Achievable AP Calculus AB

1. Limits

Our AP Calculus AB course is currently in development and is a work-in-progress.

Special limits

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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 for evaluating limits at infinity. You can use the same idea with 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, while logarithmic functions like $ln (x)$ increase more slowly than any power of $x$ .

Examples

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

Solution

(spoiler)

Answer: $0$

Direct substitution gives the indeterminate form $\frac{\infty}{\infty}$ , so we need to compare growth rates.

As $x \to \infty$ , the exponential $e^{x}$ grows faster than the polynomial $x^{3}$ . That means the denominator dominates, and the fraction gets closer and closer to $0$ .

This matches a case 1 situation from the horizontal asymptotes section: the denominator dominates, so the limit is $0$ .

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

Solution

(spoiler)

Answer: $- \infty$

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

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

So we have a very large negative number divided by a very small positive number, which becomes unbounded in the negative direction. Therefore,

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

This matches a case 3 situation: the numerator dominates because the denominator shrinks toward $0$ .

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)

As $x \to \infty$ :

In the numerator, $4^{x}$ dominates $2^{x}$ and the constant $1$ .
In the denominator, $5^{x}$ dominates the constant $- 2$ .

So the limit behaves like

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

Because $0 < \frac{4}{5} < 1$ , repeated multiplication by $\frac{4}{5}$ makes the expression shrink toward $0$ . Therefore, the limit is $0$ .

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

(spoiler)

As $x \to - \infty$ , we have $2^{x} \to 0$ , $4^{x} \to 0$ , and $5^{x} \to 0$ . Substituting those end behaviors into the expression gives

$x \to - \infty lim \frac{2 ^{x} + 4 ^{x} + 1}{5 ^{x} - 2} = \frac{0 + 0 + 1}{0 - 2} = - \frac{1}{2} .$

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

(spoiler)

Answer: $0$

Rewrite $x$ as $x^{1/2}$ . Any power of $x$ (like $x^{1/2}$ ) grows faster than $ln (x)$ as $x \to \infty$ . So the denominator dominates and the fraction approaches $0$ .

This is another case 1 situation: the denominator dominates, so the limit is $0$ .

Squeeze theorem

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

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 $a (x) \leq f (x) \leq b (x)$ near $x = 1$ , the Squeeze theorem says we just need to check whether $a (x)$ and $b (x)$ approach the same value as $x \to 1$ .

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

$x \to 1 lim (tan (π (x + 1)) - 1)$

$= tan (π (1 + 1)) - 1$

$= tan (2 π) - 1$

$= - 1$

$x \to 1 lim b (x)$ :

$x \to 1 lim (2 - ∣ x + 2∣)$

$= 2 - ∣1 + 2∣$

$= - 1$

Since both limits equal $- 1$ , and $f (x)$ stays between $a (x)$ and $b (x)$ , the limit is forced to be

$x \to 1 lim f (x) = - 1 .$

The Squeeze theorem is especially useful for limits involving oscillating trigonometric expressions.

For example, consider

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

Direct substitution would involve $cos (\frac{1}{0})$ , which is undefined, so we need a different approach.

A cosine function with amplitude 1 always stays between $- 1$ and $1$ :

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

Multiply all three parts by $x$ :

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

As $x \to 0$ , both bounding functions go to $0$ :

$x \to 0 lim (- x) = 0 and x \to 0 lim x = 0.$

So by the Squeeze theorem,

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

Special trigonometric limits

There are two standard trigonometric limits (often proved using the Squeeze theorem). In both cases, direct substitution gives 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 changes the argument of the sine function. For example,

Evaluate

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

Solution

Using the substitution $u = 3 x$ rewrites the expression so it matches the standard form $\frac{sin ( u )}{u}$ .

From $u = 3 x$ , we have $x = \frac{u}{3}$ . Also, as $x \to 0$ , we have $u \to 0$ .

Rewrite the limit:

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

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

$= 1 \times 3$

$= 3$

Give it a try:

Evaluate

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

Solution

(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 multiply by 1 in two helpful ways:

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

Group terms to form the standard limits:

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

Now apply the special limit $lim_{x \to 0} \frac{s i n ( k x )}{k x} = 1$ :

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

Challenge problems

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 satisfies

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

which is $- u$ .

