Answer
$$\mathop {\lim }\limits_{x \to c} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to c} \frac{{f'\left( x \right)}}{{g'\left( x \right)}}$$
Work Step by Step
$$\eqalign{
& {\text{L'Hopital's Rule}} \cr
& {\text{Let }}f{\text{ and }}g{\text{ be functions that are differentiable on an open }} \cr
& {\text{interval }}\left( {a,b} \right).{\text{ Assume that }}g \ne 0. \cr
& {\text{If the limit }}\frac{f}{g}{\text{ as }}x{\text{ approaches }}c{\text{ produces the indeterminate }} \cr
& {\text{form }}\frac{0}{0},{\text{ then}} \cr
& \mathop {\lim }\limits_{x \to c} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to c} \frac{{f'\left( x \right)}}{{g'\left( x \right)}} \cr} $$
