L’Hôpital’s Rule is all about limits that are not easy to evaluate using the usual methods.
For example, consider the limit $\displaystyle \lim_{x \rightarrow 0} \frac{\sin x}{x}=1.$
Evaluating Limits of the Form $\displaystyle \lim_{x \rightarrow a} \frac{f(x)}{g(x)}$ when $\displaystyle \lim_{x \rightarrow a}f(x) = \lim_{x \rightarrow a} g(x)=0$
If $f$ and $g$ are differentiable at $x=a$ and $\displaystyle \lim_{x \rightarrow a}f(x) = \lim_{x \rightarrow a} g(x)=0,$ we can approximate near $a$ using the linear approximations of $f$ and $g.$
L’Hôpital’s Rule
Suppose $f$ and $g$ are differentiable functions over an open interval containing $a,$ except possibly at $a.$ If $\displaystyle \lim_{x\rightarrow a}f(x)= 0$ and $\displaystyle \lim_{x\rightarrow a} g(x)= 0,$ then $$\lim_{x\rightarrow a}\frac{f(x)}{g(x)}= \lim_{x\rightarrow a} \frac{f'(x)}{g'(x)},$$assuming the limit on the right exists or is $\infty$ or $−\infty.$
Example
The textbook provides a geometric argument that $\displaystyle \lim_{x \rightarrow 0} \frac{\sin x}{x}=1.$
Use L’Hôpital’s Rule to evaluate this limit.
$$
\begin{array}{ll}
\displaystyle \lim_{x \rightarrow 0} \frac{\sin x}{x}&\mbox{ in $\frac{0}{0}$ form}\\
=\displaystyle \lim_{x \rightarrow 0} \frac{(\sin x)'}{x'}&\mbox{apply L'Hôpital}\\
=\displaystyle \lim_{x \rightarrow 0} \frac{\cos x}{1}&\mbox{}\\
=\displaystyle \lim_{x \rightarrow 0} \cos x &\mbox{}\\
=1 &\mbox{}\\
\end{array}
$$
Other Examples
$\displaystyle \lim_{x \rightarrow 0} \frac{3x-\sin x}{x}$
$$
\begin{array}{ll}
\displaystyle \lim_{x \rightarrow 0} \frac{3x-\sin x}{x}&\mbox{in $\frac{0}{0}$ form}\\
=\displaystyle \lim_{x \rightarrow 0}\frac{(3x-\sin x)'}{x'}&\mbox{apply L'Hôpital}\\
=\displaystyle \lim_{x \rightarrow 0}\frac{3-\cos x}{1}&\mbox{}\\
=\displaystyle \lim_{x \rightarrow 0}(3-\cos x)&\mbox{}\\
=3-1&\mbox{}\\
=2&\mbox{}\\
\end{array}
$$
$\displaystyle \lim_{x \rightarrow 0} \frac{\sqrt{x+1}-1}{x}$
$$
\begin{array}{ll}
\displaystyle \lim_{x \rightarrow 0} \frac{\sqrt{x+1}-1}{x}&\mbox{in $\frac{0}{0}$ form}\\
\displaystyle=\lim_{x \rightarrow 0} \frac{(\sqrt{x+1}-1)'}{x'}&\mbox{apply L'Hôpital}\\
\displaystyle=\lim_{x \rightarrow 0} (\sqrt{x+1}-1)'&\mbox{}\\
\displaystyle=\lim_{x \rightarrow 0} \frac{1}{2\sqrt{x+1}}&\mbox{}\\
\displaystyle=\frac{1}{2}&\\
\end{array}
$$
If at First You Don't Succeed...
If you get the indeterminate form $\displaystyle \frac{0}{0}$ again after applying L’Hôpital’s Rule, do not fret, just apply L’Hôpital’s Rule again!
