AP®MathsCollege BoardCalculus ABStudy GuidesUnit 4: Contextual Applications of DifferentiationL'Hospital's RuleL'Hospital's Rule

L'Hospital's Rule (College Board AP® Calculus AB): Study Guide

Written by: Roger B

Reviewed by: Dan Finlay

Updated on 6 May 2026

Indeterminate forms

What is an indeterminate form?

An indeterminate form is an expression that does not tell you the value of a limit
You need to be aware of two indeterminate forms
- $\frac{0}{0}$
- $\frac{\pm \infty}{\pm \infty}$
Note that indeterminate forms are labels, not numbers
- $\frac{0}{0}$ is not a number
- Avoid writing $\lim_{x \to a} f (x) = \frac{0}{0}$
The value of an indeterminate form is undefined
- Dividing by 0 always gives an undefined expression
- And note that, for example, $\frac{\infty}{\infty}$ is not equal to 1
  - $\infty$ is not a number
  - so it can't be canceled to simplify a fraction
Sometimes attempting to evaluate a limit using substitution leads to one of the indeterminate forms given above
- L'Hospital's rule provides a method for dealing with limits of that form
For example, $\lim_{x \to 0} \frac{\sin x}{x}$ leads to the indeterminate form $\frac{0}{0}$

Examiner Tips and Tricks

Note that if substitution gives limits that look like $\frac{0}{\pm \infty}$ or $\frac{\pm \infty}{0}$ , these are not indeterminate forms, and L'Hospital's rule cannot be used

In the first case, $\frac{0}{\pm \infty}$ , the limit will just be equal to $0$
In the second case, $\frac{\pm \infty}{0}$ , the limit will diverge to either $+ \infty$ or $- \infty$ depending on the behavior near the limit point
- See the 'Infinite Limits & Limits at Infinity' study guide

Examiner Tips and Tricks

Other limit methods will also sometimes work when substitution gives an indeterminate form

For example, algebraic simplification, multiplying by conjugates or multiplying by reciprocals
- See the 'Evaluating Limits Analytically' study guide

Evaluating limits using L'Hospital's rule

What is L'Hospital’s Rule?

L'Hospital's rule (sometimes written as L’Hôpital’s rule) is a method for finding the value of certain limits using calculus
- Specifically, it allows us to attempt to evaluate the limit of a quotient $\frac{f (x)}{g (x)}$
- for which attempting to evaluate the limit by substitution returns one of the indeterminate forms $\frac{0}{0}$ or $\frac{\pm \infty}{\pm \infty}$
For such a quotient function, L'Hospital's rule says that
- $\lim_{x \to a} \frac{f (x)}{g (x)} = \lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)}$
In plain language, this means you can take the derivatives of the numerator and denominator and attempt to evaluate the limit again in that form

Examiner Tips and Tricks

Notice that the formula includes the ratio of the derivatives. This is different to $\frac{d}{d x} (\frac{f (x)}{g (x)})$ . If you are using the quotient rule, then it is likely that you have made a mistake.

How do I evaluate a limit using L’Hospital’s Rule?

STEP 1
Check that the limit of the quotient results in one of the indeterminate forms given above
- You must verify that one of the following is true:
  - $\lim_{x \to a} f (x) = 0$ and $\lim_{x \to a} g (x) = 0$
  - $\lim_{x \to a} f (x) \to \pm \infty$ and $\lim_{x \to a} g (x) \to \pm \infty$
STEP 2
Find the derivatives of the numerator and denominator of the quotient
STEP 3
Check whether the limit $\lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)}$ exists
STEP 4
If that limit does exist, then $\lim_{x \to a} \frac{f (x)}{g (x)} = \lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)}$
STEP 5
If $\lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)} = \frac{f^{'} (a)}{g^{'} (a)}$ leads to $\frac{0}{0}$ or $\frac{\pm \infty}{\pm \infty}$ then you may repeat the process by considering $\lim_{x \to a} \frac{f^{''} (x)}{g^{''} (x)}$ (and possibly higher order derivatives after that)
- As long as the limits continue giving indeterminate forms you may continue applying L’Hospital’s rule
- Each time this happens find the next set of derivatives and consider the limit again

Examiner Tips and Tricks

Before beginning to use L'Hospital's rule to evaluate a limit

Be sure to confirm that using substitution gives an indeterminate form
Otherwise L'Hospital's rule is not valid

Worked Example

Use L’Hospital’s rule to evaluate each of the following limits:

(a) $\lim_{x \to \infty} \frac{5 x + 17}{4 - 3 x}$

(b) $\lim_{x \to 0} \frac{x^{3}}{- 2 x + \sin 2 x}$

Answer:

