# short explanation of l'Hôpital's rule

#### idontknow

Given limit $$\displaystyle \lim_{x\rightarrow \infty} \frac{x}{\sqrt{1+x^2 }}$$, applying l'Hôpital's rule it repeats itself but also turns into an equation.

$$\displaystyle 0<L=\lim_{x\rightarrow \infty} \frac{x'}{(\sqrt{1+x^2 })'}=\lim_{x\rightarrow \infty } \frac{\sqrt{1+x^2 }}{x}=\frac{1}{L} \;$$ ; $$\displaystyle \; L^2 =1 \: \Rightarrow L=1$$.

Also if the limit involves the factorial then l'Hôpital's rule can make it easier since $$\displaystyle x!=\Gamma (x+1) \approx x^x e^{-x} \sqrt{2\pi x } \;$$ , as $$\displaystyle x\rightarrow \infty$$.

Example : $$\displaystyle l=\lim_{n\rightarrow \infty} \frac{n^n }{n!} =\lim_{n\rightarrow \infty} \frac{n^n }{\Gamma(n+1) } = \lim_{n\rightarrow \infty} \frac{n^n }{n^n e^{-n} \sqrt{2\pi n }}=\lim_{n\rightarrow \infty} \frac{e^n }{\sqrt{2\pi n }}$$.
$$\displaystyle l=\frac{1}{\sqrt{2\pi }}\cdot \lim_{n=t^2 \rightarrow \infty} \frac{(e^{t^2})'}{t'}=\frac{1}{\sqrt{2\pi }}\cdot \lim_{n=t^2 \rightarrow \infty} 2te^{t^2} =\infty$$.

#### idontknow

Another one that turns out to be an equation:
 $L= \displaystyle\lim_{x \to \infty}\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} = \lim_{x \to \infty}\frac{1-e^{-2x}}{1+e^{-2x}} = \frac{1+0}{1+0} = 1$.
With l'Hôpital's rule: $$\displaystyle 0\leq L= \lim_{x \to \infty}\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}= \lim_{x \to \infty}\frac{(e^{x}-e^{-x})'}{(e^{x}+e^{-x})'}= \lim_{x \to \infty}\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}=\frac{1}{L} \; \Rightarrow L=1.$$

#### romsek

Math Team
why does L'Hopital's rule work in general?

topsquark

#### idontknow

for $$\displaystyle \lim_{x\rightarrow a } \dfrac{f(x)}{g(x)}=\dfrac{0 }{0}$$ then $$\displaystyle f(x)=f(x)-f(a)=(x-a)f'(a).$$ same for g(x) .