L'Hospital's Rule — AP® Calculus AB Lesson

Part 1 · Hook

Why this matters

Some limits look like

\frac{0}{0}

: both the top and the bottom vanish at the same point. Who wins? Near that point each function behaves like its tangent line, so the race is decided by their slopes. L'Hospital's rule turns that idea into a one-line method, and the exam rewards it when you first show the form.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. Which limit has an indeterminate form?

$\lim_{x \to 0}\frac{\sin x}{x}$
$\lim_{x \to 0}\frac{x + 1}{x + 2}$
$\lim_{x \to 1}\frac{x^2}{x}$
$\lim_{x \to 2}\frac{1}{x}$

Show the answer

$\sin 0 = 0$ and the bottom is 0: the form $\frac{0}{0}$ .

Correct: $\lim_{x \to 0}\frac{\sin x}{x}$ :
$\lim_{x \to 0}\frac{x + 1}{x + 2}$ :
$\lim_{x \to 1}\frac{x^2}{x}$ :
$\lim_{x \to 2}\frac{1}{x}$ :

2. What is $\frac{d}{dx}e^{3x}$ ?

$3e^{3x}$
$e^{3x}$
$3xe^{3x - 1}$
$e^3$

Show the answer

The chain rule: $e^{3x} \cdot 3$ .

Correct: $3e^{3x}$ :
$e^{3x}$ :
$3xe^{3x - 1}$ :
$e^3$ :

Part 4 · See it

See it first

Near 0,

e^x - 1

looks like

y = x

and

\sin(2x)

looks like

y = 2x

, so their quotient approaches

\frac{1}{2}

, the ratio of the slopes. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

$\lim\frac{f(x)}{g(x)}$ has the form $\frac{0}{0}$ (or $\frac{\infty}{\infty}$ )plugging in tells nothing: the answer could be any number, or no limit at all
Near $a$ , with $f(a) = g(a) = 0$ , each function is close to its tangent line $\frac{f(x)}{g(x)}$ $\approx \frac{f'(a)(x - a)}{g'(a)(x - a)}$ $= \frac{f'(a)}{g'(a)}$
L'Hospital's rule makes this exact $\lim\frac{f(x)}{g(x)}$ $= \lim\frac{f'(x)}{g'(x)}$ when the second limit exists
If the new limit is still indeterminatecheck the form again and apply the rule again

Part 6 · Key ideas

Key ideas

First show the form: write $\lim f(x) = 0$ and $\lim g(x) = 0$ (or both infinite). Without it the rule does not apply.
Differentiate the top and the bottom separately. It is not the quotient rule.
If the form is not indeterminate, just evaluate: L'Hospital's rule would give a wrong answer.
Never write $\frac{0}{0}$ as a value with an equals sign; describe the form in words or with two limits.

Part 7 · Misconception

A common mistake

The wrong idea: $\lim_{x \to 1}\frac{x^2 + 1}{x + 1}$ $= \lim_{x \to 1}\frac{2x}{1}$ $= 2$ .

What actually happens: The form is $\frac{2}{2}$ , not indeterminate, so the rule does not apply. The limit is just $\frac{2}{2} = 1$ . Always check the form first.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

Values of $f$ , $g$ and their derivatives

Selected values of the twice-differentiable functions f and g and their derivatives
$x$	$f(x)$	$f'(x)$	$g(x)$	$g'(x)$
1	3	1	-2	4
2	0	5	0	-2
3	-1	2	6	3

1. What is $\lim_{x \to 2} \frac{f(x)}{g(x)}$ ?

$-\frac{5}{2}$
$0$
$-\frac{2}{5}$
The limit does not exist

Show the answer

$f$ and $g$ are differentiable, so continuous: $\lim_{x \to 2} f(x) = 0$ and $\lim_{x \to 2} g(x) = 0$ . By L'Hospital's rule the limit is $\lim_{x \to 2}\frac{f'(x)}{g'(x)}$ , and since $f$ and $g$ are twice differentiable, $f'$ and $g'$ are continuous, so this equals $\frac{f'(2)}{g'(2)} = -\frac{5}{2}$ .

Correct: $-\frac{5}{2}$ : Both limits are 0, so the limit is $\frac{f'(2)}{g'(2)} = \frac{5}{-2}$ .
$0$ : $\frac{0}{0}$ is not 0; it is an indeterminate form that needs more work.
$-\frac{2}{5}$ : The derivatives go in the same places as the functions: $f'$ on top.
The limit does not exist: A $\frac{0}{0}$ form can have a limit; L'Hospital's rule finds it.

