L'Hôpital's Rule

You will probably have already considered limits in work on the sum of a geometric series, rational functions and asymptotes and differentiation from first principles. In this page, we will take the idea of limits further. L'Hôpital's rule gives us a technique to evaluate limits of indeterminate forms \(\large \frac{0}{0}\)and \(\large \frac{∞}{∞}\). The method for doing this is actually fairly straight forward, it requires you to be confident with differentiation techniques. In the examination, you might also be required to evaluate limits using Maclaurin Series. Check out that method on the page for Maclaurin Series.

Key Concepts

On this page, you should learn about

The indeterminate forms \(\large \frac{0}{0}\)and \(\large \frac{∞}{∞}\).
Evaluation of limits in the form \(\lim\limits_{x\rightarrow a} \frac{f(x)}{g(x)}\)

Summary

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Test Yourself

Here is a quiz about the skills learned on this page

START QUIZ!

L'Hôpital's Rule 1/1

Use l’Hôpital’s rule to determine the value of \(\large \lim\limits_{x\rightarrow 0}\frac{\sin x}{x}\)

limit =

\(\large \lim\limits_{x\rightarrow 0}\frac{\sin x}{x}=\frac{0}{0}\\ \large \lim\limits_{x\rightarrow 0}\frac{\sin x}{x}= \lim\limits_{x\rightarrow 0}\frac{\frac{\mathrm{d}}{\mathrm{d}x}(\sin x)}{\frac{\mathrm{d}}{\mathrm{d}x}(x)}=\large \lim\limits_{x\rightarrow 0}\frac{\cos x}{1}=\frac{1}{1}=1\)

Use l’Hôpital’s rule to determine the value of \(\large \lim\limits_{x\rightarrow \infty}\frac{2x}{e^x}\)

limit =

\(\large \lim\limits_{x\rightarrow \infty}\frac{2x}{e^x}=\frac{\infty}{\infty}\\ \large \lim\limits_{x\rightarrow \infty}\frac{2x}{e^x}= \lim\limits_{x\rightarrow \infty}\frac{\frac{\mathrm{d}}{\mathrm{d}x}(2x)}{\frac{\mathrm{d}}{\mathrm{d}x}(e^x)}=\large \lim\limits_{x\rightarrow \infty}\frac{2}{e^x}=\frac{2}{\infty}=0\)

Use l’Hôpital’s rule to determine the value of \(\large \lim\limits_{x\rightarrow 0}\frac{\ln(1+x)}{x}\)

limit =

\(\large \lim\limits_{x\rightarrow 0}\frac{\ln(1+x)}{x}=\frac{0}{0}\\ \large \lim\limits_{x\rightarrow 0}\frac{\ln(1+x)}{x}= \lim\limits_{x\rightarrow 0}\frac{\frac{\mathrm{d}}{\mathrm{d}x}(\ln(1+x))}{\frac{\mathrm{d}}{\mathrm{d}x}(x)}=\large \lim\limits_{x\rightarrow 0}\frac{\frac{1}{1+x}}{1}=\frac{1}{1}=1\)

Use l’Hôpital’s rule to determine the value of \(\large \lim\limits_{x\rightarrow 0}\frac{e^x-1}{x^2+x}\)

limit =

\(\large \lim\limits_{x\rightarrow 0}\frac{e^x-1}{x^2+x}=\frac{0}{0}\\ \large \lim\limits_{x\rightarrow 0}\frac{e^x-1}{x^2+x}= \lim\limits_{x\rightarrow 0}\frac{\frac{\mathrm{d}}{\mathrm{d}x}(e^x-1)}{\frac{\mathrm{d}}{\mathrm{d}x}(x^2+x)}=\large \lim\limits_{x\rightarrow 0}\frac{e^x}{2x+1}=\frac{1}{1}=1\)

Use l’Hôpital’s rule to determine the value of \(\large \lim\limits_{x\rightarrow 1}\frac{\ln x}{x-1}\)

limit =

\(\large \lim\limits_{x\rightarrow 1}\frac{\ln x}{x-1}=\frac{0}{0}\\ \large \lim\limits_{x\rightarrow 1}\frac{\ln x}{x-1}= \lim\limits_{x\rightarrow 1}\frac{\frac{\mathrm{d}}{\mathrm{d}x}(\ln x)}{\frac{\mathrm{d}}{\mathrm{d}x}(x-1)}=\large \lim\limits_{x\rightarrow 1}\frac{\frac{1}{x}}{1}=\frac{1}{1}=1\)

