4.5 L'Hopital's rule

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4. Contextual uses

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L'Hopital's rule

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What you’ll learn

How to apply L’Hopital’s rule on limits with indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$
How to handle other indeterminate forms by rewriting or using logarithms

Another application of derivatives is evaluating limits that produce the indeterminate forms $\frac{0}{0}$ or $\frac{\infty}{\infty}$ .

In these cases, L’Hopital’s rule lets you replace the original quotient with a new one formed by differentiating the numerator and denominator separately, then trying direct substitution again. This process can be repeated until direct substitution no longer gives $\frac{0}{0}$ or $\frac{\infty}{\infty}$ .

L’Hopital’s rule:

$x \to a lim f (x) = x \to a lim g (x) = 0 or x \to a lim f (x) = x \to a lim g (x) = \infty$

then

$x \to a lim \frac{f ( x )}{g ( x )} = x \to a lim \frac{f ^{'} ( x )}{g ^{'} ( x )}$

Sidenote

Not the quotient rule

Don’t use the quotient rule. With L’Hopital’s rule, differentiate the numerator and denominator separately to form the new quotient and re-check the limit.

1. Form $0/0$

Example 1. Use L’Hopital’s rule to evaluate

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

(spoiler)

Recall the special trig limit $x \to 0 lim \frac{sin ( x )}{x} = 1$ .

Confirm this using L’Hopital’s rule:

The derivative of the numerator, $sin (x)$ , is $cos (x)$ .
The derivative of the denominator, $x$ , is $1$ .

Therefore,

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

Now direct substitution works, and the limit is $1$ .

Example 2. Evaluate

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

(spoiler)

Applying L’Hopital’s rule,

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

Example 3. Evaluate

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

(spoiler)

Applying L’Hopital’s rule,

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

2. Form $\infty/\infty$

L’Hopital’s rule can also be applied when direct substitution results in the indeterminate form $\frac{\infty}{\infty}$ .

Example 1. Evaluate

$x \to \infty lim \frac{x ^{3} + 5}{e ^{x}}$

(spoiler)

$= x \to \infty lim \frac{3 x ^{2}}{e ^{x}}$

Direct substitution still gives $\frac{\infty}{\infty}$ , so apply L’Hopital’s rule again.

$= x \to \infty lim \frac{6 x}{e ^{x}}$

Applying it again,

$= x \to \infty lim \frac{6}{e ^{x}}$

Now the denominator grows without bound, so the limit is $0$ .

Example 2. Find

$x \to \infty lim \frac{ln ( e ^{x} + x )}{x}$

(spoiler)

Applying L’Hopital’s rule,

$= x \to \infty lim \frac{\frac{1}{e ^{x} + x} \cdot ( e ^{x} + 1 )}{1}$

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

Applying L’Hopital’s rule again until it’s no longer needed,

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

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

Since the $e^{x}$ terms cancel, the limit is $1$ .

3. Form $0 \times \infty$

L’Hopital’s rule applies only to quotients. If you have an indeterminate product $f (x) \times g (x)$ , rewrite it as a fraction, either as

$\frac{f ( x )}{1/ g ( x )} or \frac{g ( x )}{1/ f ( x )}$

Choose the version whose denominator is easier to differentiate or leads to a useful form.

Determine

$x \to 0^{+} lim x cot (x)$

(spoiler)

This can be rewritten as either

$x \to 0^{+} lim \frac{cot ( x )}{1/ x}$

$x \to 0^{+} lim \frac{x}{1/ cot ( x )}$

$= x \to 0^{+} lim \frac{x}{tan ( x )}$

Try applying L’Hopital’s rule to the first one:

$x \to 0^{+} lim \frac{cot ( x )}{1/ x}$

$x \to 0^{+} lim \frac{- csc ^{2} ( x )}{- 1/ x ^{2}}$

This approach doesn’t simplify the indeterminate behavior in a helpful way, because differentiating $1/ x$ keeps producing powers of $x$ in the denominator.

Applying L’Hopital’s rule to the second option:

$x \to 0^{+} lim \frac{x}{tan ( x )}$

$= x \to 0^{+} lim \frac{1}{sec ^{2} ( x )}$

$= 1$

4. Form $\infty - \infty$

$\infty - \infty$ is an indeterminate form because the limit is not necessarily $0$ . In this case, rewrite the expression as a single fraction or product. Common strategies include:

Combining with a common denominator
Rationalizing radicals with conjugates
Using properties of logarithms

Find

$x \to 1^{+} lim \frac{x}{x - 1} - \frac{1}{ln ( x )}$

(spoiler)

Combine into one rational expression.

