.

Evaluate

$x \to 0 lim \frac{sin ( 3 x )}{x}$

(spoiler)

In order to use the special trig limit, the denominator must match the argument of sine, or $3 x$ . The trick is to multiply by $1$ in the form of $\frac{3}{3}$ :

$= = = x \to 0 lim \frac{sin ( 3 x )}{x} \cdot \frac{3}{3} x \to 0 lim \frac{sin ( 3 x )}{3 x} \cdot 3 1 \cdot 3 3$

Evaluate

$x \to 0 lim \frac{sin ( 3 x )}{sin ( 4 x )}$

(spoiler)

Here, there is no $x$ outside the sine functions, so we create the special limit forms by multiplying by $\frac{3 x}{3 x}$ and $\frac{4 x}{4 x}$ .

Start by rewriting the expression as a product:

$x \to 0 lim \frac{sin ( 3 x )}{1} \times \frac{1}{sin ( 4 x )}$

Now create the special limit form:

$x \to 0 lim \frac{sin ( 3 x )}{1} \times \frac{3 x}{3 x} \times \frac{1}{sin ( 4 x )} \times \frac{4 x}{4 x}$

Rearrange the expression to form the special limits:

As a general shortcut,

$x \to 0 lim \frac{sin ( a x )}{b x} = \frac{a}{b}$

and

$x \to 0 lim \frac{a x}{sin ( b x )} = \frac{a}{b}$

Challenge problem

$x \to \infty lim x sin (\frac{2}{x}) =$

(spoiler)

Direct substitution gives the indeterminate form $\infty \times 0$ .

First, rewrite the expression as

$x \to \infty lim \frac{sin ( \frac{2}{x} )}{\frac{1}{x}}$

Next, make a substitution by letting $\frac{2}{x} = u$ . Then $\frac{1}{x} = \frac{1}{2} u$ .

As $x$ approaches $\infty, u$ approaches $0$ , and the limit rewritten in terms of $u$ is

$u \to 0 lim \frac{sin ( u )}{\frac{1}{2} u} = u \to 0 lim 2 \cdot \frac{sin ( u )}{u} = 2 \cdot 1 = 2$

Note that substitution doesn’t change the function. We’ve only changed the form it’s written in so that the special limit can be applied.

Achievable AP Calculus AB

1. Limits

Our AP Calculus AB course is now in "early access" - get 50% off for a limited time.

Special limits

9 min read

Font

Discuss

Feedback

What you’ll learn

Limits at infinity for exponential and logarithmic functions
Applying the Squeeze theorem to find limits of oscillating functions
Special trigonometric limits involving sine and cosine

Growth of exponential and logarithmic functions

Dominant term analysis can be extended to exponential and logarithmic functions by comparing their growth rates to algebraic functions (such as polynomials or roots).

As $x \to \infty$ , the growth rates of these functions can be ranked in the following order:

$Logarithmic < Algebraic < Exponential$

This comparison is specifically for large positive values of $x$ . In expressions containing a mix of these functions, the faster-growing function dominates the end behavior.

Example 1: $x \to \infty$

Evaluate:

a) $x \to \infty lim \frac{x ^{1000}}{e ^{x}}$

b) $x \to \infty lim \frac{x}{ln x}$

c) $x \to \infty lim \frac{2 ^{x}}{5 ^{x}}$

Solutions

a) $x \to \infty lim \frac{x ^{1000}}{e ^{x}}$

(spoiler)

Direct substitution gives the indeterminate form $\frac{\infty}{\infty}$ .

Even with the massive power of $x$ , the exponential function $e^{x}$ in the denominator eventually dominates the polynomial $x^{1000}$ in the numerator for large values of $x$ .

This behavior fits case 1 from the horizontal asymptotes section, and the limit is $0$ .

b) $x \to \infty lim \frac{x}{l n x}$

(spoiler)

Direct substitution gives $\frac{\infty}{\infty}$ .

Since $x$ , an algebraic function, grows faster than the natural logarithm $ln x$ , the numerator eventually dominates.

This behavior fits case 3 and the limit is $\infty$ .

c) $x \to \infty lim \frac{2 ^{x}}{5 ^{x}}$

(spoiler)

Both are exponential functions, but the denominator dominates because it has a larger base. Since this fits case 1, the limit is $0$ .

Alternatively, we can use exponent rules to rewrite the expression as a single function

$x \to \infty lim (\frac{2}{5})^{x}$

Because $0 < \frac{2}{5} < 1$ , repeated multiplication by $\frac{2}{5}$ makes the expression shrink toward $0$ . Therefore, the limit is $0$ .

