another form of the L'Hospital's rule I need some help, any Ideas?
in L'Hospital's rule, replace the assumption that $\frac{f}{g}$ tends to $\frac{0}{0}$ with the assumption that it tends to $\frac{\infty}{\infty}$. if $\frac{f'}{g'}$ tends to $L$. prove that $\frac{f}{g}$ tends to $L$ also.
 A: 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
$$\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = L$$