$x \to 1^{+} lim \frac{x}{x - 1} \cdot \frac{ln ( x )}{ln ( x )} - \frac{1}{ln ( x )} \cdot \frac{x - 1}{x - 1}$

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

Direct substitution gives $\frac{0}{0}$ , so apply L’Hopital’s rule.

$= x \to 1^{+} lim \frac{x ( 1/ x ) + ln ( x ) - 1}{( x - 1 ) ( 1/ x ) + ln ( x )}$

$= x \to 1^{+} lim \frac{ln ( x )}{1 - ( 1/ x ) + ln ( x )}$

Direct substitution still gives $\frac{0}{0}$ , so apply L’Hopital’s rule again:

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

Sidenote

Infinity + infinity

$\infty + \infty$ is not an indeterminate form - the result is $\infty$ .

Similarly, $- \infty - \infty$ becomes $- \infty$ .

5. Forms $1^{\infty}, 0^{0}, \infty^{0}$

When direct substitution produces one of these indeterminate exponential forms, rewrite the expression using logarithms and the power property.

Example 1. Determine

$x \to \infty lim (1 + \frac{1}{x})^{x}$

Direct substitution gives $1^{\infty}$ . Let

$L ln (L) = x \to \infty lim (1 + \frac{1}{x})^{x} = ln [x \to \infty lim (1 + \frac{1}{x})^{x}] = x \to \infty lim ln (1 + \frac{1}{x})^{x} = x \to \infty lim x \cdot ln (1 + \frac{1}{x})$

Direct substitution here now gives $\infty \times 0$ , so rewrite as a quotient:

(spoiler)

$x \to \infty lim \frac{ln ( 1 + \frac{1}{x} )}{\frac{1}{x}}$

Now direct substitution gives $\frac{0}{0}$ , so apply L’Hopital’s rule.

$x \to \infty lim \frac{\frac{x}{x + 1} \cdot ( - \frac{1}{x ^{2}} )}{- \frac{1}{x ^{2}}}$

After canceling $- \frac{1}{x ^{2}}$ , the limit becomes

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

Remember that this value is for $ln (L)$ , and the goal is to find the limit $L$ .

$ln (L) L = 1 = e$

Therefore

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

AP tip:

Remember this limit definition of $e$ .

Example 2. Determine

$x \to 0^{+} lim x^{\frac{1}{l n ( x )}}$

(spoiler)

Direct substitution results in $0^{0}$ .

Let

$L = x \to 0^{+} lim x^{\frac{1}{l n ( x )}}$

Taking the natural log of both sides,

$ln (L) = ln (x \to 0^{+} lim x^{\frac{1}{l n ( x )}}) = x \to 0^{+} lim ln (x^{\frac{1}{l n ( x )}}) = x \to 0^{+} lim \frac{1}{ln ( x )} \cdot ln (x) = 1$

In this case, L’Hopital’s rule was not needed.

Since $ln (L) = 1$ , then $L = e$ , so

$x \to 0^{+} lim x^{\frac{1}{l n ( x )}} = e$

Example 3. Evaluate

$x \to \infty lim x^{1/ x}$

(spoiler)

Direct substitution gives $\infty^{0}$ . Let

$L = x \to \infty lim x^{1/ x}$

Then

$ln (L) = ln (x \to \infty lim x^{1/ x}) = x \to \infty lim \frac{1}{x} \cdot ln (x) = x \to \infty lim \frac{ln ( x )}{x}$

Direct substitution now gives $\frac{\infty}{\infty}$ . We can either apply L’Hopital’s rule or use dominant term analysis from section 1.5 to conclude that

$x \to \infty lim \frac{ln ( x )}{x} = 0$

Since $ln (L) = 0$ , then $L = 1$ , so

$x \to 0^{+} lim x^{x} = 1$

In the following example, direct substitution does not lead to an indeterminate form.

$x \to \infty lim (1 - \frac{1}{x})^{c o s (x)}$

(spoiler)

Because $cos (x)$ oscillates and does not approach a single value as $x \to \infty$ , this expression does not approach a single exponential form like $1^{\infty}$ . Therefore this limit does not exist.