Example 2: $x \to - \infty$

As $x$ approaches $- \infty$ , basic exponential functions such as $e^{x}$ approach $0$ instead, which leads to different end behavior.

Evaluate:

$x \to - \infty lim \frac{x + 1}{e ^{x}}$

(spoiler)

As $x \to - \infty$ , the pieces behave like this:

Numerator: $(x + 1) \to - \infty$
Denominator: $e^{x} \to 0^{+}$ (a very small positive number)

Therefore,

$x \to - \infty lim \frac{x + 1}{e ^{x}} = - \infty$

This fits case 3: the numerator dominates because the denominator shrinks toward $0$ .

Let

$f (x) = \frac{2 ^{x} + 6 ^{x} + 1}{5 ^{x} - 2}$

Evaluate:

a) $x \to \infty lim f (x)$

b) $x \to - \infty lim f (x)$

Answers

(spoiler)

a) $\infty$
b) $- \frac{1}{2}$

Solutions

a) $x \to \infty lim f (x)$

(spoiler)

As $x \to \infty$ :

In the numerator, $6^{x}$ dominates $2^{x}$ and the constant $1$ .
In the denominator, $5^{x}$ dominates the constant $- 2$ .

So for large $x$ , the expression can be reduced to

$x \to \infty lim f (x) \approx x \to \infty lim \frac{6 ^{x}}{5 ^{x}} = x \to \infty lim (\frac{6}{5})^{x}$

This fits case 3. You may also consider that because the base ( $\frac{6}{5}$ ) is greater than $1$ , repeated multiplication causes the expression to grow without bound, and the limit is $\infty$ .

b) $x \to - \infty lim f (x)$

(spoiler)

As $x \to - \infty$ , the exponential functions $2^{x}, 6^{x},$ and $5^{x}$ all approach $0$ . Substituting those end behaviors into the expression gives

$x \to - \infty lim \frac{2 ^{x} + 6 ^{x} + 1}{5 ^{x} - 2} = \frac{0 + 0 + 1}{0 - 2} = - \frac{1}{2}$

Squeeze theorem

Also called the sandwich theorem, the Squeeze theorem helps you evaluate limits by trapping a function between two others whose limits are already known.

Squeeze theorem:

If $g (x) \leq f (x) \leq h (x)$ for all $x$ near $a$ , and

$x \to a lim g (x) = x \to a lim h (x) = L,$

then

$x \to a lim f (x) = L$

Example

Let $g$ and $h$ be the functions defined by $g (x) = 2 x^{2} + 4 x + 1$ and $h (x) = x^{3} - 3 x + 8$ .

If $f$ is a function such that $g (x) \leq f (x) \leq h (x)$ for all $x > 0$ , find $x \to 1 lim f (x)$ .

(spoiler)

Since $f (x)$ is squeezed between the bounding functions $g (x)$ and $h (x)$ as $x \to 1$ , evaluate the limits of the bounds:

$x \to 1 lim g (x) = x \to 1 lim (2 x^{2} + 4 x + 1) = 7$

and

$x \to 1 lim h (x) = x \to 1 lim (x^{3} - 3 x + 8) = 7$

Because the two limits are equal, $x \to 1 lim f (x) = 7$ by the Squeeze theorem.

Bounded functions at infinity

Consider the limit

$x \to \infty lim \frac{sin x}{x}$

Trigonometric functions like $sin x$ and $cos x$ oscillate indefinitely between $- 1$ and $1$ , so direct substitution does not result in an indeterminate form.

Since the denominator grows without bound while the numerator remains bounded, dominant term analysis suggests that the fraction approaches $0$ .

This can be proven using the Squeeze theorem. Since $sin x$ is bounded for all $x$ ,

$- 1 \leq sin x \leq 1$

Dividing each part of the inequality by $x$ gives:

$- \frac{1}{x} \leq \frac{sin x}{x} \leq \frac{1}{x}$

The limits of both bounding functions equal $0$ as $x$ approaches $\infty$ . Therefore, by the Squeeze theorem,

$x \to \infty lim \frac{sin x}{x} = 0$

Special trigonometric limits

If $x$ approaches $0$ instead of $\infty$ , the previous approach no longer applies. Although the following limits can be proven using the Squeeze theorem, you are only expected to memorize them:

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

Their reciprocals are the same:

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

AP tip:

Know these standard trig limits well.

Examples

A common variant changes the argument of the sine function.