Other Examples
$\displaystyle \lim_{x \rightarrow 0} \frac{\sqrt{x+1}-1-x/2}{x^2}$
$$
\begin{array}{ll}
\displaystyle \lim_{x \rightarrow 0} \frac{\sqrt{x+1}-1-x/2}{x^2}&\mbox{in $\frac{0}{0}$ form}\\
\displaystyle=\lim_{x \rightarrow 0} \frac{\frac{1}{2}(x+1)^{-1/2}-1/2}{2x}&\mbox{still in $\frac{0}{0}$ form!}\\
\displaystyle=\lim_{x \rightarrow 0} \frac{-\frac{1}{4}(x+1)^{-3/2}}{2}&\mbox{}\\
\displaystyle=-\frac{1}{8}&\mbox{}\\
\end{array}
$$
$\displaystyle \lim_{x \rightarrow 0} \frac{x-\sin x}{x^3}$
$$
\begin{array}{ll}
\displaystyle \lim_{x \rightarrow 0} \frac{x-\sin x}{x^3}&\mbox{ in $\frac{0}{0}$ form}\\
\displaystyle=\lim_{x \rightarrow 0} \frac{1-\cos x}{3x^2}&\mbox{ still in $\frac{0}{0}$ form!}\\
\displaystyle=\lim_{x \rightarrow 0} \frac{\sin x}{6x}&\mbox{still in $\frac{0}{0}$ form!!}\\
\displaystyle=\lim_{x \rightarrow 0} \frac{\cos x}{6}&\mbox{}\\
\displaystyle=\frac{1}{6}&\mbox{}\\
\end{array}
$$
A Word of Caution
Consider the example $\displaystyle \lim_{x \rightarrow 0}\frac{1-\cos x}{x+x^2}=\lim_{x \rightarrow 0}\frac{\sin x}{1+2x}.$
It is oh so very tempting to differentiate again and say that $$\displaystyle \lim_{x \rightarrow 0}\frac{1-\cos x}{x+x^2}=\lim_{x \rightarrow 0}\frac{\sin x}{1+2x}=\lim_{x \rightarrow 0}\frac{\cos x}{2}=\frac{1}{2},$$ but this is WRONG!
(Not wrong enough for angry kitty to make an appearance though. Whew!)
A Word of Caution
Notice that we could have simply evaluated the limit after our first round of differentiation $$\displaystyle \lim_{x \rightarrow 0}\frac{1-\cos x}{x+x^2}=\lim_{x \rightarrow 0}\frac{\sin x}{1+2x}=\frac{0}{1}=0.$$
Great News!
L’Hôpital’s Rule works for one-sided limits too!
Examples
$\displaystyle \lim_{x \rightarrow 0^{+}} \frac{\sin x}{x^2}$
$$
\begin{array}{ll}
\displaystyle \lim_{x \rightarrow 0^{+}} \frac{\sin x}{x^2}&\mbox{in $\frac{0}{0}$ form}\\
\displaystyle=\lim_{x \rightarrow 0^{+}} \frac{\cos x}{2x}&\mbox{apply L'Hôpital}\\
\displaystyle=\infty&\mbox{}\\
\end{array}
$$
$\displaystyle \lim_{x \rightarrow 0^{-}} \frac{\sin x}{x^2}$
$$
\begin{array}{ll}
\displaystyle \lim_{x \rightarrow 0^{-}} \frac{\sin x}{x^2}&\mbox{in $\frac{0}{0}$ form}\\
\displaystyle=\lim_{x \rightarrow 0^{-}} \frac{\cos x}{2x}&\mbox{apply L'Hôpital}\\
\displaystyle=-\infty&\mbox{}\\
\end{array}
$$
More Great News!
L’Hôpital’s Rule also works for the indeterminate form $\displaystyle \frac{\infty}{\infty}.$
Examples
$\displaystyle \lim_{x \rightarrow (\pi/2)^-} \frac{\sec x }{1+\tan x}$
$$
\begin{array}{ll}
\displaystyle \lim_{x \rightarrow (\pi/2)^-} \frac{\sec x }{1+\tan x}&\mbox{in $\frac{\infty}{\infty}$ form}\\
=\displaystyle \lim_{x \rightarrow (\pi/2)^-} \frac{\sec x \tan x}{\sec^2 x}&\mbox{apply L'Hôpital}\\
=\displaystyle \lim_{x \rightarrow (\pi/2)^-} \frac{\tan x}{\sec x}&\mbox{simplify}\\
=\displaystyle \lim_{x \rightarrow (\pi/2)^-} \sin x&\mbox{simplify}\\
=1&\mbox{}\\
\end{array}
$$
$\displaystyle \lim_{x \rightarrow \infty} \frac{\ln x}{2 \sqrt{x}}$
$$
\begin{array}{ll}
\displaystyle \lim_{x \rightarrow \infty} \frac{\ln x}{2 \sqrt{x}}&\mbox{in $\frac{\infty}{\infty}$ form}\\
\displaystyle=\lim_{x \rightarrow \infty} \frac{\frac{1}{x}}{2 \left(\frac{1}{2}x^{-1/2}\right)}&\mbox{apply L'Hôpital}\\
\displaystyle=\lim_{x \rightarrow \infty} \frac{1}{\sqrt{x}}&\mbox{simplify}\\
=0
\end{array}
$$
$\displaystyle \lim_{x \rightarrow \infty} \frac{e^x}{x^2}$
$$
\begin{array}{ll}
\displaystyle \lim_{x \rightarrow \infty} \frac{e^x}{x^2}&\mbox{in $\frac{\infty}{\infty}$ form}\\
\displaystyle=\lim_{x \rightarrow \infty} \frac{e^x}{2x}&\mbox{still in $\frac{\infty}{\infty}$ form!}\\
\displaystyle=\lim_{x \rightarrow \infty} \frac{e^x}{2}&\mbox{}\\
\displaystyle=\infty&\mbox{}\\
\end{array}
$$
The Form $0 \cdot \infty$
If you ever run into the indeterminate form $0\cdot \infty,$ we can often massage it into either a $\displaystyle \frac{0}{0}$ form or $\displaystyle \frac{\infty}{\infty}$ form.