Use L’Hopital’s rule only on quotients, when the indeterminate form is $\frac{0}{0}$ or $\pm \frac{\infty}{\infty}$ .
For the other indeterminate forms, here is a summary for what to do:

Form	What to do
$0 \cdot \infty$	Rewrite as fraction
$\infty - \infty$	Combine
$0^{0}, \infty^{0}, 1^{\infty}$	Use logarithms

L'Hopital's rule

What you’ll learn

How to apply L’Hopital’s rule on limits with indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$
How to handle other indeterminate forms by rewriting or using logarithms

Another application of derivatives is evaluating limits that produce the indeterminate forms $\frac{0}{0}$ or $\frac{\infty}{\infty}$ .

L’Hopital’s rule:

$x \to a lim f (x) = x \to a lim g (x) = 0 or x \to a lim f (x) = x \to a lim g (x) = \infty$

then

$x \to a lim \frac{f ( x )}{g ( x )} = x \to a lim \frac{f ^{'} ( x )}{g ^{'} ( x )}$

Sidenote

Not the quotient rule

Don’t use the quotient rule. With L’Hopital’s rule, differentiate the numerator and denominator separately to form the new quotient and re-check the limit.

1. Form $0/0$

Example 1. Use L’Hopital’s rule to evaluate

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

(spoiler)

Recall the special trig limit $x \to 0 lim \frac{sin ( x )}{x} = 1$ .

Confirm this using L’Hopital’s rule:

The derivative of the numerator, $sin (x)$ , is $cos (x)$ .
The derivative of the denominator, $x$ , is $1$ .

Therefore,

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

Now direct substitution works, and the limit is $1$ .

Example 2. Evaluate

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

(spoiler)

Applying L’Hopital’s rule,

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

Example 3. Evaluate

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

(spoiler)

Applying L’Hopital’s rule,

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

2. Form $\infty/\infty$

L’Hopital’s rule can also be applied when direct substitution results in the indeterminate form $\frac{\infty}{\infty}$ .

Example 1. Evaluate

$x \to \infty lim \frac{x ^{3} + 5}{e ^{x}}$

(spoiler)

$= x \to \infty lim \frac{3 x ^{2}}{e ^{x}}$

Direct substitution still gives $\frac{\infty}{\infty}$ , so apply L’Hopital’s rule again.

$= x \to \infty lim \frac{6 x}{e ^{x}}$

Applying it again,

$= x \to \infty lim \frac{6}{e ^{x}}$

Now the denominator grows without bound, so the limit is $0$ .

Example 2. Find

$x \to \infty lim \frac{ln ( e ^{x} + x )}{x}$

(spoiler)

Applying L’Hopital’s rule,

$= x \to \infty lim \frac{\frac{1}{e ^{x} + x} \cdot ( e ^{x} + 1 )}{1}$

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

Applying L’Hopital’s rule again until it’s no longer needed,

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

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

Since the $e^{x}$ terms cancel, the limit is $1$ .

3. Form $0 \times \infty$

L’Hopital’s rule applies only to quotients. If you have an indeterminate product $f (x) \times g (x)$ , rewrite it as a fraction, either as

$\frac{f ( x )}{1/ g ( x )} or \frac{g ( x )}{1/ f ( x )}$

Choose the version whose denominator is easier to differentiate or leads to a useful form.

Determine

$x \to 0^{+} lim x cot (x)$

(spoiler)

This can be rewritten as either

$x \to 0^{+} lim \frac{cot ( x )}{1/ x}$

$x \to 0^{+} lim \frac{x}{1/ cot ( x )}$

$= x \to 0^{+} lim \frac{x}{tan ( x )}$

Try applying L’Hopital’s rule to the first one:

$x \to 0^{+} lim \frac{cot ( x )}{1/ x}$

$x \to 0^{+} lim \frac{- csc ^{2} ( x )}{- 1/ x ^{2}}$

This approach doesn’t simplify the indeterminate behavior in a helpful way, because differentiating $1/ x$ keeps producing powers of $x$ in the denominator.

Applying L’Hopital’s rule to the second option:

$x \to 0^{+} lim \frac{x}{tan ( x )}$

$= x \to 0^{+} lim \frac{1}{sec ^{2} ( x )}$

$= 1$

4. Form $\infty - \infty$

$\infty - \infty$ is an indeterminate form because the limit is not necessarily $0$ . In this case, rewrite the expression as a single fraction or product. Common strategies include:

Combining with a common denominator
Rationalizing radicals with conjugates
Using properties of logarithms

Find

$x \to 1^{+} lim \frac{x}{x - 1} - \frac{1}{ln ( x )}$

(spoiler)

Combine into one rational expression.