Then the limit becomes

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

Evaluate

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

Solution

(spoiler)

First, factor the denominator:

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

Now separate out a $\frac{sin ( 2 x )}{2 x}$ factor:

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

Apply the special limit and then substitute $x = 0$ into the remaining factor:

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

$= \frac{1}{4}$

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 tricky or oscillating functions
Special trigonometric limits involving sine and cosine

Growth of exponential and logarithmic functions

Examples

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

Solution

(spoiler)

Answer: $0$

Direct substitution gives the indeterminate form $\frac{\infty}{\infty}$ , so we need to compare growth rates.

As $x \to \infty$ , the exponential $e^{x}$ grows faster than the polynomial $x^{3}$ . That means the denominator dominates, and the fraction gets closer and closer to $0$ .

This matches a case 1 situation from the horizontal asymptotes section: the denominator dominates, so the limit is $0$ .

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

Solution

(spoiler)

Answer: $- \infty$

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

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

So we have a very large negative number divided by a very small positive number, which becomes unbounded in the negative direction. Therefore,

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

This matches a case 3 situation: the numerator dominates because the denominator shrinks toward $0$ .

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)

As $x \to \infty$ :

In the numerator, $4^{x}$ dominates $2^{x}$ and the constant $1$ .
In the denominator, $5^{x}$ dominates the constant $- 2$ .

So the limit behaves like

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

Because $0 < \frac{4}{5} < 1$ , repeated multiplication by $\frac{4}{5}$ makes the expression shrink toward $0$ . Therefore, the limit is $0$ .

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

(spoiler)

As $x \to - \infty$ , we have $2^{x} \to 0$ , $4^{x} \to 0$ , and $5^{x} \to 0$ . Substituting those end behaviors into the expression gives

$x \to - \infty lim \frac{2 ^{x} + 4 ^{x} + 1}{5 ^{x} - 2} = \frac{0 + 0 + 1}{0 - 2} = - \frac{1}{2} .$

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

(spoiler)

Answer: $0$

Rewrite $x$ as $x^{1/2}$ . Any power of $x$ (like $x^{1/2}$ ) grows faster than $ln (x)$ as $x \to \infty$ . So the denominator dominates and the fraction approaches $0$ .

This is another case 1 situation: the denominator dominates, so the limit is $0$ .

Squeeze theorem

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

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 $a (x) \leq f (x) \leq b (x)$ near $x = 1$ , the Squeeze theorem says we just need to check whether $a (x)$ and $b (x)$ approach the same value as $x \to 1$ .

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

$x \to 1 lim (tan (π (x + 1)) - 1)$

$= tan (π (1 + 1)) - 1$

$= tan (2 π) - 1$

$= - 1$

$x \to 1 lim b (x)$ :

$x \to 1 lim (2 - ∣ x + 2∣)$

$= 2 - ∣1 + 2∣$

$= - 1$

Since both limits equal $- 1$ , and $f (x)$ stays between $a (x)$ and $b (x)$ , the limit is forced to be

$x \to 1 lim f (x) = - 1 .$

The Squeeze theorem is especially useful for limits involving oscillating trigonometric expressions.

For example, consider

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

Direct substitution would involve $cos (\frac{1}{0})$ , which is undefined, so we need a different approach.

A cosine function with amplitude 1 always stays between $- 1$ and $1$ :

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

Multiply all three parts by $x$ :

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

As $x \to 0$ , both bounding functions go to $0$ :

$x \to 0 lim (- x) = 0 and x \to 0 lim x = 0.$

So by the Squeeze theorem,

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

Special trigonometric limits

There are two standard trigonometric limits (often proved using the Squeeze theorem). In both cases, direct substitution gives 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 changes the argument of the sine function. For example,

Evaluate

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

Solution

Using the substitution $u = 3 x$ rewrites the expression so it matches the standard form $\frac{sin ( u )}{u}$ .

From $u = 3 x$ , we have $x = \frac{u}{3}$ . Also, as $x \to 0$ , we have $u \to 0$ .

Rewrite the limit:

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

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

$= 1 \times 3$

$= 3$

Give it a try:

Evaluate

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

Solution

(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 multiply by 1 in two helpful ways:

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

Group terms to form the standard limits:

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

Now apply the special limit $lim_{x \to 0} \frac{s i n ( k x )}{k x} = 1$ :

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

Challenge problems

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 satisfies

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

which is $- u$ .

