Question

if $f$ and $g$ are differentiable at $a\in R$ such that $f\left(a\right)=g\left(a\right)=0$ and $g\left(a\right)\ne 0$ then show that ${lim}_{x\to a}\frac{f\left(x\right)}{g\left(x\right)}=\frac{f^{\prime }\left(a\right)}{g^{\prime }\left(a\right)}$.

Answer 1

$\underset{x\to a}{lim}\frac{f\left(x\right)}{g\left(x\right)}$ ($\frac{0}{0}$ form)

Using L' Hospital rule

$=\underset{x\to a}{lim}\frac{f^{\prime }\left(x\right)}{g^{\prime }\left(x\right)}$

$=\frac{f^{\prime }\left(a\right)}{g^{\prime }\left(a\right)}$