$x \to 1^{+} lim \frac{x}{x - 1} \cdot \frac{ln ( x )}{ln ( x )} - \frac{1}{ln ( x )} \cdot \frac{x - 1}{x - 1}$

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

Direct substitution gives $\frac{0}{0}$ , so apply L’Hopital’s rule.

$= x \to 1^{+} lim \frac{x ( 1/ x ) + ln ( x ) - 1}{( x - 1 ) ( 1/ x ) + ln ( x )}$

$= x \to 1^{+} lim \frac{ln ( x )}{1 - ( 1/ x ) + ln ( x )}$

Direct substitution still gives $\frac{0}{0}$ , so apply L’Hopital’s rule again:

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

Sidenote

Infinity + infinity

$\infty + \infty$ is not an indeterminate form - the result is $\infty$ .

Similarly, $- \infty - \infty$ becomes $- \infty$ .

5. Forms $1^{\infty}, 0^{0}, \infty^{0}$

When direct substitution produces one of these indeterminate exponential forms, rewrite the expression using logarithms and the power property.

Example 1. Determine

$x \to \infty lim (1 + \frac{1}{x})^{x}$

Direct substitution gives $1^{\infty}$ . Let

$L ln (L) = x \to \infty lim (1 + \frac{1}{x})^{x} = ln [x \to \infty lim (1 + \frac{1}{x})^{x}] = x \to \infty lim ln (1 + \frac{1}{x})^{x} = x \to \infty lim x \cdot ln (1 + \frac{1}{x})$

Direct substitution here now gives $\infty \times 0$ , so rewrite as a quotient:

(spoiler)

$x \to \infty lim \frac{ln ( 1 + \frac{1}{x} )}{\frac{1}{x}}$

Now direct substitution gives $\frac{0}{0}$ , so apply L’Hopital’s rule.

$x \to \infty lim \frac{\frac{x}{x + 1} \cdot ( - \frac{1}{x ^{2}} )}{- \frac{1}{x ^{2}}}$

After canceling $- \frac{1}{x ^{2}}$ , the limit becomes

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

Remember that this value is for $ln (L)$ , and the goal is to find the limit $L$ .

$ln (L) L = 1 = e$

Therefore

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

AP tip:

Remember this limit definition of $e$ .

Example 2. Determine

$x \to 0^{+} lim x^{\frac{1}{l n ( x )}}$

(spoiler)

Direct substitution results in $0^{0}$ .

Let

$L = x \to 0^{+} lim x^{\frac{1}{l n ( x )}}$

Taking the natural log of both sides,

$ln (L) = ln (x \to 0^{+} lim x^{\frac{1}{l n ( x )}}) = x \to 0^{+} lim ln (x^{\frac{1}{l n ( x )}}) = x \to 0^{+} lim \frac{1}{ln ( x )} \cdot ln (x) = 1$

In this case, L’Hopital’s rule was not needed.

Since $ln (L) = 1$ , then $L = e$ , so

$x \to 0^{+} lim x^{\frac{1}{l n ( x )}} = e$

Example 3. Evaluate

$x \to \infty lim x^{1/ x}$

(spoiler)

Direct substitution gives $\infty^{0}$ . Let

$L = x \to \infty lim x^{1/ x}$

Then

$ln (L) = ln (x \to \infty lim x^{1/ x}) = x \to \infty lim \frac{1}{x} \cdot ln (x) = x \to \infty lim \frac{ln ( x )}{x}$

Direct substitution now gives $\frac{\infty}{\infty}$ . We can either apply L’Hopital’s rule or use dominant term analysis from section 1.5 to conclude that

$x \to \infty lim \frac{ln ( x )}{x} = 0$

Since $ln (L) = 0$ , then $L = 1$ , so

$x \to 0^{+} lim x^{x} = 1$

In the following example, direct substitution does not lead to an indeterminate form.

$x \to \infty lim (1 - \frac{1}{x})^{c o s (x)}$

(spoiler)

Achievable AP Calculus AB

4. Contextual uses

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

L'Hopital's rule

8 min read

Font

Discuss

Feedback

What you’ll learn

How to apply L’Hopital’s rule on limits with indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$
How to handle other indeterminate forms by rewriting or using logarithms

Another application of derivatives is evaluating limits that produce the indeterminate forms $\frac{0}{0}$ or $\frac{\infty}{\infty}$ .

L’Hopital’s rule:

$x \to a lim f (x) = x \to a lim g (x) = 0 or x \to a lim f (x) = x \to a lim g (x) = \infty$

then

$x \to a lim \frac{f ( x )}{g ( x )} = x \to a lim \frac{f ^{'} ( x )}{g ^{'} ( x )}$

Sidenote

Not the quotient rule

Don’t use the quotient rule. With L’Hopital’s rule, differentiate the numerator and denominator separately to form the new quotient and re-check the limit.

