What you’ll learn:

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

Growth of exponential and logarithmic functions

The previous page introduced dominant term analysis as a technique to evaluate limits at infinity. This approach can also be applied to exponential and logarithmic functions by comparing their growth rates to those of polynomials.

As you move from left to right on their graphs, exponential functions like $e^{x}$ grow faster than any polynomial (visualize the graph), while logarithmic functions increase more slowly than any power of $x$ .

Examples

1. Evaluate $x \to \infty lim \frac{x ^{3}}{e ^{x}}$ .

Solution

(spoiler)

Answer: $0$

Direct substitution results in the indeterminate form $\frac{\infty}{\infty}$ so we need to analyze the behavior of the function.

Because exponential functions grow at a faster rate than polynomials, the $e^{x}$ in the denominator dominates and the numerator is negligible as $x \to \infty$ .

This is like a case 1 situation from the horizontal asymptotes section on the previous page, and the entire function tends toward $0$ the bigger $x$ gets.

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

Solution

(spoiler)

Answer: $- \infty$

As $x$ approaches $- \infty$ , $e^{x}$ approaches 0 so $e^{x}$ does not dominate in this case. Though not entirely proper, this trick of “plugging in” $- \infty$ for $x$ can help with understanding each piece more clearly:

$(x + 1) \to (- \infty + 1)$ :
- Approaches $- \infty$
$e^{x} \to e^{- \infty} = \frac{1}{e ^{\infty}}$ :
- Approaches $0^{+}$ (very small positive number)

$(- \infty)$ divided by a very small positive number results in $- \infty$ . This is like a case 3 situation where the numerator dominates.

3. Let

$f (x) = \frac{2 ^{x} + 4 ^{x} + 1}{5 ^{x} - 2}$

Evaluate:

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

Answers

(spoiler)

a) $0$
b) $- 1/2$

Solutions

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

(spoiler)

In the numerator, $4^{x}$ dominates (grows faster than $2^{x}$ ).
In the denominator, $5^{x}$ dominates.

Then

$x \to \infty lim f (x) \approx x \to \infty lim \frac{4 ^{x}}{5 ^{x}}$

$= x \to \infty lim (\frac{4}{5})^{x}$

But as $x \to \infty$ , the fraction $\frac{4}{5}$ is continuously multiplied by itself, and the overall value is decreasing. Therefore. the limit is $0$ .

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

(spoiler)

This time, neither $4^{x}$ nor $5^{x}$ dominate because as $x \to - \infty$ , both of those approach 0, as does $2^{x}$ . Then the expression can be reduced to

$x \to \infty lim \frac{0 - 0 + 1}{0 - 2}$

$= - \frac{1}{2}$

4. Find $x \to \infty lim \frac{ln ( x )}{x}$

(spoiler)

Answer: $0$

$x = x^{1/2}$ is a power function and still grows more quickly than $ln (x)$ . For large $x$ , the denominator dominates and $ln (x)$ is negligible. This is like a case 1 situation and the function shrinks to 0.

Squeeze theorem

Also known as the sandwich theorem, this theorem helps evaluate limits that seem complicated at first glance by “squeezing” the function between two others with known limits. The theorem is:

Squeeze theorem

If, for all $x$ near $a$ ,

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

and

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

then

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

Example

Suppose $a (x) = tan (π (x + 1)) - 1$ and $b (x) = 2 - ∣ x + 2∣$ .

If $f (x)$ is a function such that $a (x) \leq f (x) \leq b (x)$ for $0 \leq x \leq 2$ , find $x \to 1 lim f (x)$ .

Solution

(spoiler)

Because we’re given that $f (x)$ is “squeezed” between $a (x)$ and $b (x)$ for all $x$ near the point of interest $x = 1$ , we just need to find if $x \to 1 lim a (x)$ and $x \to 1 lim b (x)$ both equal the same value.

$lim_{x \to 1} a (x)$ :

$x \to 1 lim tan (π (x + 1)) - 1$

$= tan (2 π) - 1$

$= - 1$

$x \to 1 lim b (x)$

$lim x \to 1 2 - ∣ x + 2∣$

$= 2 - ∣1 + 2∣$

$= - 1$

Since $x \to 1 lim a (x) = x \to 1 lim b (x) = - 1$ , and $f (x)$ is squeezed between them, $x \to 1 lim f (x)$ is “forced” to be $- 1$ .

