TITLE: How to prove l'Hospital's rule for $\infty/\infty$
QUESTION [6 upvotes]: I'm having trouble with this l'Hospital's rule wiki page(the proof of l'Hospital's rule):
http://en.wikipedia.org/wiki/LHospital%27s_rule
Well, in the case where the limit looks like $0/0$, it's quite easy to understand.
On the contrary, the other case( $\infty/\infty$ case) is puzzling. Notations like 'liminf', 'limsup', etc. confuse me further.(Can't understand)
It would be of great help if the proof can be explained step by step.

REPLY [13 votes]: The case $\frac{0}{0}$ is an immediate consequence of Cauchy's Mean value Theorem.
$\frac{\infty}{\infty}$ can also be proven the same way, but it is a little more technical since you have to be careful with the interval where you apply this Theorem. 
Lets see if I can remember it:
Proof for $\frac{\infty}{\infty}$
Let $\lim_{x \to c} f(x) =\lim_{x \to c} g(x) =+\infty$ (the other cases can be obtained from this by replaceing $f,g$ by $\pm f, \pm g$. 
Assume $\lim_{x \to c} \frac{f'(x)}{g'(x)}=l$ and that $g'$ doesn't vanish near $c$.
I will prove that $\lim_{x \to c^-} \frac{f(x)}{g(x)}=l$, the other one side limit is identical.
Let $\epsilon >0$. Then, there exists a $\delta>0$ such that 
$$\left| \frac{f'(x)}{g'(x)}- l \right| < \epsilon $$
for all $c- \delta < x <c$.
By Cauchy Mean Value Theorem, for each $c- \delta < x <c$ there exists some $y_x \in (c_\epsilon, x)$ such that
$$\frac{f(x)-f(c- \delta)}{g(x)-g(c-\delta)} = \frac{f'(y_x)}{g'(y_x)}$$
Therefore, for each $\epsilon >0$, there exists some $\delta$ such that for all $c-\delta < x <c$ we have
$$ \left| \frac{f(x)-f(c- \delta)}{g(x)-g(c-\delta)}- l \right| < \epsilon $$
Now, use the fact that $\lim_{x \to c} f(x) =\lim_{x \to c} g(x) =+\infty$ to prove that
$$\lim_{x \to c} \left( \frac{f(x)-f(c- \delta)}{g(x)-g(c-\delta)} -\frac{f(x)}{g(x)} \right) =0$$
Therefore, there exists some $\delta' < \delta$ so that for all $c- \delta' < x <c$ we have
$$  \left| \frac{f(x)-f(c- \delta)}{g(x)-g(c-\delta)} -\frac{f(x)}{g(x)} \right| < \epsilon \,.$$
Combining the two inequalities you get for $c- \delta' < x <c$:
$$ \left| \frac{f(x)}{g(x)} - l \right| < 2\epsilon \,.$$
Added: To prove
$$\lim_{x \to c} \left( \frac{f(x)-f(c- \delta)}{g(x)-g(c-\delta)} -\frac{f(x)}{g(x)} \right) =0$$
Note that 
$$ \frac{f(x)-f(c- \delta)}{g(x)-g(c-\delta)} -\frac{f(x)}{g(x)} = \frac{f(x)g(c-\delta)-g(x)f(c-\delta)}{g(x)\left( g(x)-g(c-\delta)\right)}\\
=\frac{f(x)g(c-\delta)}{g(x)\left( g(x)-g(c-\delta)\right)}-\frac{f(c-\delta)}{g(x)-g(c-\delta)}$$
It is clear that the second fraction goes to $0$, the first fraction requires just a bit of effort.
Note that at this point in the argument, $\epsilon$ and $\delta$ are fixed.
Now, for all $c-\delta < x <c$ we have
$$ \left| \frac{f(x)-f(c- \delta)}{g(x)-g(c-\delta)}- l \right| < \epsilon$$
therefore, for all $c-\delta < x <c$ we have
$$|f(x)| \leq |f(x)-f(c- \delta)| +|f(c- \delta)| < (\epsilon+ |l|) |g(x)-g(c-\delta)|+|f(c- \delta)| \,.$$
Using this inequality, you get immediately that the first fraction also goes to $0$.