1. Form $0/0$

Example 1. Use L’Hopital’s rule to evaluate

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

(spoiler)

Recall the special trig limit $x \to 0 lim \frac{sin ( x )}{x} = 1$ .

Confirm this using L’Hopital’s rule:

The derivative of the numerator, $sin (x)$ , is $cos (x)$ .
The derivative of the denominator, $x$ , is $1$ .

Therefore,

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

Now direct substitution works, and the limit is $1$ .

Example 2. Evaluate

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

(spoiler)

Applying L’Hopital’s rule,

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

Example 3. Evaluate

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

(spoiler)

Applying L’Hopital’s rule,

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

2. Form $\infty/\infty$

L’Hopital’s rule can also be applied when direct substitution results in the indeterminate form $\frac{\infty}{\infty}$ .

Example 1. Evaluate

$x \to \infty lim \frac{x ^{3} + 5}{e ^{x}}$

(spoiler)

$= x \to \infty lim \frac{3 x ^{2}}{e ^{x}}$

Direct substitution still gives $\frac{\infty}{\infty}$ , so apply L’Hopital’s rule again.

$= x \to \infty lim \frac{6 x}{e ^{x}}$

Applying it again,

$= x \to \infty lim \frac{6}{e ^{x}}$

Now the denominator grows without bound, so the limit is $0$ .

Example 2. Find

$x \to \infty lim \frac{ln ( e ^{x} + x )}{x}$

(spoiler)

Applying L’Hopital’s rule,

$= x \to \infty lim \frac{\frac{1}{e ^{x} + x} \cdot ( e ^{x} + 1 )}{1}$

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

Applying L’Hopital’s rule again until it’s no longer needed,

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

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

Since the $e^{x}$ terms cancel, the limit is $1$ .

3. Form $0 \times \infty$

L’Hopital’s rule applies only to quotients. If you have an indeterminate product $f (x) \times g (x)$ , rewrite it as a fraction, either as

$\frac{f ( x )}{1/ g ( x )} or \frac{g ( x )}{1/ f ( x )}$

Choose the version whose denominator is easier to differentiate or leads to a useful form.

Determine

$x \to 0^{+} lim x cot (x)$

(spoiler)

This can be rewritten as either

$x \to 0^{+} lim \frac{cot ( x )}{1/ x}$

$x \to 0^{+} lim \frac{x}{1/ cot ( x )}$

$= x \to 0^{+} lim \frac{x}{tan ( x )}$

Try applying L’Hopital’s rule to the first one:

$x \to 0^{+} lim \frac{cot ( x )}{1/ x}$

$x \to 0^{+} lim \frac{- csc ^{2} ( x )}{- 1/ x ^{2}}$

This approach doesn’t simplify the indeterminate behavior in a helpful way, because differentiating $1/ x$ keeps producing powers of $x$ in the denominator.

Applying L’Hopital’s rule to the second option:

$x \to 0^{+} lim \frac{x}{tan ( x )}$

$= x \to 0^{+} lim \frac{1}{sec ^{2} ( x )}$

$= 1$

4. Form $\infty - \infty$

$\infty - \infty$ is an indeterminate form because the limit is not necessarily $0$ . In this case, rewrite the expression as a single fraction or product. Common strategies include:

Combining with a common denominator
Rationalizing radicals with conjugates
Using properties of logarithms

Find

$x \to 1^{+} lim \frac{x}{x - 1} - \frac{1}{ln ( x )}$

(spoiler)

Combine into one rational expression.

$x \to 1^{+} lim \frac{x}{x - 1} \cdot \frac{ln ( x )}{ln ( x )} - \frac{1}{ln ( x )} \cdot \frac{x - 1}{x - 1}$

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

Direct substitution gives $\frac{0}{0}$ , so apply L’Hopital’s rule.

$= x \to 1^{+} lim \frac{x ( 1/ x ) + ln ( x ) - 1}{( x - 1 ) ( 1/ x ) + ln ( x )}$

$= x \to 1^{+} lim \frac{ln ( x )}{1 - ( 1/ x ) + ln ( x )}$

Direct substitution still gives $\frac{0}{0}$ , so apply L’Hopital’s rule again:

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

Sidenote

Infinity + infinity

$\infty + \infty$ is not an indeterminate form - the result is $\infty$ .

