# L'Hôpital's Rule
###### tags: `Calculus`
#### <proof>
$$
\lim_{x\rightarrow p}f(x)=\lim_{x\rightarrow p}g(x)=0\\Let\;\widetilde f(x)\left\{\begin{array}{l}f(x)\;,\;x\neq p\\0\;,\;x=p\end{array}\Rightarrow\widetilde f\;is\;continuous\;up\;to\;p\right.\\Let\;\widetilde g(x)\left\{\begin{array}{l}g(x)\;,\;x\neq p\\0\;,\;x=p\end{array}\Rightarrow\widetilde g\;is\;continuous\;up\;to\;p\right.\\\frac{f(x)}{g(x)}=\frac{\widetilde f(x)}{\widetilde g(x)}=\frac{\widetilde f(x)-\widetilde f(p)}{\widetilde g(x)-\widetilde g(p)}\overset{Cauchy's\;M.V.T}{\underset{as\;x\rightarrow p,\;\widehat x\rightarrow p}=}\frac{\widetilde f'(\widehat x)}{\widetilde g'(\widehat x)}=\frac{f'(\widehat x)}{g'(\widehat x)}=L
$$