Evaluate

$x \to 0 lim \frac{sin ( 3 x )}{x}$

(spoiler)

In order to use the special trig limit, the denominator must match the argument of sine, or $3 x$ . The trick is to multiply by $1$ in the form of $\frac{3}{3}$ :

$= = = x \to 0 lim \frac{sin ( 3 x )}{x} \cdot \frac{3}{3} x \to 0 lim \frac{sin ( 3 x )}{3 x} \cdot 3 1 \cdot 3 3$

Evaluate

$x \to 0 lim \frac{sin ( 3 x )}{sin ( 4 x )}$

(spoiler)

Here, there is no $x$ outside the sine functions, so we create the special limit forms by multiplying by $\frac{3 x}{3 x}$ and $\frac{4 x}{4 x}$ .

Start by rewriting the expression as a product:

$x \to 0 lim \frac{sin ( 3 x )}{1} \times \frac{1}{sin ( 4 x )}$

Now create the special limit form:

$x \to 0 lim \frac{sin ( 3 x )}{1} \times \frac{3 x}{3 x} \times \frac{1}{sin ( 4 x )} \times \frac{4 x}{4 x}$

Rearrange the expression to form the special limits:

As a general shortcut,

$x \to 0 lim \frac{sin ( a x )}{b x} = \frac{a}{b}$

and

$x \to 0 lim \frac{a x}{sin ( b x )} = \frac{a}{b}$

Challenge problem

$x \to \infty lim x sin (\frac{2}{x}) =$

(spoiler)

Direct substitution gives the indeterminate form $\infty \times 0$ .

First, rewrite the expression as

$x \to \infty lim \frac{sin ( \frac{2}{x} )}{\frac{1}{x}}$

Next, make a substitution by letting $\frac{2}{x} = u$ . Then $\frac{1}{x} = \frac{1}{2} u$ .

As $x$ approaches $\infty, u$ approaches $0$ , and the limit rewritten in terms of $u$ is

$u \to 0 lim \frac{sin ( u )}{\frac{1}{2} u} = u \to 0 lim 2 \cdot \frac{sin ( u )}{u} = 2 \cdot 1 = 2$

Note that substitution doesn’t change the function. We’ve only changed the form it’s written in so that the special limit can be applied.

Growth of exponential and logarithmic functions

Exponential functions ( $e^{x}$ ) outgrow any polynomial as $x \to \infty$
Logarithmic functions ( $ln (x)$ ) grow slower than any power of $x$ as $x \to \infty$
For limits at infinity:
- If denominator grows faster (e.g., $e^{x}$ ), limit $\to 0$
- If numerator grows faster as denominator $\to 0$ , limit $\to \pm \infty$
- Compare dominant terms for large $x$

Squeeze theorem

Used to evaluate limits by bounding $f (x)$ between two functions with known limits
If $g (x) \leq f (x) \leq h (x)$ and $lim_{x \to a} g (x) = lim_{x \to a} h (x) = L$ , then $lim_{x \to a} f (x) = L$
Especially useful for oscillating functions and trigonometric limits

Special trigonometric limits

$lim_{x \to 0} \frac{s i n ( x )}{x} = 1$
$lim_{x \to 0} \frac{1 - c o s ( x )}{x} = 0$
For $lim_{x \to 0} \frac{s i n ( k x )}{x} = k$
- Use substitution or multiply/divide to match standard forms

Example strategies and challenge problems

Use substitution to rewrite limits in standard forms (e.g., $u = k x$ )
Factor denominators to isolate special limit forms
For limits like $\frac{s i n ( a x )}{s i n ( b x )}$ as $x \to 0$ , answer is $\frac{a}{b}$ by matching to $\frac{s i n ( a x )}{a x} \cdot \frac{b x}{s i n ( b x )}$

Special limits

What you’ll learn

Limits at infinity for exponential and logarithmic functions
Applying the Squeeze theorem to find limits of oscillating functions
Special trigonometric limits involving sine and cosine

Growth of exponential and logarithmic functions

Dominant term analysis can be extended to exponential and logarithmic functions by comparing their growth rates to algebraic functions (such as polynomials or roots).

As $x \to \infty$ , the growth rates of these functions can be ranked in the following order:

$Logarithmic < Algebraic < Exponential$

This comparison is specifically for large positive values of $x$ . In expressions containing a mix of these functions, the faster-growing function dominates the end behavior.

Example 1: $x \to \infty$

Evaluate:

a) $x \to \infty lim \frac{x ^{1000}}{e ^{x}}$

b) $x \to \infty lim \frac{x}{ln x}$

c) $x \to \infty lim \frac{2 ^{x}}{5 ^{x}}$

Solutions

a) $x \to \infty lim \frac{x ^{1000}}{e ^{x}}$

(spoiler)

Direct substitution gives the indeterminate form $\frac{\infty}{\infty}$ .