Similarly, $- \infty - \infty$ becomes $- \infty$ .

5. Forms $1^{\infty}, 0^{0}, \infty^{0}$

When direct substitution produces one of these indeterminate exponential forms, rewrite the expression using logarithms and the power property.

Example 1. Determine

$x \to \infty lim (1 + \frac{1}{x})^{x}$

Direct substitution gives $1^{\infty}$ . Let

$L ln (L) = x \to \infty lim (1 + \frac{1}{x})^{x} = ln [x \to \infty lim (1 + \frac{1}{x})^{x}] = x \to \infty lim ln (1 + \frac{1}{x})^{x} = x \to \infty lim x \cdot ln (1 + \frac{1}{x})$

Direct substitution here now gives $\infty \times 0$ , so rewrite as a quotient:

(spoiler)

$x \to \infty lim \frac{ln ( 1 + \frac{1}{x} )}{\frac{1}{x}}$

Now direct substitution gives $\frac{0}{0}$ , so apply L’Hopital’s rule.

$x \to \infty lim \frac{\frac{x}{x + 1} \cdot ( - \frac{1}{x ^{2}} )}{- \frac{1}{x ^{2}}}$

After canceling $- \frac{1}{x ^{2}}$ , the limit becomes

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

Remember that this value is for $ln (L)$ , and the goal is to find the limit $L$ .

$ln (L) L = 1 = e$

Therefore

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

AP tip:

Remember this limit definition of $e$ .

Example 2. Determine

$x \to 0^{+} lim x^{\frac{1}{l n ( x )}}$

(spoiler)

Direct substitution results in $0^{0}$ .

Let

$L = x \to 0^{+} lim x^{\frac{1}{l n ( x )}}$

Taking the natural log of both sides,

$ln (L) = ln (x \to 0^{+} lim x^{\frac{1}{l n ( x )}}) = x \to 0^{+} lim ln (x^{\frac{1}{l n ( x )}}) = x \to 0^{+} lim \frac{1}{ln ( x )} \cdot ln (x) = 1$

In this case, L’Hopital’s rule was not needed.

Since $ln (L) = 1$ , then $L = e$ , so

$x \to 0^{+} lim x^{\frac{1}{l n ( x )}} = e$

Example 3. Evaluate

$x \to \infty lim x^{1/ x}$

(spoiler)

Direct substitution gives $\infty^{0}$ . Let

$L = x \to \infty lim x^{1/ x}$

Then

$ln (L) = ln (x \to \infty lim x^{1/ x}) = x \to \infty lim \frac{1}{x} \cdot ln (x) = x \to \infty lim \frac{ln ( x )}{x}$

Direct substitution now gives $\frac{\infty}{\infty}$ . We can either apply L’Hopital’s rule or use dominant term analysis from section 1.5 to conclude that

$x \to \infty lim \frac{ln ( x )}{x} = 0$

Since $ln (L) = 0$ , then $L = 1$ , so

$x \to 0^{+} lim x^{x} = 1$

In the following example, direct substitution does not lead to an indeterminate form.

$x \to \infty lim (1 - \frac{1}{x})^{c o s (x)}$

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Use L’Hopital’s rule only on quotients, when the indeterminate form is $\frac{0}{0}$ or $\pm \frac{\infty}{\infty}$ .
For the other indeterminate forms, here is a summary for what to do:

Form	What to do
$0 \cdot \infty$	Rewrite as fraction
$\infty - \infty$	Combine
$0^{0}, \infty^{0}, 1^{\infty}$	Use logarithms

L'Hopital's rule

What you’ll learn

How to apply L’Hopital’s rule on limits with indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$
How to handle other indeterminate forms by rewriting or using logarithms

Another application of derivatives is evaluating limits that produce the indeterminate forms $\frac{0}{0}$ or $\frac{\infty}{\infty}$ .

L’Hopital’s rule:

$x \to a lim f (x) = x \to a lim g (x) = 0 or x \to a lim f (x) = x \to a lim g (x) = \infty$

then

$x \to a lim \frac{f ( x )}{g ( x )} = x \to a lim \frac{f ^{'} ( x )}{g ^{'} ( x )}$

Sidenote

Not the quotient rule

Don’t use the quotient rule. With L’Hopital’s rule, differentiate the numerator and denominator separately to form the new quotient and re-check the limit.

1. Form $0/0$

Example 1. Use L’Hopital’s rule to evaluate

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

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Recall the special trig limit $x \to 0 lim \frac{sin ( x )}{x} = 1$ .