Then the limit becomes

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

Evaluate

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

Solution

(spoiler)

First, factor the denominator:

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

Now separate out a $\frac{sin ( 2 x )}{2 x}$ factor:

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

Apply the special limit and then substitute $x = 0$ into the remaining factor:

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

$= \frac{1}{4}$

Key points

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 $\frac{s i n ( a x )}{a x} \cdot \frac{b x}{s i n ( b x )}$

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

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

Solution

(spoiler)

Answer: $0$

Direct substitution gives the indeterminate form $\frac{\infty}{\infty}$ , so we need to compare growth rates.

As $x \to \infty$ , the exponential $e^{x}$ grows faster than the polynomial $x^{3}$ . That means the denominator dominates, and the fraction gets closer and closer to $0$ .

This matches a case 1 situation from the horizontal asymptotes section: the denominator dominates, so the limit is $0$ .

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

Solution

(spoiler)

Answer: $- \infty$

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

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

So we have a very large negative number divided by a very small positive number, which becomes unbounded in the negative direction. Therefore,

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

This matches a case 3 situation: the numerator dominates because the denominator shrinks toward $0$ .

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)

As $x \to \infty$ :

In the numerator, $4^{x}$ dominates $2^{x}$ and the constant $1$ .
In the denominator, $5^{x}$ dominates the constant $- 2$ .

So the limit behaves like

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

Because $0 < \frac{4}{5} < 1$ , repeated multiplication by $\frac{4}{5}$ makes the expression shrink toward $0$ . Therefore, the limit is $0$ .

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

(spoiler)

As $x \to - \infty$ , we have $2^{x} \to 0$ , $4^{x} \to 0$ , and $5^{x} \to 0$ . Substituting those end behaviors into the expression gives

$x \to - \infty lim \frac{2 ^{x} + 4 ^{x} + 1}{5 ^{x} - 2} = \frac{0 + 0 + 1}{0 - 2} = - \frac{1}{2} .$

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

(spoiler)

Answer: $0$

Rewrite $x$ as $x^{1/2}$ . Any power of $x$ (like $x^{1/2}$ ) grows faster than $ln (x)$ as $x \to \infty$ . So the denominator dominates and the fraction approaches $0$ .

This is another case 1 situation: the denominator dominates, so the limit is $0$ .

Squeeze theorem

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

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 $a (x) \leq f (x) \leq b (x)$ near $x = 1$ , the Squeeze theorem says we just need to check whether $a (x)$ and $b (x)$ approach the same value as $x \to 1$ .

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

$x \to 1 lim (tan (π (x + 1)) - 1)$

$= tan (π (1 + 1)) - 1$

$= tan (2 π) - 1$

$= - 1$

$x \to 1 lim b (x)$ :

$x \to 1 lim (2 - ∣ x + 2∣)$

$= 2 - ∣1 + 2∣$

$= - 1$

Since both limits equal $- 1$ , and $f (x)$ stays between $a (x)$ and $b (x)$ , the limit is forced to be

$x \to 1 lim f (x) = - 1 .$

The Squeeze theorem is especially useful for limits involving oscillating trigonometric expressions.

For example, consider

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

Direct substitution would involve $cos (\frac{1}{0})$ , which is undefined, so we need a different approach.

A cosine function with amplitude 1 always stays between $- 1$ and $1$ :

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

Multiply all three parts by $x$ :

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

As $x \to 0$ , both bounding functions go to $0$ :

$x \to 0 lim (- x) = 0 and x \to 0 lim x = 0.$

So by the Squeeze theorem,

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

Special trigonometric limits

There are two standard trigonometric limits (often proved using the Squeeze theorem). In both cases, direct substitution gives 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 changes the argument of the sine function. For example,

Evaluate

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

Solution

Using the substitution $u = 3 x$ rewrites the expression so it matches the standard form $\frac{sin ( u )}{u}$ .

From $u = 3 x$ , we have $x = \frac{u}{3}$ . Also, as $x \to 0$ , we have $u \to 0$ .

Rewrite the limit:

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

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

$= 1 \times 3$

$= 3$

Give it a try:

Evaluate

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

Solution

(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 multiply by 1 in two helpful ways:

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

Group terms to form the standard limits:

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

Now apply the special limit $lim_{x \to 0} \frac{s i n ( k x )}{k x} = 1$ :

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

Challenge problems

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 satisfies

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

which is $- u$ .

Then the limit becomes

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

Evaluate

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

Solution

(spoiler)

First, factor the denominator:

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

Now separate out a $\frac{sin ( 2 x )}{2 x}$ factor:

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

Apply the special limit and then substitute $x = 0$ into the remaining factor:

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

$= \frac{1}{4}$