Even with the massive power of $x$ , the exponential function $e^{x}$ in the denominator eventually dominates the polynomial $x^{1000}$ in the numerator for large values of $x$ .

This behavior fits case 1 from the horizontal asymptotes section, and the limit is $0$ .

b) $x \to \infty lim \frac{x}{l n x}$

(spoiler)

Direct substitution gives $\frac{\infty}{\infty}$ .

Since $x$ , an algebraic function, grows faster than the natural logarithm $ln x$ , the numerator eventually dominates.

This behavior fits case 3 and the limit is $\infty$ .

c) $x \to \infty lim \frac{2 ^{x}}{5 ^{x}}$

(spoiler)

Both are exponential functions, but the denominator dominates because it has a larger base. Since this fits case 1, the limit is $0$ .

Alternatively, we can use exponent rules to rewrite the expression as a single function

$x \to \infty lim (\frac{2}{5})^{x}$

Because $0 < \frac{2}{5} < 1$ , repeated multiplication by $\frac{2}{5}$ makes the expression shrink toward $0$ . Therefore, the limit is $0$ .

Example 2: $x \to - \infty$

As $x$ approaches $- \infty$ , basic exponential functions such as $e^{x}$ approach $0$ instead, which leads to different end behavior.

Evaluate:

$x \to - \infty lim \frac{x + 1}{e ^{x}}$

(spoiler)

As $x \to - \infty$ , the pieces behave like this:

Numerator: $(x + 1) \to - \infty$
Denominator: $e^{x} \to 0^{+}$ (a very small positive number)

Therefore,

$x \to - \infty lim \frac{x + 1}{e ^{x}} = - \infty$

This fits case 3: the numerator dominates because the denominator shrinks toward $0$ .

Let

$f (x) = \frac{2 ^{x} + 6 ^{x} + 1}{5 ^{x} - 2}$

Evaluate:

a) $x \to \infty lim f (x)$

b) $x \to - \infty lim f (x)$

Answers

(spoiler)

a) $\infty$
b) $- \frac{1}{2}$

Solutions

a) $x \to \infty lim f (x)$

(spoiler)

As $x \to \infty$ :

In the numerator, $6^{x}$ dominates $2^{x}$ and the constant $1$ .
In the denominator, $5^{x}$ dominates the constant $- 2$ .

So for large $x$ , the expression can be reduced to

$x \to \infty lim f (x) \approx x \to \infty lim \frac{6 ^{x}}{5 ^{x}} = x \to \infty lim (\frac{6}{5})^{x}$

This fits case 3. You may also consider that because the base ( $\frac{6}{5}$ ) is greater than $1$ , repeated multiplication causes the expression to grow without bound, and the limit is $\infty$ .

b) $x \to - \infty lim f (x)$

(spoiler)

As $x \to - \infty$ , the exponential functions $2^{x}, 6^{x},$ and $5^{x}$ all approach $0$ . Substituting those end behaviors into the expression gives

$x \to - \infty lim \frac{2 ^{x} + 6 ^{x} + 1}{5 ^{x} - 2} = \frac{0 + 0 + 1}{0 - 2} = - \frac{1}{2}$

Squeeze theorem

Also called the sandwich theorem, the Squeeze theorem helps you evaluate limits by trapping a function between two others whose limits are already known.

Squeeze theorem:

If $g (x) \leq f (x) \leq h (x)$ for all $x$ near $a$ , and

$x \to a lim g (x) = x \to a lim h (x) = L,$

then

$x \to a lim f (x) = L$

Example

Let $g$ and $h$ be the functions defined by $g (x) = 2 x^{2} + 4 x + 1$ and $h (x) = x^{3} - 3 x + 8$ .

If $f$ is a function such that $g (x) \leq f (x) \leq h (x)$ for all $x > 0$ , find $x \to 1 lim f (x)$ .

(spoiler)

Since $f (x)$ is squeezed between the bounding functions $g (x)$ and $h (x)$ as $x \to 1$ , evaluate the limits of the bounds:

$x \to 1 lim g (x) = x \to 1 lim (2 x^{2} + 4 x + 1) = 7$

and

$x \to 1 lim h (x) = x \to 1 lim (x^{3} - 3 x + 8) = 7$

Because the two limits are equal, $x \to 1 lim f (x) = 7$ by the Squeeze theorem.

Bounded functions at infinity

Consider the limit

$x \to \infty lim \frac{sin x}{x}$

Trigonometric functions like $sin x$ and $cos x$ oscillate indefinitely between $- 1$ and $1$ , so direct substitution does not result in an indeterminate form.