Use l’Hôpital’s rule to determine the value of \(\large \lim\limits_{x\rightarrow 0}\frac{e^x-e^{-x}}{\sin x}\)

limit =

\(\large \lim\limits_{x\rightarrow 0}\frac{e^x-e^{-x}}{\sin x}=\frac{0}{0}\\ \large \lim\limits_{x\rightarrow 0}\frac{e^x-e^{-x}}{\sin x}= \lim\limits_{x\rightarrow 0}\frac{\frac{\mathrm{d}}{\mathrm{d}x}(e^x-e^{-x})}{\frac{\mathrm{d}}{\mathrm{d}x}(\sin x)}=\large \lim\limits_{x\rightarrow 0}\frac{e^x+e^{-x}}{\cos x}=\frac{1+1}{1}=2\)

Use l’Hôpital’s rule to determine the value of \(\large \lim\limits_{x\rightarrow \infty}\frac{\ln x}{e^x}\)

limit =

\(\large \lim\limits_{x\rightarrow \infty}\frac{\ln x}{e^x}=\frac{\infty}{\infty}\\ \large \lim\limits_{x\rightarrow \infty}\frac{\ln x}{e^x}= \lim\limits_{x\rightarrow \infty}\frac{\frac{\mathrm{d}}{\mathrm{d}x}(\ln x)}{\frac{\mathrm{d}}{\mathrm{d}x}(e^x)}=\large \lim\limits_{x\rightarrow \infty}\frac{\frac{1}{x}}{e^x}=\frac{0}{\infty}=0\)

Use l’Hôpital’s rule to determine the value of \(\large \lim\limits_{x\rightarrow 0}\frac{\arcsin x}{\tan x}\)

limit =

\(\large \lim\limits_{x\rightarrow 0}\frac{\arcsin x}{\tan x}=\frac{0}{0}\\ \large \lim\limits_{x\rightarrow 0}\frac{\arcsin x}{\tan x}= \lim\limits_{x\rightarrow 0}\frac{\frac{\mathrm{d}}{\mathrm{d}x}(\arcsin x)}{\frac{\mathrm{d}}{\mathrm{d}x}(\tan x)}=\large \lim\limits_{x\rightarrow 0}\frac{\frac{1}{\sqrt{1-x^2}}}{\sec ^2x}=\frac{1}{1}=1\)

Use l’Hôpital’s rule to determine the value of \(\large \lim\limits_{x\rightarrow \infty}\frac{x^2}{e^x+1}\)

limit =

\(\large \lim\limits_{x\rightarrow \infty}\frac{x^2}{e^x+1}=\frac{\infty}{\infty}\\ \large \lim\limits_{x\rightarrow \infty}\frac{x^2}{e^x+1}= \lim\limits_{x\rightarrow \infty}\frac{\frac{\mathrm{d}}{\mathrm{d}x}(x^2)}{\frac{\mathrm{d}}{\mathrm{d}x}(e^x+1)}=\large \lim\limits_{x\rightarrow \infty}\frac{2 x}{e^x}=\frac{\infty}{\infty}\)

We apply l'Hôpital's rule again

\(\large \lim\limits_{x\rightarrow \infty}\frac{2 x}{e^x}= \lim\limits_{x\rightarrow \infty}\frac{\frac{\mathrm{d}}{\mathrm{d}x}(2x)}{\frac{\mathrm{d}}{\mathrm{d}x}(e^x)}=\large \lim\limits_{x\rightarrow \infty}\frac{2 }{e^x}=\frac{2}{\infty}=0\)

Use l’Hôpital’s rule to determine the value of \(\large \lim\limits_{x\rightarrow 0}\frac{2\sin x-\sin 2x}{x^3}\)

limit =

\(\large \lim\limits_{x\rightarrow 0}\frac{2\sin x-\sin 2x}{x^3}=\frac{0}{0}\\ \large \lim\limits_{x\rightarrow 0}\frac{2\sin x-\sin 2x}{x^3}= \lim\limits_{x\rightarrow 0}\frac{\frac{\mathrm{d}}{\mathrm{d}x}(2\sin x-\sin 2x)}{\frac{\mathrm{d}}{\mathrm{d}x}(x^3)}=\large \lim\limits_{x\rightarrow 0}\frac{2\cos x-2\cos 2x}{3x^2}=\frac{2-2}{0}=\frac{0}{0}\)