Examples
$\displaystyle \lim_{x \rightarrow \infty} \left(x \sin \frac{1}{x}\right)$
$$
\begin{array}{lll}
\displaystyle \lim_{x \rightarrow \infty} \left(x \sin \frac{1}{x}\right)
&=\displaystyle \lim_{x \rightarrow \infty} \frac{\sin \frac{1}{x}}{1/x}&\mbox{massage $\infty \cdot 0$ into $\frac{0}{0}$ form}\\
&=\displaystyle \lim_{x \rightarrow \infty} \frac{(\cos \frac{1}{x})\cdot \frac{-1}{x^2}}{-1/x^2}&\mbox{using L'Hôpital's rule}\\
&=\displaystyle \lim_{x \rightarrow \infty} \cos \frac{1}{x}&\\
&=1&
\end{array}
$$
$\displaystyle \lim_{x \rightarrow 0^{+}} \sqrt{x}\ln x$
$$
\begin{array}{lll}
\displaystyle \lim_{x \rightarrow 0^{+}} \sqrt{x}\ln x &=\displaystyle \lim_{x \rightarrow 0^{+}} \frac{\ln x}{1/\sqrt{x}}&\mbox{massage $0 \cdot \infty$ into $\frac{\infty}{\infty}$ form}\\
&=\displaystyle \lim_{x \rightarrow 0^{+}} \frac{\ln x}{x^{-1/2}}&\mbox{rewrite to use power rule}\\
&=\displaystyle \lim_{x \rightarrow 0^{+}} \frac{1/x}{-\frac{1}{2}x^{-3/2}}&\mbox{using L'Hôpital's rule}\\
&=\displaystyle -2\lim_{x \rightarrow 0^{+}} \frac{x^{3/2}}{x}&\mbox{}\\
&=\displaystyle -2\lim_{x \rightarrow 0^{+}} x^{1/2}&\mbox{}\\
&=\displaystyle -2\cdot 0&\mbox{}\\
&=0&\mbox{}\\
\end{array}
$$
The Form $\infty - \infty$
If you ever run into the indeterminate form $\infty - \infty,$ again we can often massage it into either a $\displaystyle \frac{0}{0}$ form or $\displaystyle \frac{\infty}{\infty}$ form.
Example: $\,\,\,\,\displaystyle \lim_{x \rightarrow 0} \left(\frac{1}{\sin x}-\frac{1}{x}\right)$
$$
\begin{array}{lll}
\displaystyle \lim_{x \rightarrow 0} \left(\frac{1}{\sin x}-\frac{1}{x}\right) &=\displaystyle \lim_{x \rightarrow 0} \left(\frac{x}{x\sin x}-\frac{\sin x}{x\sin x}\right)&\mbox{in $\infty-\infty$ form}\\
&=\displaystyle \lim_{x \rightarrow 0} \frac{x-\sin x}{x\sin x}&\mbox{now in $\frac{0}{0}$ form!}\\
&=\displaystyle \lim_{x \rightarrow 0} \frac{1-\cos x}{(x)'(\sin x)+(x)(\sin x)'}&\mbox{apply L'Hôpital}\\
&=\displaystyle \lim_{x \rightarrow 0} \frac{1-\cos x}{\sin x+x\cos x}&\mbox{still in $\frac{0}{0}$ form!}\\
&=\displaystyle \lim_{x \rightarrow 0} \frac{\sin x}{\cos x+(x)'(\cos x)+(x)(\cos x)'}&\mbox{apply L'Hôpital again!}\\
&=\displaystyle \lim_{x \rightarrow 0} \frac{\sin x}{\cos x+\cos x-x\sin x}&\mbox{}\\
&=\displaystyle \frac{0}{2}&\\
&=\displaystyle 0&\\
\end{array}
$$
Other Indeterminate Forms: $1^{\infty},$ $0^0,$ and $\infty^0$
Example: $\,\,\,\,\displaystyle \lim_{x \rightarrow 0^+} \left(1+x\right)^{1/x}.$
This is a $1^{\infty}$ indeterminate form.
