L'Hôpital's rule

Warm-up

Three quick questions from earlier lessons.

Discover

Near $x = 0$ , $\sin(3x) \approx 3x$ and $\tan x \approx x$ : each is close to its tangent line. What does $\dfrac{\sin(3x)}{\tan x}$ approach as $x \to 0$ ?

3
0

Key idea

For $0/0$ (or $\infty/\infty$ ), compare the rates instead of the values.

L'Hôpital's rule: if

f(a) = g(a) = 0

(or both tend to

\pm\infty

), then

\displaystyle\lim_{x \to a} \dfrac{f(x)}{g(x)} = \lim_{x \to a} \dfrac{f'(x)}{g'(x)}

, when the second limit exists. Differentiate top and bottom separately, not with the quotient rule.

Worked example

Find $\displaystyle\lim_{x \to 0} \dfrac{e^{2x} - 1}{\sin x}$ .

Form

\dfrac{0}{0}

Check the form: e⁰ − 1 = 0 and sin 0 = 0.

\displaystyle\lim_{x \to 0} \dfrac{2e^{2x}}{\cos x}

Differentiate top and bottom.

= \dfrac{2}{1} = 2

Now substitute.

Your turn, guided

Your turn: $\displaystyle\lim_{x \to 1} \dfrac{\ln x}{x - 1}$ .

\displaystyle\lim_{x \to 1} \dfrac{1/x}{1}

0/0: differentiate.

= 1

Substitute x = 1.

Practice ★

Find $\displaystyle\lim_{x \to 0} \dfrac{e^{ 5x} - 1}{x}$ .
Find $\displaystyle\lim_{x \to 0} \dfrac{e^{ 4x} - 1}{x}$ .
Find $\displaystyle\lim_{x \to 0} \dfrac{e^{ 2x} - 1}{x}$ .

Watch out!

Practice ★★

Find $\displaystyle\lim_{x \to 0} \dfrac{1 - \cos(6x)}{x^2}$ .
Find $\displaystyle\lim_{x \to 0} \dfrac{1 - \cos(5x)}{x^2}$ .

Find $\displaystyle\lim_{x \to 1} \dfrac{x^{ 7 } - 1}{\ln x}$ .
Find $\displaystyle\lim_{x \to 1} \dfrac{x^{ 8 } - 1}{\ln x}$ .

Exit ticket

Three questions to check you have got it.

Find $\displaystyle\lim_{x \to 0} \dfrac{e^{ 5x} - 1}{x}$ .
Find $\displaystyle\lim_{x \to 0} \dfrac{e^{ 4x} - 1}{x}$ .
Find $\displaystyle\lim_{x \to 0} \dfrac{e^{ 2x} - 1}{x}$ .

Summary card

L'Hôpital's rule

Check the form is $0/0$ or $\infty/\infty$ .
Differentiate top and bottom separately.
Take the new limit (repeat if still $0/0$ ).

Watch out: no indeterminate form, no L'Hôpital.

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