We apply l'Hôpital's rule again

\(\large \lim\limits_{x\rightarrow 0}\frac{2\cos x-2\cos 2x}{3x^2}=\lim\limits_{x\rightarrow 0}\frac{\frac{\mathrm{d}}{\mathrm{d}x}(2\cos x-2\cos 2x)}{\frac{\mathrm{d}}{\mathrm{d}x}(3x^2)}=\lim\limits_{x\rightarrow 0}\frac{-2\sin x+4\sin 2x}{6x}=\frac{0}{0}\)

We apply l'Hôpital's rule again

\(\large\lim\limits_{x\rightarrow 0}\frac{-2\sin x+4\sin 2x}{6x}=\lim\limits_{x\rightarrow 0}\frac{\frac{\mathrm{d}}{\mathrm{d}x}(-2\sin x+4\sin 2x)}{\frac{\mathrm{d}}{\mathrm{d}x}(6x)}=\lim\limits_{x\rightarrow 0}\frac{-2\cos x+8\cos 2x}{6}=\frac{-2+8}{6}=\frac{6}{6}=1\)

Exam-style Questions

Question 1

HL moderate No calculator

Use l’Hôpital’s rule to determine the value of \(\large \lim\limits_{x\rightarrow 0}( x \ln x)\)

Hint

Write the limit in the correct form \(\large \lim\limits_{x\rightarrow 0} (x \ln x)= \lim\limits_{x\rightarrow 0} \frac{\ln x}{\frac{1}{x}}\)

Full Solution

\(\large \lim\limits_{x\rightarrow 0} (x \ln x)= \lim\limits_{x\rightarrow 0} \frac{\ln x}{\frac{1}{x}}=\frac{-\infty}{\infty}\)

This is in the indeterminate form

Let \(\large f(x) = \ln x\) and \(\large g(x) = x^{-1}\)

Then, \(\large f'(x) = \frac{1}{x}\) and \(\large g'(x) = -x^{-2}=-\frac{1}{x^2}\)

Therefore,

\( \lim\limits_{x\rightarrow 0} \frac{\ln x}{\frac{1}{x}}= \lim\limits_{x\rightarrow 0} \frac{\frac{1}{x}}{-\frac{1}{x^2}}\)

We can simplify this

\(\lim\limits_{x\rightarrow 0} \frac{\frac{1}{x}}{-\frac{1}{x^2}}=\lim\limits_{x\rightarrow 0} \frac{-x^2}{x}=\lim\limits_{x\rightarrow 0}(-x)=0\)

Therefore, \(\large \lim\limits_{x\rightarrow 0} (x \ln x)= 0\)

Question 2

HL moderate No calculator

Use l’Hôpital’s rule to determine the value of \(\large \lim\limits_{x\rightarrow 0}\frac{e^{x^2}-1}{\sin x^2}\)

Hint

When you apply l'Hôpital's Rule, simplify your answer.

Full Solution

\(\large \lim\limits_{x\rightarrow 0}\frac{e^{x^2}-1}{\sin x^2}=\frac{1-1}{0}=\frac{0}{0}\)

This is in the indeterminate form

Let \(\large f(x) = e^{x^2}-1\) and \(\large g(x) =\sin x^2\)

Then, \(\large f'(x) = 2xe^{ x^2}\) and \(\large g'(x) = 2x\cos x^2\)

Therefore,

\(\large \lim\limits_{x\rightarrow 0}\frac{e^{x^2}-1}{\sin x^2}= \lim\limits_{x\rightarrow 0}\frac{2xe^{x^2}}{2x\cos x^2}\)

We can simplify this

\(\large \lim\limits_{x\rightarrow 0}\frac{e^{x^2}}{\cos x^2}=\frac{1}{1}=1\)

Question 3

HL difficult No calculator

Use l’Hôpital’s rule to determine the value of \(\large \lim\limits_{x\rightarrow 0}\frac{1-\cos x^2}{x^4}\)

Hint

You will need to apply l'Hôpital's rule twice.

