Blog · Calculus guides · 23 October 2026 · 4 min read

L'Hôpital's Rule: Steps, Examples & Practice Problems

Q: When can you use L'Hôpital's rule?

Only when direct substitution gives 0/0 or (±∞)/(±∞). Forms like 0·∞, ∞ - ∞, 1^∞, 0^0 and ∞^0 need rearranging into one of those two first (the last three by taking logs).

L'Hôpital's rule says that if a limit of $\frac{f ( x )}{g ( x )}$ comes out as $\frac{0}{0}$ or $\frac{\infty}{\infty}$ , you can replace it with the limit of $\frac{f ^{'} ( x )}{g ^{'} ( x )}$ , differentiating the top and the bottom separately. If the new limit exists, it is the answer. For example, $lim_{x \to 0} \frac{s i n ( 3 x )}{2 x} = lim_{x \to 0} \frac{3 c o s ( 3 x )}{2} = \frac{3}{2}$ .

The formula

x \to a lim \frac{f ( x )}{g ( x )} = x \to a lim \frac{f ^{'} ( x )}{g ^{'} ( x )} when the left side is \frac{0}{0} or \frac{\pm \infty}{\pm \infty}

Named after the Marquis de L'Hôpital, who published it in 1696. He had paid Johann Bernoulli for his discoveries, so it is arguably the first calculus result that was bought rather than found.

When to use it

Use L'Hôpital's rule when substituting the limit point gives $\frac{0}{0}$ or $\frac{\infty}{\infty}$ , and the usual tricks do not apply easily: there is a trig function, an exponential or a logarithm in the mix, so you cannot just factor and cancel. Always substitute first. If you get an actual number, that is the limit and you are finished. If you get something like $\frac{5}{0}$ , the limit is infinite or does not exist, and the rule does not apply. Other indeterminate forms, $0 \cdot \infty$ and $\infty - \infty$ , can be rearranged into a fraction first, and then the rule works on them too.

How to do it, step by step

Substitute first. Plug the limit point into the top and the bottom. Only carry on if you get $\frac{0}{0}$ or $\frac{\infty}{\infty}$ .
Differentiate the top and the bottom separately. This is not the quotient rule. Top's derivative over bottom's derivative, nothing else.
Substitute again. If you get a number, that is your answer.
Repeat if it is still indeterminate. If the new fraction is still $\frac{0}{0}$ , apply the rule again. Check every time; the moment it stops being indeterminate, stop.

Worked examples

Example 1: A trig limit at zero

Easy

Find:

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

Substitute: $sin 0 = 0$ on top and $0$ on the bottom, so it is $\frac{0}{0}$ and the rule applies.

The top differentiates to $3 cos (3 x)$ (the chain rule puts the 3 in front) and the bottom to $2$ .

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

Answer

x \to 0 lim \frac{sin ( 3 x )}{2 x} = \frac{3}{2}

Example 2: A logarithm at one

Medium

Find:

x \to 1 lim \frac{ln x}{x ^{2} - 1}

Substitute: $ln 1 = 0$ and $1 - 1 = 0$ . It is $\frac{0}{0}$ again.

Top: $\frac{d}{d x} ln x = \frac{1}{x}$ . Bottom: $\frac{d}{d x} (x^{2} - 1) = 2 x$ .

x \to 1 lim \frac{ln x}{x ^{2} - 1} = x \to 1 lim \frac{1/ x}{2 x} = \frac{1}{2}

Factoring would not have helped here, since $ln x$ has nothing to cancel with $(x - 1) (x + 1)$ . This is exactly the kind of limit the rule is for.

Answer

x \to 1 lim \frac{ln x}{x ^{2} - 1} = \frac{1}{2}

Example 3: Zero times infinity

Harder

Find:

x \to 0^{+} lim x ln x

As $x \to 0^{+}$ , $x \to 0$ but $ln x \to - \infty$ . That is $0 \cdot \infty$ , which is indeterminate, but it is not a fraction, so the rule cannot be used yet.

Rewrite it as a fraction by moving $x$ to the bottom as $\frac{1}{x}$ . Now it is $\frac{- \infty}{\infty}$ :

x \to 0^{+} lim x ln x = x \to 0^{+} lim \frac{ln x}{1/ x}

Differentiate the top and the bottom: $\frac{1}{x}$ over $- \frac{1}{x ^{2}}$ , which simplifies to $- x$ .

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

Which way you rewrite matters. Putting $ln x$ on the bottom instead, as $\frac{x}{1/ l n x}$ , also gives $\frac{0}{0}$ , but its derivatives get worse every round. If the rule makes things messier, flip the fraction.

Answer

x \to 0^{+} lim x ln x = 0

Your turn

Three to try, checked by the same machine that grades the game. Type the answer however you like; any correct form counts.