The Squeeze theorem is particularly useful for limits involving more complicated trigonometric functions.

For example, to find $x \to 0 lim x cos (\frac{1}{x})$ , notice that direct substitution results in $cos (\frac{1}{0})$ as a component (function is undefined at $x = 0$ ) and the previous techniques covered don’t suffice to resolve this.

Instead, use the fact that the output values of a cosine function of amplitude 1 are always squeezed between -1 and 1:

$- 1 \leq cos (\frac{1}{x}) \leq 1$

Then multiplying by $x$ ,

$- x \leq x cos (\frac{1}{x}) \leq x$

Since $x \to 0 lim (- x) = 0$ and $x \to 0 lim x = 0$ , by the Squeeze Theorem:

$x \to 0 lim x cos (\frac{1}{x}) = 0$

Special trigonometric limits

There are two special trig limits that can be proved using the squeeze theorem (although the proof won’t be shown here) since direct substitution results in the indeterminate form $\frac{0}{0}$ .

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

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

AP tip:

Know these two standard trig limits well.

A common variation of these involves a change in the argument of the sine function. For example,

1. Evaluate

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

Solution

Using the substitution $u = 3 x$ will help you rewrite the expression in a form that more clearly mirrors $\frac{sin ( x )}{x}$ . It’s important to note that from the substitution, $x = \frac{u}{3}$ and as $x \to 0$ , $u \to 0$ as well. Then the limit can be rewritten as:

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

$= u \to 0 lim \frac{sin ( u )}{u} \times 3$

$= 1 \times 3$

$= 3$

Give it a try:

2. Evaluate

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

Solution

(spoiler)

This one is different because there is no variable $x$ outside of the sine arguments. Instead, we’ll have to create our own. The limit can be separated into

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

With a bit of algebraic manipulation (multiplying by a giant one), we can turn each into the special limit:

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

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

$= 1 \times 1 \times \frac{3}{4} = \frac{3}{4}$

Challenge problems

1. Evaluate

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

Solution

(spoiler)

Using substitution, let the argument $2 - x = u .$

As $x \to 2, u \to 0$ .

Notice that the denominator

$(x - 2) = - (2 - x)$

which is $- u$ .

Then the limit becomes

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

2. Evaluate

$x \to 0 lim \frac{sin ( 2 x )}{6 x ^{2} + 4 x}$

Solution

(spoiler)

First, factor the denominator:

$x \to 0 lim \frac{sin ( 2 x )}{2 x ( 3 x + 4 )}$

Separate into $\frac{sin ( u )}{u}$ form:

$= x \to 0 lim \frac{sin ( 2 x )}{2 x} \cdot \frac{1}{3 x + 4}$

$= 1 \cdot \frac{1}{3 ( 0 ) + 4}$

$= \frac{1}{4}$

What you’ll learn:

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

Growth of exponential and logarithmic functions

Examples

1. Evaluate $x \to \infty lim \frac{x ^{3}}{e ^{x}}$ .

Solution

(spoiler)

Answer: $0$

Direct substitution results in the indeterminate form $\frac{\infty}{\infty}$ so we need to analyze the behavior of the function.

Because exponential functions grow at a faster rate than polynomials, the $e^{x}$ in the denominator dominates and the numerator is negligible as $x \to \infty$ .

This is like a case 1 situation from the horizontal asymptotes section on the previous page, and the entire function tends toward $0$ the bigger $x$ gets.

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

Solution

(spoiler)

Answer: $- \infty$

$(x + 1) \to (- \infty + 1)$ :
- Approaches $- \infty$
$e^{x} \to e^{- \infty} = \frac{1}{e ^{\infty}}$ :
- Approaches $0^{+}$ (very small positive number)

$(- \infty)$ divided by a very small positive number results in $- \infty$ . This is like a case 3 situation where the numerator dominates.

3. Let

$f (x) = \frac{2 ^{x} + 4 ^{x} + 1}{5 ^{x} - 2}$

Evaluate:

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

Answers

(spoiler)

a) $0$
b) $- 1/2$

Solutions

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

(spoiler)

In the numerator, $4^{x}$ dominates (grows faster than $2^{x}$ ).
In the denominator, $5^{x}$ dominates.

Then

$x \to \infty lim f (x) \approx x \to \infty lim \frac{4 ^{x}}{5 ^{x}}$

$= x \to \infty lim (\frac{4}{5})^{x}$

But as $x \to \infty$ , the fraction $\frac{4}{5}$ is continuously multiplied by itself, and the overall value is decreasing. Therefore. the limit is $0$ .