The second application will need simplifying

Full Solution

\(\large \lim\limits_{x\rightarrow 0}\frac{1-\cos x^2}{x^4}=\frac{1-1}{0}=\frac{0}{0}\)

This is in the indeterminate form

Let \(\large f(x) = 1-\cos x^2\) and \(\large g(x) = x^4\)

Then, \(\large f'(x) = 2x\sin x^2\) and \(\large g'(x) = 4x^3\)

Therefore,

\(\large \lim\limits_{x\rightarrow 0}\frac{1-\cos x^2}{x^4}= \lim\limits_{x\rightarrow 0}\frac{2x\sin x^2}{4x^3}\)

We can simplify this

\(\large \lim\limits_{x\rightarrow 0}\frac{\sin x^2}{2x^2}=\frac{0}{0}\)

We can apply l'Hôpital's rule again

Let \(\large f(x) = \sin x^2\) and \(\large g(x) = 2x^2\)

Then, \(\large f'(x) = 2x\cos x^2\) and \(\large g'(x) = 4x\)

Therefore, \(\large \lim\limits_{x\rightarrow 0}\frac{\sin x^2}{2x^2}= \lim\limits_{x\rightarrow 0}\frac{2x\cos x^2}{4x}\)

We can simplify this

\(\large \lim\limits_{x\rightarrow 0}\frac{\cos x^2}{2}=\frac{1}{2}\)

Therefore, \(\large \lim\limits_{x\rightarrow 0}\frac{1-\cos x^2}{x^4}=\frac{1}{2}\)

Question 4

HL difficult No calculator

Use l’Hôpital’s rule to determine the value of \(\large \lim\limits_{x\rightarrow 0}\frac{6\tan x-6x}{x^3}\)

Hint

You will need to apply l'Hôpital's rule three times.

Ensure that you simplify the limit if it is possible.

Full Solution

\(\large \lim\limits_{x\rightarrow 0}\frac{6\tan x-6x}{x^3}=\frac{0-0}{0}=\frac{0}{0}\)

This is in the indeterminate form

Let \(\large f(x) = 6\tan x-6x\) and \(\large g(x) = x^3\)

Then, \(\large f'(x) = 6 \sec ^2x-6\) and \(\large g'(x) = 3x^2\)

Therefore,

\(\large \lim\limits_{x\rightarrow 0}\frac{6\tan x-6x}{x^3}= \lim\limits_{x\rightarrow 0}\frac{\frac{6}{(\cos x)^2}-6}{3x^2}=\frac{0}{0}\)

We can apply l'Hôpital's rule again

Let

\(\large f(x) = \frac{6}{(\cos x)^2}-6\\ \large f(x) =6(\cos x)^{-2}-6\) and \(\large g(x) = 3x^2\)

Then,

\(\large f'(x) =6(-2)(-\sin x)(\cos x)^{-3}\\ \large f'(x)=\frac{12\sin x}{(\cos x)^3}\) and \(\large g'(x) = 6x\)

Therefore, \(\large \lim\limits_{x\rightarrow 0}\frac{6\tan x-6x}{x^3}= \lim\limits_{x\rightarrow 0}\frac{\frac{12\sin x}{(\cos x)^3}}{6x}\)

We can simplify this

\(\large \lim\limits_{x\rightarrow 0}\frac{2\sin x}{x(\cos x)^3}\)

We can apply l'Hôpital's rule again

Let

\(\large f(x) = 2\sin x\) and \(\large g(x) = x(\cos x)^3\)

Then,

\(\large f'(x) =2\cos x\) and \(\large g'(x) = 1(\cos x)^3+x\cdot3(-\sin x)(\cos x)^2\\ \large g'(x) = (\cos x)^3-3x\sin x(\cos x)^2\)

Therefore, \(\large \lim\limits_{x\rightarrow 0}\frac{6\tan x-6x}{x^3}=\lim\limits_{x\rightarrow 0}\frac{2\cos x}{ (\cos x)^3-3x\sin x(\cos x)^2}=\frac{2}{1-0}=2\)

\(\large \lim\limits_{x\rightarrow 0}\frac{6\tan x-6x}{x^3}=2\)

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