Problem 1

Easy

Find the limit:

x \to 0 lim \frac{e ^{4 x} - 1}{2 x}

Problem 2

Medium

Find the limit:

x \to 0 lim \frac{ln ( 1 + 5 x )}{tan ( 2 x )}

Problem 3

Harder

Find the limit:

x \to 0 lim \frac{e ^{x} - 1 - x}{x ^{2}}

Common mistakes

Using it when the limit is not indeterminate. $lim_{x \to 0} \frac{c o s x}{x + 1}$ is just $\frac{1}{1} = 1$ . Differentiating anyway gives $\frac{- s i n 0}{1} = 0$ , which is wrong. Substitute first, every time.
Using the quotient rule instead. You differentiate the top and bottom separately. The quotient rule is for differentiating a fraction, which is a different question.
Not checking before the second round. After one application, substitute again. Applying the rule to something that is no longer $\frac{0}{0}$ gives a wrong answer with total confidence.
Forgetting the chain rule on the top or bottom. $\frac{d}{d x} sin (3 x) = 3 cos (3 x)$ . Lose that 3 and you get $\frac{1}{2}$ instead of $\frac{3}{2}$ .

Practice it now

Five minutes, no account

L'Hôpital limits arrive in a Sprint at difficulty 5 to 7: sines, tangents, $e^{a x} - 1$ , $1 - cos (a x)$ and logarithms over a multiple of $x$ at zero, plus $\frac{l n x}{x ^{c} - 1}$ at one. Earlier, at 3 and 4, come $\frac{0}{0}$ fractions of polynomials, where the rule works too but factoring is usually quicker.

Play a Sprint See it on the formula sheet

FAQ

When can you use L'Hôpital's rule?: Only when direct substitution gives $\frac{0}{0}$ or $\frac{\pm \infty}{\pm \infty}$ . Forms like $0 \cdot \infty$ , $\infty - \infty$ , $1^{\infty}$ , $0^{0}$ and $\infty^{0}$ need rearranging into one of those two first (the last three by taking logs).
Can you use L'Hôpital's rule twice?: Yes, as often as you need, as long as you check it is still indeterminate each time. $lim_{x \to 0} \frac{1 - c o s x}{x ^{2}}$ goes to $\frac{s i n x}{2 x}$ , still $\frac{0}{0}$ , then to $\frac{c o s x}{2}$ , which is $\frac{1}{2}$ .
Do you need L'Hôpital's rule for limits at infinity?: It works there, but for rational functions it is overkill. Comparing the highest powers is faster: see limits at infinity. In our answer data players get 92% of those right without it.
Are L'Hôpital limits harder than ordinary limits?: Not in practice. Players get 84% of L'Hôpital problems right, against 77% for limits done by plain substitution, which look easier and are not. Once you have checked the $\frac{0}{0}$ , the rule is mechanical, which is a lot of its charm. For polynomial fractions, though, factoring and cancelling is usually quicker.
Where can I find L'Hôpital's rule practice problems?: Right on this page: "Your turn" has three L'Hôpital's rule practice problems, easy, medium and harder, checked as you type by the same checker that grades Asymptote, with the answer and the first step if you miss. For more practice against the clock, a five-minute Sprint is free and needs no account.

L'Hôpital's Rule: Steps, Examples & Practice Problems

When to use it

Use L'Hôpital's rule when substituting the limit point gives

\frac{0}{0}

\frac{\infty}{\infty}

, and the usual tricks do not apply easily: there is a trig function, an exponential or a logarithm in the mix, so you cannot just factor and cancel. Always substitute first. If you get an actual number, that is the limit and you are finished. If you get something like

\frac{5}{0}

, the limit is infinite or does not exist, and the rule does not apply. Other indeterminate forms,

0 \cdot \infty

and

\infty - \infty

, can be rearranged into a fraction first, and then the rule works on them too.

How to do it, step by step

Substitute first. Plug the limit point into the top and the bottom. Only carry on if you get

\frac{0}{0}

\frac{\infty}{\infty}

Differentiate the top and the bottom separately. This is not the quotient rule. Top's derivative over bottom's derivative, nothing else.

Substitute again. If you get a number, that is your answer.

Repeat if it is still indeterminate. If the new fraction is still

\frac{0}{0}

, apply the rule again. Check every time; the moment it stops being indeterminate, stop.

Worked examples

Example 1: A trig limit at zero

Easy

Find:

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

Substitute: $sin 0 = 0$ on top and $0$ on the bottom, so it is $\frac{0}{0}$ and the rule applies.

The top differentiates to $3 cos (3 x)$ (the chain rule puts the 3 in front) and the bottom to $2$ .

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

Answer

x \to 0 lim \frac{sin ( 3 x )}{2 x} = \frac{3}{2}

Example 2: A logarithm at one

Medium

Find:

x \to 1 lim \frac{ln x}{x ^{2} - 1}

Substitute: $ln 1 = 0$ and $1 - 1 = 0$ . It is $\frac{0}{0}$ again.