(a)

This limit could also be found by 'multiplying by reciprocals', but the question says to use L'Hospital's rule

First check that substitution gives an indeterminate form, so L'Hospital's rule is valid

$\lim_{x \to \infty} (5 x + 17) \to \infty$ and $\lim_{x \to \infty} (4 - 3 x) \to - \infty$

$\lim_{x \to \infty} \frac{5 x + 17}{4 - 3 x}$ leads to $\frac{\infty}{- \infty}$ which is an indeterminate form

Find the derivatives of the numerator and denominator

$\frac{d}{d x} (5 x + 17) = 5 \frac{d}{d x} (4 - 3 x) = - 3$

Apply L'Hospital's Rule, $\lim_{x \to a} \frac{f (x)}{g (x)} = \lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)}$

$\lim_{x \to \infty} \frac{5 x + 17}{4 - 3 x} = \lim_{x \to \infty} \frac{5}{- 3}$

There's no x in that final expression, so x going to infinity doesn't matter!

$\lim_{x \to \infty} \frac{5 x + 17}{4 - 3 x} = - \frac{5}{3}$

(b)

First check that substitution gives an indeterminate form, so L'Hospital's rule is valid

$\lim_{x \to 0} x^{3} = 0^{3} = 0$

$\lim_{x \to 0} (- 2 x + \sin 2 x) = - 2 \cdot 0 + \sin (2 \cdot 0) = 0$

$\lim_{x \to 0} \frac{x^{3}}{- 2 x + \sin 2 x}$ leads to $\frac{0}{0}$ which is an indeterminate form

Find the derivatives of the numerator and denominator

$\frac{d}{d x} (x^{3}) = 3 x^{2} \frac{d}{d x} (- 2 x + \sin 2 x) = - 2 + 2 \cos 2 x$

Apply L'Hospital's Rule, $\lim_{x \to a} \frac{f (x)}{g (x)} = \lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)}$

$\begin{array}{rcl} \lim_{x \to 0} \frac{x^{3}}{- 2 x + \sin 2 x} & = & \lim_{x \to 0} \frac{3 x^{2}}{- 2 + 2 \cos 2 x} \end{array}$

Show that this is another indeterminate form

$\lim_{x \to 0} 3 x^{2} = 3 \cdot 0^{2} = 0$

$\lim_{x \to 0} (- 2 + 2 \cos 2 x) = - 2 + 2 \cos (2 \cdot 0) = 0$

This leads to $\begin{array}{rcl} \frac{0}{0} \end{array}$ again

Apply L'Hospital's rule again

$\frac{d}{d x} (3 x^{2}) = 6 x \frac{d}{d x} (- 2 + 2 \cos 2 x) = - 4 \sin 2 x$

$\begin{array}{rcl} \lim_{x \to 0} \frac{3 x^{2}}{- 2 + 2 \cos 2 x} & = & \lim_{x \to 0} \frac{6 x}{- 4 \sin 2 x} \end{array}$

Show that this is also an indeterminate form

$\lim_{x \to 0} 6 x = 6 \cdot 0 = 0$

$\lim_{x \to 0} (- 4 \sin 2 x) = - 4 \sin (2 \cdot 0) = 0$

This leads to $\begin{array}{rcl} \frac{0}{0} \end{array}$ again

Apply L'Hospital's rule again

$\frac{d}{d x} (6 x) = 6 \frac{d}{d x} (- 4 \sin 2 x) = - 8 \cos 2 x$

$\begin{array}{rcl} \lim_{x \to 0} \frac{6 x}{- 4 \sin 2 x} & = & \lim_{x \to 0} \frac{6}{- 8 \cos 2 x} \end{array}$

Check that this is not an indeterminate form

$\lim_{x \to 0} 6 = 6$

$\lim_{x \to 0} (- 8 \cos 2 x) = - 8 \cos (2 \cdot 0) = - 8$

Find the limit by substitution

$\begin{array}{rcl} \lim_{x \to 0} \frac{6}{- 8 \cos 2 x} & = & - \frac{6}{8} \end{array}$

Simplify the fraction

$\lim_{x \to 0} \frac{x^{3}}{- 2 x + \sin 2 x} = - \frac{3}{4}$

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L'Hospital's Rule (College Board AP® Calculus AB): Study Guide

Indeterminate forms

What is an indeterminate form?

Examiner Tips and Tricks

Examiner Tips and Tricks

Evaluating limits using L'Hospital's rule

What is L'Hospital’s Rule?

Examiner Tips and Tricks

How do I evaluate a limit using L’Hospital’s Rule?

Examiner Tips and Tricks

Worked Example

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Roger B

Reviewer: Dan Finlay