2. Find $\lim_{x \to 2} \frac{f(x)}{x^2 - 4}$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Both the top and the bottom approach 0 (form $\frac{0}{0}$ ). L'Hospital's rule: $\lim_{x \to 2}\frac{f'(x)}{2x}$ $= \frac{5}{4}$ .

Answer: $\frac{5}{4}$ ≈ 1.250

3. What is $\lim_{x \to 1} \frac{f(x)}{g(x)}$ ?

$\frac{1}{4}$
$-\frac{3}{2}$
$-\frac{2}{3}$
The limit does not exist

Show the answer

Check the form first: $f(1) = 3$ and $g(1) = -2$ , so the quotient is continuous at 1 and the limit is $\frac{3}{-2}$ . L'Hospital's rule does not apply.

$\frac{1}{4}$ : That is $\frac{f'(1)}{g'(1)}$ . L'Hospital's rule needs a $\frac{0}{0}$ or $\frac{\infty}{\infty}$ form, which this is not.
Correct: $-\frac{3}{2}$ : The form is $\frac{3}{-2}$ , not indeterminate, so substitute.
$-\frac{2}{3}$ : That is $\frac{g(1)}{f(1)}$ , the quotient upside down.
The limit does not exist: Both functions are continuous and $g(1) \ne 0$ , so the quotient is continuous at 1.

4. To use L'Hospital's rule for $\lim_{x \to 2} \frac{f(x)}{g(x)}$ , what must be shown first?

$f'(2)$ and $g'(2)$ are both nonzero
$\frac{f(x)}{g(x)}$ is differentiable at $x = 2$
$\lim_{x \to 2} f(x) = 0$ and $\lim_{x \to 2} g(x) = 0$
$f(2) > g(2)$

Show the answer

L'Hospital's rule applies to a limit of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ . Write both limits: $\lim_{x \to 2} f(x) = 0$ and $\lim_{x \to 2} g(x) = 0$ .

$f'(2)$ and $g'(2)$ are both nonzero: $f'(2)$ may be 0; the rule needs $g'$ nonzero near 2, and the indeterminate form.
$\frac{f(x)}{g(x)}$ is differentiable at $x = 2$ : The quotient is not even defined at 2.
Correct: $\lim_{x \to 2} f(x) = 0$ and $\lim_{x \to 2} g(x) = 0$ : The hypothesis: both limits are 0 (or both infinite). Here continuity gives $f(2) = g(2) = 0$ .
$f(2) > g(2)$ : Comparing the values says nothing about the form of the limit.

5. Find $\lim_{x \to 0}\frac{e^{3x} - 1}{x}$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Form $\frac{0}{0}$ . L'Hospital's rule: $\lim_{x \to 0}\frac{3e^{3x}}{1}$ $= 3$ .

Answer: 3

6. Which statement correctly justifies using L'Hospital's rule for $\lim_{x \to 0}\frac{e^x - 1 - x}{x^2}$ ?

$\frac{e^x - 1 - x}{x^2}$ equals $\frac{0}{0}$ at $x = 0$
$\lim_{x \to 0}(e^x - 1 - x)$ $= 0$ and $\lim_{x \to 0}x^2 = 0$
The derivatives of the top and bottom exist
The function is continuous at $x = 0$

Show the answer

The rule's hypothesis is a $\frac{0}{0}$ (or $\frac{\infty}{\infty}$ ) form, shown by writing both limits. Avoid writing an equals sign before $\frac{0}{0}$ , as if it were a number.

$\frac{e^x - 1 - x}{x^2}$ equals $\frac{0}{0}$ at $x = 0$ : Never write $\frac{0}{0}$ as a value; state the two limits.
Correct: $\lim_{x \to 0}(e^x - 1 - x)$ $= 0$ and $\lim_{x \to 0}x^2 = 0$ : Both limits are shown to be 0: the form $\frac{0}{0}$ .
The derivatives of the top and bottom exist: Differentiability is needed, but so is the indeterminate form.
The function is continuous at $x = 0$ : It is not even defined at 0; the form of the limit is what matters.

Part 9 · Summary

Summary

For a limit of the form

\frac{0}{0}

or

\frac{\infty}{\infty}

, L'Hospital's rule gives

\lim\frac{f(x)}{g(x)}

= \lim\frac{f'(x)}{g'(x)}

, derivatives taken separately. Show both limits first; repeat if the new limit is still indeterminate; and evaluate directly when the form is not indeterminate.

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

Values of ff, gg and their derivatives

Summary

What comes next

Connections

Values of $f$ , $g$ and their derivatives