Show that : $ \frac{f'(x)}{g'(x)}\xrightarrow[x \to a]{} l \implies \frac{f(x)}{g(x)}\xrightarrow[x \to a]{} l\quad (l\in \mathbb{\bar{R}}) $ Let $f$,$g\in \mathcal{C}((a,b),\mathbb{R})$ are differentiable on $(a,b)$


we assume that $f(a)=g(a)=0$ and $g'\neq  0  $ on $(a,b)$ 


Show that $$ \frac{f'(x)}{g'(x)}\xrightarrow[x \to a]{} l \implies \frac{f(x)}{g(x)}\xrightarrow[x \to a]{} l\quad  (l\in \mathbb{\bar{R}}) $$

My thoughts

by Cauchy's mean value theorem : $\exists\ C\in (a,b)$ such that $\left(f(b)-f(a)\right)g'(c)=\left(g(a)-g(b)\right)f'(c)$

since $g'\neq 0$ then $\dfrac{f(b) - f(a)}{g(b) - g(a)} = \dfrac{f'(c)}{g'(c)}.$ 

or $f(a)=g(a)=0$ then $\dfrac{f(b)}{g(b)} = \dfrac{f'(c)}{g'(c)}.$

i want to use L'Hôpitals rule but i'm stuck
Clearly 
$$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f(x) - f(a)}{g(x) -g(a)}.$$

For each $x &gt; a$, use Cauchy's Mean Value Theorem on $(a,x)$ to say that there exists $c = c(x) \in (a,x)$ such that $\frac{f'(c)}{g'(c)} = \frac{f(x)-f(a)}{g(x) - g(a)}$. Notice that $c(x) \to a$ as $x \to a$, so that

$$\lim_{x \to a}\frac{f(x) - f(a)}{g(x) -g(a)} = \lim_{x\to a} \frac{f'(c(x))}{g'(c(x))} = \lim_{c \to a} \frac{f'(c)}{g'(c)} = l$$
As $f(a)=0$ &amp; $g(a)=0$ &amp; $g'(a)\not =0$ so by L'Hospital rule,$$\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}=l.$$