We will first consider the a different limit, $\displaystyle \lim_{x \rightarrow 0^+} \ln\left(\left(1+x\right)^{1/x}\right):$
$$
\begin{array}{lll}
\displaystyle \lim_{x \rightarrow 0^+} \ln\left(\left(1+x\right)^{1/x}\right)&=\displaystyle \lim_{x \rightarrow 0^+} \frac{1}{x}\ln\left(1+x\right)&\mbox{}\\
&=\displaystyle \lim_{x \rightarrow 0^+} \frac{\ln\left(1+x\right)}{x}&\mbox{now in $\frac{0}{0}$ form!}\\
&=\displaystyle \lim_{x \rightarrow 0^+} \frac{\frac{1}{x+1}}{1}&\mbox{using L'Hôpital!}\\
&=\displaystyle \lim_{x \rightarrow 0^+} \frac{1}{x+1}&\mbox{}\\
&=1&\\
\end{array}
$$
We can now evaluate the original limit by using the Composite Function Theorem:
$$
\begin{array}{lll}
&\displaystyle \lim_{x \rightarrow 0^+} \ln\left(\left(1+x\right)^{1/x}\right)=1&\mbox{}\\
\implies & \ln\left( \displaystyle \lim_{x \rightarrow 0^+} \left(1+x\right)^{1/x}\right)=1&\mbox{ by the Composite Function Theorem}\\
\implies & \displaystyle \lim_{x \rightarrow 0^+} \left(1+x\right)^{1/x}=e&\mbox{}\\
\end{array}
$$
Other Indeterminate Forms: $1^{\infty},$ $0^0,$ and $\infty^0$
Theorem: If $\displaystyle \lim_{x \rightarrow a} \ln f(x)=L,$ then $$\lim_{x \rightarrow a} f(x)=e^L.$$ In this case, $a$ may be finite of infinite.
Example: $\,\,\,\,\displaystyle \lim_{x \rightarrow \infty} x^{1/x}.$
This is an $\infty^0$ indeterminate form
$$
\begin{array}{lll}
\displaystyle \lim_{x \rightarrow \infty} \ln\left(x^{1/x}\right)&=\displaystyle \lim_{x \rightarrow \infty} \frac{1}{x}\ln\left(x\right)&\mbox{}\\
&=\displaystyle \lim_{x \rightarrow \infty} \frac{\ln\left(x\right)}{x}&\mbox{now in $\frac{\infty}{\infty}$ form!}\\
&=\displaystyle \lim_{x \rightarrow \infty} \frac{1/x}{1}&\mbox{applying L'Hôpital!}\\
&=0&\mbox{}\\
\end{array}
$$
Then,
$$
\begin{array}{lll}
&\displaystyle \lim_{x \rightarrow \infty} \ln\left(x^{1/x}\right)=0&\mbox{}\\
\implies & \ln\left(\displaystyle \lim_{x \rightarrow \infty} \left(x^{1/x}\right)\right)=0&\mbox{ by the Composite Function Theorem}\\
\end{array}
$$
Therefore, $\displaystyle \lim_{x \rightarrow \infty} x^{1/x}=e^0=1.$
And one more thing...
Dire Warning
WHEN YOU DO THIS
$\displaystyle \lim_{x \rightarrow 0} \frac{\sin x}{x}=\frac{0}{0}$
YOU MAKE KITTY ANGRY!!!!!!!!!!!!!!!!!!
Please don't treat indeterminate forms like numbers. They aren't. Plus, it makes angry kitty even angrier.