Confirm this using L’Hopital’s rule:

The derivative of the numerator, $sin (x)$ , is $cos (x)$ .
The derivative of the denominator, $x$ , is $1$ .

Therefore,

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

Now direct substitution works, and the limit is $1$ .

Example 2. Evaluate

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

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Applying L’Hopital’s rule,

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

Example 3. Evaluate

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

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Applying L’Hopital’s rule,

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

2. Form $\infty/\infty$

L’Hopital’s rule can also be applied when direct substitution results in the indeterminate form $\frac{\infty}{\infty}$ .

Example 1. Evaluate

$x \to \infty lim \frac{x ^{3} + 5}{e ^{x}}$

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$= x \to \infty lim \frac{3 x ^{2}}{e ^{x}}$

Direct substitution still gives $\frac{\infty}{\infty}$ , so apply L’Hopital’s rule again.

$= x \to \infty lim \frac{6 x}{e ^{x}}$

Applying it again,

$= x \to \infty lim \frac{6}{e ^{x}}$

Now the denominator grows without bound, so the limit is $0$ .

Example 2. Find

$x \to \infty lim \frac{ln ( e ^{x} + x )}{x}$

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Applying L’Hopital’s rule,

$= x \to \infty lim \frac{\frac{1}{e ^{x} + x} \cdot ( e ^{x} + 1 )}{1}$

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

Applying L’Hopital’s rule again until it’s no longer needed,

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

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

Since the $e^{x}$ terms cancel, the limit is $1$ .

3. Form $0 \times \infty$

L’Hopital’s rule applies only to quotients. If you have an indeterminate product $f (x) \times g (x)$ , rewrite it as a fraction, either as

$\frac{f ( x )}{1/ g ( x )} or \frac{g ( x )}{1/ f ( x )}$

Choose the version whose denominator is easier to differentiate or leads to a useful form.

Determine

$x \to 0^{+} lim x cot (x)$

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This can be rewritten as either

$x \to 0^{+} lim \frac{cot ( x )}{1/ x}$

$x \to 0^{+} lim \frac{x}{1/ cot ( x )}$

$= x \to 0^{+} lim \frac{x}{tan ( x )}$

Try applying L’Hopital’s rule to the first one:

$x \to 0^{+} lim \frac{cot ( x )}{1/ x}$

$x \to 0^{+} lim \frac{- csc ^{2} ( x )}{- 1/ x ^{2}}$

This approach doesn’t simplify the indeterminate behavior in a helpful way, because differentiating $1/ x$ keeps producing powers of $x$ in the denominator.

Applying L’Hopital’s rule to the second option:

$x \to 0^{+} lim \frac{x}{tan ( x )}$

$= x \to 0^{+} lim \frac{1}{sec ^{2} ( x )}$

$= 1$

4. Form $\infty - \infty$

$\infty - \infty$ is an indeterminate form because the limit is not necessarily $0$ . In this case, rewrite the expression as a single fraction or product. Common strategies include:

Combining with a common denominator
Rationalizing radicals with conjugates
Using properties of logarithms

Find

$x \to 1^{+} lim \frac{x}{x - 1} - \frac{1}{ln ( x )}$

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Combine into one rational expression.

$x \to 1^{+} lim \frac{x}{x - 1} \cdot \frac{ln ( x )}{ln ( x )} - \frac{1}{ln ( x )} \cdot \frac{x - 1}{x - 1}$

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

Direct substitution gives $\frac{0}{0}$ , so apply L’Hopital’s rule.

$= x \to 1^{+} lim \frac{x ( 1/ x ) + ln ( x ) - 1}{( x - 1 ) ( 1/ x ) + ln ( x )}$

$= x \to 1^{+} lim \frac{ln ( x )}{1 - ( 1/ x ) + ln ( x )}$

Direct substitution still gives $\frac{0}{0}$ , so apply L’Hopital’s rule again:

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

Sidenote

Infinity + infinity

$\infty + \infty$ is not an indeterminate form - the result is $\infty$ .

Similarly, $- \infty - \infty$ becomes $- \infty$ .

5. Forms $1^{\infty}, 0^{0}, \infty^{0}$

When direct substitution produces one of these indeterminate exponential forms, rewrite the expression using logarithms and the power property.