Top: $\frac{d}{d x} ln x = \frac{1}{x}$ . Bottom: $\frac{d}{d x} (x^{2} - 1) = 2 x$ .

x \to 1 lim \frac{ln x}{x ^{2} - 1} = x \to 1 lim \frac{1/ x}{2 x} = \frac{1}{2}

Factoring would not have helped here, since $ln x$ has nothing to cancel with $(x - 1) (x + 1)$ . This is exactly the kind of limit the rule is for.

Answer

x \to 1 lim \frac{ln x}{x ^{2} - 1} = \frac{1}{2}

Example 3: Zero times infinity

Harder

Find:

x \to 0^{+} lim x ln x

As $x \to 0^{+}$ , $x \to 0$ but $ln x \to - \infty$ . That is $0 \cdot \infty$ , which is indeterminate, but it is not a fraction, so the rule cannot be used yet.

Rewrite it as a fraction by moving $x$ to the bottom as $\frac{1}{x}$ . Now it is $\frac{- \infty}{\infty}$ :

x \to 0^{+} lim x ln x = x \to 0^{+} lim \frac{ln x}{1/ x}

Differentiate the top and the bottom: $\frac{1}{x}$ over $- \frac{1}{x ^{2}}$ , which simplifies to $- x$ .

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

Answer

x \to 0^{+} lim x ln x = 0

Your turn

Three to try, checked by the same machine that grades the game. Type the answer however you like; any correct form counts.

Problem 1

Easy

Find the limit:

x \to 0 lim \frac{e ^{4 x} - 1}{2 x}

Problem 2

Medium

Find the limit:

x \to 0 lim \frac{ln ( 1 + 5 x )}{tan ( 2 x )}

Problem 3

Harder

Find the limit:

x \to 0 lim \frac{e ^{x} - 1 - x}{x ^{2}}

Common mistakes

Using it when the limit is not indeterminate.

lim_{x \to 0} \frac{c o s x}{x + 1}

is just

\frac{1}{1} = 1

. Differentiating anyway gives

\frac{- s i n 0}{1} = 0

, which is wrong. Substitute first, every time.

Using the quotient rule instead. You differentiate the top and bottom separately. The quotient rule is for differentiating a fraction, which is a different question.

Not checking before the second round. After one application, substitute again. Applying the rule to something that is no longer

\frac{0}{0}

gives a wrong answer with total confidence.

Forgetting the chain rule on the top or bottom.

\frac{d}{d x} sin (3 x) = 3 cos (3 x)

. Lose that 3 and you get

\frac{1}{2}

instead of

\frac{3}{2}

Practice it now

Five minutes, no account

Play a Sprint See it on the formula sheet

FAQ

When can you use L'Hôpital's rule?

Only when direct substitution gives

\frac{0}{0}

\frac{\pm \infty}{\pm \infty}

. Forms like

0 \cdot \infty

\infty - \infty

1^{\infty}

0^{0}

and

\infty^{0}

need rearranging into one of those two first (the last three by taking logs).

Can you use L'Hôpital's rule twice?

Yes, as often as you need, as long as you check it is still indeterminate each time.

lim_{x \to 0} \frac{1 - c o s x}{x ^{2}}

goes to

\frac{s i n x}{2 x}

, still

\frac{0}{0}

, then to

\frac{c o s x}{2}

, which is

\frac{1}{2}

Do you need L'Hôpital's rule for limits at infinity?

It works there, but for rational functions it is overkill. Comparing the highest powers is faster: see limits at infinity. In our answer data players get 92% of those right without it.

Are L'Hôpital limits harder than ordinary limits?

Not in practice. Players get 84% of L'Hôpital problems right, against 77% for limits done by plain substitution, which look easier and are not. Once you have checked the

\frac{0}{0}

, the rule is mechanical, which is a lot of its charm. For polynomial fractions, though, factoring and cancelling is usually quicker.

Where can I find L'Hôpital's rule practice problems?

Right on this page: "Your turn" has three L'Hôpital's rule practice problems, easy, medium and harder, checked as you type by the same checker that grades Asymptote, with the answer and the first step if you miss. For more practice against the clock, a five-minute Sprint is free and needs no account.

L'Hôpital's Rule: Steps, Examples & Practice Problems

The formula

When to use it

How to do it, step by step

Worked examples

Example 1: A trig limit at zero

Example 2: A logarithm at one

Example 3: Zero times infinity

Your turn

Common mistakes

FAQ

Keep going

L'Hôpital's Rule: Steps, Examples & Practice Problems

The formula

When to use it

How to do it, step by step

Worked examples

Example 1: A trig limit at zero

Example 2: A logarithm at one

Example 3: Zero times infinity

Your turn

Common mistakes

FAQ

Keep going