Let $x_0\in \R$ $I$ be an interval containing $x_0$ and let $f$ and $g$ be two differentiable functions defined on $I\setminus\{x_0\}$ with $g'(x)\neq 0$ for all $x\in I$ Suppose that $$ \lim_{x\to x_0} f(x) = 0, \quad \lim_{x\to x_0} g(x)=0 $$ and that $$ \lim_{x\to x_0} \frac{f'(x)}{g'(x)}=m. $$

We want to prove that hence $g(x)\neq 0$ for all $x\in I\setminus\{x_0\}$ and $$ \lim_{x\to x_0} \frac{f(x)}{g(x)}=m. $$

First of all (with little abuse of notation) we suppose that $f$ and $g$ are defined also in the point $x_0$ by $f(x_0)=0$ and $g(x_0)=0$ The resulting functions are continuous in $x_0$ and hence in the whole interval $I$

Let us first prove that $g(x)\neq 0$ for all $x\in I\setminus\{x_0\}$ If by contradiction $g(\bar x)=0$ since we also have $g(x_0)=0$ by Rolle's Theorem we get that $g'(\xi)=0$ for some $\xi\in (x_0,\bar x)$ which is against our hypotheses.

Consider now any sequence $x_n\to x_0$ with $x_n\in I\setminus\{x_0\}$ By Cauchy's mean value Theorem there exists a sequence $x'_n$ such that $$ \frac{f(x_n)}{g(x_n)} = \frac{f(x_n)-f(x_0)}{g(x_n)-g(x_0)} = \frac{f'(x'_n)}{g'(x'_n)}. $$ But as $x_n\to x_0$ and since $x'_n \in (x_0,x_n)$ we get that $x'_n\to x_0$ and hence $$ \lim_{n\to\infty} \frac{f(x_n)}{g(x_n)} = \lim_{n\to\infty} \frac{f'(x_n)}{g'(x_n)} = \lim_{x\to x_0} \frac{f'(x)}{g'(x)} = m. $$

Since this is true for any given sequence $x_n\to x_0$ we conclude that $$ \lim_{x\to x_0} \frac{f(x)}{g(x)} = m. $$