This is used to calculate limits of the form $\lim _{x \rightarrow c} \frac{f(-x)}{g(x)}$, where $c$ is an extended real number, ie. $c$ is either a real number or $\pm \infty$.

Special case: Suppose $f(x)$ and $g(x)$ are continuously differentiable functions in some interval I containing $C$.

Also, assume that $f(c)=g(c)=0$.

And, $~ g^{\prime}(c) \neq 0$.

Then,

$\begin{aligned}\frac{f(x)}{g(x)} & =\frac{f(x)-f(c)}{g(x)-g(c)} \quad(\because f(c)=0=g(c)) \\& =\frac{\frac{f(x)-f(c)}{x-c}}{\frac{g(x)-g(c)}{x-c}} \quad \text { if } \quad x \in I, x \neq c .\end{aligned}$

Now,

$\lim _{x \rightarrow c} \frac{f(x)-f(c)}{x-c}=f^{\prime}(c)$

$\begin{aligned}& \text { and } \lim _{x \rightarrow c} \frac{g(x)-g(c)}{x-c}=g^{\prime}(c) \neq 0 \\\therefore \quad & \lim _{x \rightarrow c} \frac{\left(\frac{f(x)-f(c)}{x-c}\right)}{\frac{g(x)-g(c)}{x-c}}=\frac{f^{\prime}(c)}{g^{\prime}(c)}\end{aligned}$

$\therefore \quad \lim _{x \rightarrow c} \frac{f(x)}{g(x)}=\frac{f^{\prime}(c)}{g^{\prime}(c)}=\frac{\lim _{x \rightarrow c} f^{\prime}(x)}{\lim _{x \rightarrow c} g^{\prime}(x)}$

because $f^{\prime}(x)$ & $g^{\prime}(x)$ are assumed to be continuous at $x=c$.

Hence, $\lim _{x \rightarrow c} \frac{f(x)}{g(x)}=\lim _{x \rightarrow c} \frac{f^{\prime}(x)}{g^{\prime}(x)}$

The above rule, which is valid for more general cases, is known as the L'Hôpital's rule.

L'Hopital's : French mathematician (pronounced is lopital)

Suppose $\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c} g(x)=0$ or $\pm \infty$.

$\left(i.e. \lim _{x \rightarrow c} \frac{f(x)}{g(x)} \text{is of the form} \frac{0}{0} \text{or} \frac{\infty}{\infty} \right)$

Also, assume that $\lim _{x \rightarrow c} \frac{f^{\prime}(x)}{g^{\prime}(x)}$ exists and $g^{\prime}(x) \neq 0 \quad \forall x$ in the interval I except possibly at $x=c$.

Then, $\lim _{x \rightarrow c} \frac{f(x)}{g(x)}$ exist and

$\lim _{x \rightarrow c} \frac{f(x)}{g(x)}=\lim _{x \rightarrow c} \frac{f^{\prime}(x)}{g^{\prime}(x)} \text {. }$

Solution:

$\lim _{x \rightarrow 0} \frac{\frac{d}{dx} \sin x}{\frac{d}{d x} x}=\lim _{x \rightarrow 0} \frac{\cos x}{1}=\frac{\cos 0}{1}=1 \text {. }$

By the L'Hopital's rule,

$\lim _{x \rightarrow 0} \frac{\sin x}{x}=1 \text {. }$

Note that here we used the fact that

$\frac{d}{d x}(\sin x)=\cos x \text {. }$

$\begin{aligned}& \lim _{x \rightarrow 0} \frac{e^x-1-x}{x^2} \quad \text { ( } \frac{0}{0} \text { form) } \\& \quad \\& \text{Solution:} \\& \quad \\& =\lim _{x \rightarrow 0} \frac{e^x-1}{2 x} \quad \text { (by L'Höpital) } \\& =\lim _{x \rightarrow 0} \frac{e^x}{2} \quad \text { (by L'Hopital's) }\\& =\frac{e^0}{2}=\frac{1}{2} .\end{aligned}$

Remark: We might have to apply L'Hópitals rule several times in order to calculate the limit.

$\lim _{x \rightarrow \infty} \frac{e^x+e^{-x}}{e^{-x}-e^{-x}} \quad\left(\frac{\infty}{\infty}\right. \text { form) }$

$L^{\prime} H=\lim _{x \rightarrow \infty} \frac{e^x-e^{-x}}{e^x+e^{-x}} \quad \text { ( } \frac{\infty}{\infty} \text { form) }$

$L^{\prime} H = \lim _{x \rightarrow \infty} \frac{e^x+e^{-x}}{e^x-e^{-x}} \text { which is the original limit itself. }$

So we wont be able to calculate the limit by applying L'Hopital's rule directly.

However, if we put $e^x=y$, the as $x \rightarrow \infty, y \rightarrow \infty$.