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

(spoiler)

This time, neither $4^{x}$ nor $5^{x}$ dominate because as $x \to - \infty$ , both of those approach 0, as does $2^{x}$ . Then the expression can be reduced to

$x \to \infty lim \frac{0 - 0 + 1}{0 - 2}$

$= - \frac{1}{2}$

4. Find $x \to \infty lim \frac{ln ( x )}{x}$

(spoiler)

Answer: $0$

Squeeze theorem

Also known as the sandwich theorem, this theorem helps evaluate limits that seem complicated at first glance by “squeezing” the function between two others with known limits. The theorem is:

Squeeze theorem

If, for all $x$ near $a$ ,

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

and

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

then

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

Example

Suppose $a (x) = tan (π (x + 1)) - 1$ and $b (x) = 2 - ∣ x + 2∣$ .

If $f (x)$ is a function such that $a (x) \leq f (x) \leq b (x)$ for $0 \leq x \leq 2$ , find $x \to 1 lim f (x)$ .

Solution

(spoiler)

$lim_{x \to 1} a (x)$ :

$x \to 1 lim tan (π (x + 1)) - 1$

$= tan (2 π) - 1$

$= - 1$

$x \to 1 lim b (x)$

$lim x \to 1 2 - ∣ x + 2∣$

$= 2 - ∣1 + 2∣$

$= - 1$

Since $x \to 1 lim a (x) = x \to 1 lim b (x) = - 1$ , and $f (x)$ is squeezed between them, $x \to 1 lim f (x)$ is “forced” to be $- 1$ .

The Squeeze theorem is particularly useful for limits involving more complicated trigonometric functions.

Instead, use the fact that the output values of a cosine function of amplitude 1 are always squeezed between -1 and 1:

$- 1 \leq cos (\frac{1}{x}) \leq 1$

Then multiplying by $x$ ,

$- x \leq x cos (\frac{1}{x}) \leq x$

Since $x \to 0 lim (- x) = 0$ and $x \to 0 lim x = 0$ , by the Squeeze Theorem:

$x \to 0 lim x cos (\frac{1}{x}) = 0$

Special trigonometric limits

There are two special trig limits that can be proved using the squeeze theorem (although the proof won’t be shown here) since direct substitution results in the indeterminate form $\frac{0}{0}$ .

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

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

AP tip:

Know these two standard trig limits well.

A common variation of these involves a change in the argument of the sine function. For example,

1. Evaluate

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

Solution

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

$= u \to 0 lim \frac{sin ( u )}{u} \times 3$

$= 1 \times 3$

$= 3$

Give it a try:

2. Evaluate

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

Solution

(spoiler)

This one is different because there is no variable $x$ outside of the sine arguments. Instead, we’ll have to create our own. The limit can be separated into

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

With a bit of algebraic manipulation (multiplying by a giant one), we can turn each into the special limit:

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

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

$= 1 \times 1 \times \frac{3}{4} = \frac{3}{4}$

Challenge problems

1. Evaluate

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

Solution

(spoiler)

Using substitution, let the argument $2 - x = u .$

As $x \to 2, u \to 0$ .

Notice that the denominator

$(x - 2) = - (2 - x)$

which is $- u$ .

Then the limit becomes

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

2. Evaluate

$x \to 0 lim \frac{sin ( 2 x )}{6 x ^{2} + 4 x}$

Solution

(spoiler)

First, factor the denominator:

$x \to 0 lim \frac{sin ( 2 x )}{2 x ( 3 x + 4 )}$

Separate into $\frac{sin ( u )}{u}$ form:

$= x \to 0 lim \frac{sin ( 2 x )}{2 x} \cdot \frac{1}{3 x + 4}$

$= 1 \cdot \frac{1}{3 ( 0 ) + 4}$

$= \frac{1}{4}$

Special limits

What you’ll learn:

Growth of exponential and logarithmic functions

Examples

Solution

Solution

Answers

Solutions

Squeeze theorem

Example

Solution

Special trigonometric limits

Solution

Solution

Challenge problems

Solution

Solution

Special limits

What you’ll learn:

Growth of exponential and logarithmic functions

Examples

Solution

Solution

Answers

Solutions

Squeeze theorem

Example

Solution

Special trigonometric limits

Solution

Solution

Challenge problems

Solution

Solution