Since the denominator grows without bound while the numerator remains bounded, dominant term analysis suggests that the fraction approaches $0$ .

This can be proven using the Squeeze theorem. Since $sin x$ is bounded for all $x$ ,

$- 1 \leq sin x \leq 1$

Dividing each part of the inequality by $x$ gives:

$- \frac{1}{x} \leq \frac{sin x}{x} \leq \frac{1}{x}$

The limits of both bounding functions equal $0$ as $x$ approaches $\infty$ . Therefore, by the Squeeze theorem,

$x \to \infty lim \frac{sin x}{x} = 0$

Special trigonometric limits

If $x$ approaches $0$ instead of $\infty$ , the previous approach no longer applies. Although the following limits can be proven using the Squeeze theorem, you are only expected to memorize them:

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

Their reciprocals are the same:

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

AP tip:

Know these standard trig limits well.

Examples

A common variant changes the argument of the sine function.

Evaluate

$x \to 0 lim \frac{sin ( 3 x )}{x}$

(spoiler)

In order to use the special trig limit, the denominator must match the argument of sine, or $3 x$ . The trick is to multiply by $1$ in the form of $\frac{3}{3}$ :

$= = = x \to 0 lim \frac{sin ( 3 x )}{x} \cdot \frac{3}{3} x \to 0 lim \frac{sin ( 3 x )}{3 x} \cdot 3 1 \cdot 3 3$

Evaluate

$x \to 0 lim \frac{sin ( 3 x )}{sin ( 4 x )}$

(spoiler)

Here, there is no $x$ outside the sine functions, so we create the special limit forms by multiplying by $\frac{3 x}{3 x}$ and $\frac{4 x}{4 x}$ .

Start by rewriting the expression as a product:

$x \to 0 lim \frac{sin ( 3 x )}{1} \times \frac{1}{sin ( 4 x )}$

Now create the special limit form:

$x \to 0 lim \frac{sin ( 3 x )}{1} \times \frac{3 x}{3 x} \times \frac{1}{sin ( 4 x )} \times \frac{4 x}{4 x}$

Rearrange the expression to form the special limits:

As a general shortcut,

$x \to 0 lim \frac{sin ( a x )}{b x} = \frac{a}{b}$

and

$x \to 0 lim \frac{a x}{sin ( b x )} = \frac{a}{b}$

Challenge problem

$x \to \infty lim x sin (\frac{2}{x}) =$

(spoiler)

Direct substitution gives the indeterminate form $\infty \times 0$ .

First, rewrite the expression as

$x \to \infty lim \frac{sin ( \frac{2}{x} )}{\frac{1}{x}}$

Next, make a substitution by letting $\frac{2}{x} = u$ . Then $\frac{1}{x} = \frac{1}{2} u$ .

As $x$ approaches $\infty, u$ approaches $0$ , and the limit rewritten in terms of $u$ is

$u \to 0 lim \frac{sin ( u )}{\frac{1}{2} u} = u \to 0 lim 2 \cdot \frac{sin ( u )}{u} = 2 \cdot 1 = 2$

Note that substitution doesn’t change the function. We’ve only changed the form it’s written in so that the special limit can be applied.

Special limits

What you’ll learn

Growth of exponential and logarithmic functions

Example 1: x→∞

Solutions

Example 2: x→−∞

Answers

Solutions

Squeeze theorem

Example

Bounded functions at infinity

Special trigonometric limits

Examples

Challenge problem

Special limits

What you’ll learn

Growth of exponential and logarithmic functions

Example 1: x→∞

Solutions

Example 2: x→−∞

Answers

Solutions

Squeeze theorem

Example

Bounded functions at infinity

Special trigonometric limits

Examples

Challenge problem

Special limits

What you’ll learn

Growth of exponential and logarithmic functions

Example 1: x→∞

Solutions

Example 2: x→−∞

Answers

Solutions

Squeeze theorem

Example

Bounded functions at infinity

Special trigonometric limits

Examples

Challenge problem

Special limits

What you’ll learn

Growth of exponential and logarithmic functions

Example 1: x→∞

Solutions

Example 2: x→−∞

Answers

Solutions

Squeeze theorem

Example

Bounded functions at infinity

Special trigonometric limits

Examples

Challenge problem

Example 1: $x \to \infty$

Example 2: $x \to - \infty$

Example 1: $x \to \infty$

Example 2: $x \to - \infty$

Example 1: $x \to \infty$

Example 2: $x \to - \infty$

Example 1: $x \to \infty$

Example 2: $x \to - \infty$