$\begin{aligned}& \quad \frac{e^x+e^{-x}}{e^{-x}-e^{-x}}=\frac{y+\frac{1}{y}}{y-\frac{1}{y}}=\frac{y^2+1}{y^2-1} \\& \therefore \quad \lim _{x \rightarrow \infty} \frac{e^{-x}+e^{-x}}{e^x-e^{-x}}=\lim _{y \rightarrow \infty} \frac{y^2+1}{y^2-1}=\lim _{y \rightarrow \infty} \frac{1+\frac{1}{y^2}}{1-\frac{1}{y}} \\&=\frac{1+0}{1-0}=1 .\end{aligned}$

Or, we could use L'Hopital's rule to write

$\lim _{y \rightarrow \infty} \frac{y^2+1}{y^2-1} \stackrel{L^{\prime} H}{=} \lim _{y \rightarrow \infty} \frac{2 y}{2 y}=1 \text {. }$

$\begin{aligned}& \lim _{x \rightarrow \infty} \frac{\sqrt{x}+\frac{1}{\sqrt{x}}}{\sqrt{x}-\frac{1}{\sqrt{x}}} \quad\left(\frac{\infty}{\infty}\right. \text { form) } \\& \stackrel{L^{\prime} H}{=} \lim _{x \rightarrow \infty} \frac{\frac{1}{2 \sqrt{x}}+\frac{-1}{2} x^{-3 / 2}}{\frac{1}{2 \sqrt{x}}+\frac{1}{2} x^{-3 / 2}} \quad (\frac{0}{0}\text{from})\\ & \lim _{x \rightarrow \infty} \frac{-\frac{1}{4} x^{-3/2} +\frac{3}{4} x^{-5/ 2}}{-\frac{1}{4} x^{-3/2}-\frac{3}{4} x^{-5 / 2}} \quad (\frac{0}{0} \text { form) }) \\&\end{aligned}$

=... becomes more & more complicated.

However, $\begin{aligned} \lim _{x \rightarrow \infty} \frac{\sqrt{x}+\frac{1}{\sqrt{x}}}{\sqrt{x}-\frac{1}{\sqrt{x}}}=\lim _{x \rightarrow \infty} \frac{x+1}{x-1} & \stackrel{L'H}{=} \lim _{x \rightarrow \infty} \frac{1}{1} =1 .\end{aligned}$

L'Hôpital's Rule can be used for other indeterminate forms like

$0 . \infty, \infty-\infty$, $1^{\infty}, 0^{\circ}$, etc. by somehow changing into $\frac{0}{0}$ a $\frac{\infty}{\infty}$ form.

Examples: (1) $\lim _{x \rightarrow \infty} x^2 e^{-x} \quad$ ( $\infty .0$ form)

$\begin{aligned}& =\lim _{x \rightarrow \infty} \frac{x^2}{e^{x}} \quad \text { ( } \frac{\infty}{\infty} \text { form) } \\& \stackrel{L^{\prime} H}{=} \lim _{x \rightarrow \infty} \frac{2 x}{e^x} \quad \text { ( } \frac{\infty}{\infty} \text { form) } \\& \stackrel{L^{\prime} H}{=} \lim _{x \rightarrow \infty} \frac{2}{e^x}=0 \text {. } \& \end{aligned}$

More generally, we can show that

$\lim _{x \rightarrow \infty} x^n \cdot e^{-x}=0 \text { for any positive integer n }$

$\lim _{x \rightarrow 0^{+}} x \ln x$

(here we are taking $\lim _{x \rightarrow 0^{+}} ~$ because $\ln x$ is defined only for $x>0$)

As $x \rightarrow 0+, \ln x \rightarrow-\infty \quad$ $y=\ln x$

Let's write $x \ln x=\frac{\ln x}{x^{-1}}$ ( $\frac{-\infty}{\infty}$ form $y$)

$\lim _{x \rightarrow 0^{+}} x \ln x =\lim _{x \rightarrow 0^{+}} \frac{\ln x}{x^{-1}} \\ \stackrel{L^{\prime} H}{=}\lim _{x \rightarrow 0^{+}} \frac{1 / x}{-1 / x^2}=\lim _{x \rightarrow 0+}(-x)=0 .$

$\lim _{x \rightarrow 0^{+}} x^x \quad\left(0^0\right.$ form)

let $f(x)=x^x$. Then $\ln f(x)=x \ln x$

We have seen that $\lim _{x \rightarrow 0^{+}} x \ln x=0$

$\therefore \quad \lim _{x \rightarrow 0^{+}} \ln f(x)=0$.

Now, $\quad f(x)=e^{\ln f(x)}$

$\therefore \lim _{x \rightarrow \infty} f(x)=\lim _{x \rightarrow \infty} e^{\ln f(x)} =e^{\text lim _{x \rightarrow 0^{+}} \ln f(x)}$

$=e^0=1 \text {. } \left(\because e^x\right.$ is continuous)

Remark: Note that if $\lim _{x \rightarrow c} \frac{f^{\prime}(x)}{g^{\prime}(x)}$ does not exist, we cannot conclude that $\lim _{x \rightarrow c} \frac{f(x)}{\partial(x)}$ does not exist.

Example: $f(x)=x+\sin x ; g(x)=x$

$\begin{aligned}& \lim _{x \rightarrow \infty} f(x)=\infty=\lim _{x \rightarrow \infty} g(x) \\& f^{\prime}(x)=1+\cos x, g^{\prime}(x)=1\end{aligned}$

$\lim \frac{f^{\prime}(x)}{g^{\prime}(x)}=\lim _{x \rightarrow \infty}(1+\cos x)$, which does not exist.

However,

$\text { e. } \begin{aligned}& \lim _{x \rightarrow \infty} \frac{f(x)}{g(x)}=\lim _{x \rightarrow \infty} \frac{x+\sin x}{x} \\&=\lim _{x \rightarrow \infty}\left[1+\frac{\sin x}{x}\right] \\& 0 \leq\left|\frac{\sin x}{x}\right| \leq \frac{1}{x} \rightarrow 0 \text { is } x \rightarrow \infty .\end{aligned}$

By the Sandwich the, $\lim _{x \rightarrow \infty} \frac{\sin x}{x}=0$

$\therefore \lim _{x \rightarrow \infty} \frac{f(x)}{g(x)}=1+0=1 \text {. }$