Example 1. Determine

$x \to \infty lim (1 + \frac{1}{x})^{x}$

Direct substitution gives $1^{\infty}$ . Let

$L ln (L) = x \to \infty lim (1 + \frac{1}{x})^{x} = ln [x \to \infty lim (1 + \frac{1}{x})^{x}] = x \to \infty lim ln (1 + \frac{1}{x})^{x} = x \to \infty lim x \cdot ln (1 + \frac{1}{x})$

Direct substitution here now gives $\infty \times 0$ , so rewrite as a quotient:

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$x \to \infty lim \frac{ln ( 1 + \frac{1}{x} )}{\frac{1}{x}}$

Now direct substitution gives $\frac{0}{0}$ , so apply L’Hopital’s rule.

$x \to \infty lim \frac{\frac{x}{x + 1} \cdot ( - \frac{1}{x ^{2}} )}{- \frac{1}{x ^{2}}}$

After canceling $- \frac{1}{x ^{2}}$ , the limit becomes

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

Remember that this value is for $ln (L)$ , and the goal is to find the limit $L$ .

$ln (L) L = 1 = e$

Therefore

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

AP tip:

Remember this limit definition of $e$ .

Example 2. Determine

$x \to 0^{+} lim x^{\frac{1}{l n ( x )}}$

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Direct substitution results in $0^{0}$ .

Let

$L = x \to 0^{+} lim x^{\frac{1}{l n ( x )}}$

Taking the natural log of both sides,

$ln (L) = ln (x \to 0^{+} lim x^{\frac{1}{l n ( x )}}) = x \to 0^{+} lim ln (x^{\frac{1}{l n ( x )}}) = x \to 0^{+} lim \frac{1}{ln ( x )} \cdot ln (x) = 1$

In this case, L’Hopital’s rule was not needed.

Since $ln (L) = 1$ , then $L = e$ , so

$x \to 0^{+} lim x^{\frac{1}{l n ( x )}} = e$

Example 3. Evaluate

$x \to \infty lim x^{1/ x}$

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Direct substitution gives $\infty^{0}$ . Let

$L = x \to \infty lim x^{1/ x}$

Then

$ln (L) = ln (x \to \infty lim x^{1/ x}) = x \to \infty lim \frac{1}{x} \cdot ln (x) = x \to \infty lim \frac{ln ( x )}{x}$

Direct substitution now gives $\frac{\infty}{\infty}$ . We can either apply L’Hopital’s rule or use dominant term analysis from section 1.5 to conclude that

$x \to \infty lim \frac{ln ( x )}{x} = 0$

Since $ln (L) = 0$ , then $L = 1$ , so

$x \to 0^{+} lim x^{x} = 1$

In the following example, direct substitution does not lead to an indeterminate form.

$x \to \infty lim (1 - \frac{1}{x})^{c o s (x)}$

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Key points

Use L’Hopital’s rule only on quotients, when the indeterminate form is $\frac{0}{0}$ or $\pm \frac{\infty}{\infty}$ .
For the other indeterminate forms, here is a summary for what to do:

Form	What to do
$0 \cdot \infty$	Rewrite as fraction
$\infty - \infty$	Combine
$0^{0}, \infty^{0}, 1^{\infty}$	Use logarithms

L'Hopital's rule

What you’ll learn

1. Form 0/0

2. Form ∞/∞

3. Form 0×∞

4. Form ∞−∞

5. Forms 1∞,00,∞0

L'Hopital's rule

What you’ll learn

1. Form 0/0

2. Form ∞/∞

3. Form 0×∞

4. Form ∞−∞

5. Forms 1∞,00,∞0

L'Hopital's rule

What you’ll learn

1. Form 0/0

2. Form ∞/∞

3. Form 0×∞

4. Form ∞−∞

5. Forms 1∞,00,∞0

L'Hopital's rule

What you’ll learn

1. Form 0/0

2. Form ∞/∞

3. Form 0×∞

4. Form ∞−∞

5. Forms 1∞,00,∞0

1. Form $0/0$

2. Form $\infty/\infty$

3. Form $0 \times \infty$

4. Form $\infty - \infty$

5. Forms $1^{\infty}, 0^{0}, \infty^{0}$

1. Form $0/0$

2. Form $\infty/\infty$

3. Form $0 \times \infty$

4. Form $\infty - \infty$

5. Forms $1^{\infty}, 0^{0}, \infty^{0}$

1. Form $0/0$

2. Form $\infty/\infty$

3. Form $0 \times \infty$

4. Form $\infty - \infty$

5. Forms $1^{\infty}, 0^{0}, \infty^{0}$

1. Form $0/0$

2. Form $\infty/\infty$

3. Form $0 \times \infty$

4. Form $\infty - \infty$

5. Forms $1^{\infty}, 0^{0}, \infty^{0}$