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.
On this page, you should learn about
- The indeterminate forms \(\large \frac{0}{0}\)and \(\large \frac{∞}{∞}\).
- Evaluation of limits in the form \(\lim\limits_{x\rightarrow a} \frac{f(x)}{g(x)}\)
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
\(\large\lim\limits_{x\rightarrow 0}\frac{-2\sin x+4\sin 2x}{6x}=\lim\limits_{x\rightarrow 0}\frac{\frac{\mathrm{d}}{\mathrm{d}x}(-2\sin x+4\sin 2x)}{\frac{\mathrm{d}}{\mathrm{d}x}(6x)}=\lim\limits_{x\rightarrow 0}\frac{-2\cos x+8\cos 2x}{6}=\frac{-2+8}{6}=\frac{6}{6}=1\)
Use l’Hôpital’s rule to determine the value of \(\large \lim\limits_{x\rightarrow 0}( x \ln x)\)
Hint
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, \(\large \lim\limits_{x\rightarrow 0} (x \ln x)= 0\)
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
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
\(\large \lim\limits_{x\rightarrow 0}\frac{e^{x^2}}{\cos x^2}=\frac{1}{1}=1\)
Use l’Hôpital’s rule to determine the value of \(\large \lim\limits_{x\rightarrow 0}\frac{1-\cos x^2}{x^4}\)
Hint
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, \(\large \lim\limits_{x\rightarrow 0}\frac{1-\cos x^2}{x^4}=\frac{1}{2}\)
Use l’Hôpital’s rule to determine the value of \(\large \lim\limits_{x\rightarrow 0}\frac{6\tan x-6x}{x^3}\)
Hint
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, \(\large \lim\limits_{x\rightarrow 0}\frac{6\tan x-6x}{x^3}=\lim\limits_{x\rightarrow 0}\frac{2\cos x}{ (\cos x)^3-3x\sin x(\cos x)^2}=\frac{2}{1-0}=2\)
\(\large \lim\limits_{x\rightarrow 0}\frac{6\tan x-6x}{x^3}=2\)
How much of L'Hôpital's Rule have you understood?
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