Suppose $f$ and $g$ are continuous and differentiable for all $x\neq a$, and that g' is never zero. Suppose that $\mathop {\lim }\limits_{x \to a } f\left( x \right) = \mathop {\lim }\limits_{x \to a } g\left( x \right) = \infty $
and that the limit
\mathop {\lim }\limits_{x \to a } \frac{{f'\left( x \right)}}{{g'\left( x \right)}} = L
exists. Then \mathop {\lim }\limits_{x \to a } \frac{{f'\left( x \right)}}{{g'\left( x \right)}} = \mathop {\lim }\limits_{x \to a } \frac{{f\left( x \right)}}{{g\left( x \right)}} = L
According to Cauchy we have ($\alpha < x < a$ and $\alpha < c < x$ )
\frac{{f\left( x \right) - f\left( \alpha \right)}}{{g\left( x \right) - g\left( \alpha \right)}} = \frac{{f'\left( c \right)}}{{g'\left( c \right)}}
Let's write the first member of the equality as
\frac{{f\left( x \right)}}{{g\left( x \right)}}\frac{{1 - \frac{{f\left( \alpha \right)}}{{f\left( x \right)}}}}{{1 - \frac{{g\left( \alpha \right)}}{{g\left( x \right)}}}} = \frac{{f'\left( c \right)}}{{g'\left( c \right)}}
Then
\frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{{f'\left( c \right)}}{{g'\left( c \right)}}\frac{{1 - \frac{{g\left( \alpha \right)}}{{g\left( x \right)}}}}{{1 - \frac{{f\left( \alpha \right)}}{{f\left( x \right)}}}}(\star)
From the condition \mathop {\lim }\limits_{x \to \infty } \frac{{f'\left( x \right)}}{{g'\left( x \right)}} = L we get that for any $\epsilon > 0$, we can choose $\alpha$ arbitrarely close to $a$ such that for all $x=c$ where $\alpha < c < a$, we have
\left| \frac{{f'\left( c \right)}}{{g'\left( c \right)}}-L\right| < \epsilon
L-\epsilon <\frac{{f'\left( c \right)}}{{g'\left( c \right)}} < L+\epsilon
Now, since
$\mathop {\lim }\limits_{x \to a} \frac{{1 - \frac{{g\left( \alpha \right)}}{{g\left( x \right)}}}}{{1 - \frac{{f\left( a \right)}}{{f\left( \alpha \right)}}}} = 1$
We have
$1-\epsilon < \frac{{1 - \frac{{g\left( \alpha \right)}}{{g\left( x \right)}}}}{{1 - \frac{{f\left( a \right)}}{{f\left( \alpha \right)}}}} < 1+\epsilon$
Multiplying the previous inequalities gives
\left( {L - \epsilon} \right)\left( {1 - \epsilon} \right) < \frac{{f'\left( c \right)}}{{g'\left( c \right)}}\frac{{1 - \frac{{g\left( \alpha \right)}}{{g\left( x \right)}}}}{{1 - \frac{{f\left( a \right)}}{{f\left( \alpha \right)}}}} < \left( {L + \epsilon} \right)\left( {1 + \epsilon} \right)
and because of $(\star)$ we get
$\left( {L - \epsilon} \right)\left( {1 - \epsilon} \right) < \frac{{f\left( x \right)}}{{g\left( x \right)}} < \left( {L + \epsilon} \right)\left( {1 + \epsilon} \right)$
Since $\epsilon$ is arbitrarely small, for $x$ sufficiently close to $a$ we will have
$\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = L$