How to prove l'Hospital's rule for $\infty/\infty$ 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.
 A: 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 $0$.
A: I think the following proof is a bit simpler than the one given above:
Suppose $\delta>0$ is such that $g(x)\neq 0$ for all $x\in (a-\delta,a+\delta)^{\circ}$ (which is the open interval $(a-\delta,a+\delta)$ with a removed). Now let $\epsilon>0$, and for each $x\in (a-\delta,a)$, let $\delta_x\in (0,\delta)$ be such that $\lim_{x\to a^{-}}\delta_x=0$. Since $\lim_{x\to a^{-}}g(x)=\infty$, for each $x\in (a-\delta,a)$ there exists $\beta_x\in (a-\delta_x,a)$ such that
$$
g(\beta_x)>\frac{f(a-\delta_x)}{\epsilon}\implies \left|\frac{f(a-\delta_x)}{g(\beta_x)}-0\right|<\epsilon \implies \lim_{x\to a^{-}}\frac{f(a-\delta_x)}{g(\beta_x)}=0.
$$
By a similar argument $\beta_x$ can be chosen so that we also have $\lim_{x\to a^{-}}\frac{g(a-\delta_x)}{g(\beta_x)}=0$. Note that this implies $g(\beta_x)-g(a-\delta_x)\neq 0$ for $x$ sufficiently close to $a$. Now by the Cauchy Mean Value Theorem, there exists an $\alpha_x\in (a-\delta_x,\beta_x)$ such that
$$
(f(\beta_x)-f(a-\delta_x))g'(\alpha_x)=(g(\beta_x)-g(a-\delta_x))f'(\alpha_x)\implies \frac{f'(\alpha_x)}{g'(\alpha_x)}=\frac{f(\beta_x)-f(a-\delta_x)}{g(\beta_x)-g(a-\delta_x)}.
$$
We then have
$$
\lim_{x\to a^{-}}\frac{f(\beta_x)-f(a-\delta_x)}{g(\beta_x)-g(a-\delta_x)}=\lim_{x\to a^{-}}\frac{f'(\alpha_x)}{g'(\alpha_x)}=\lim_{x\to a^{-}}\frac{f'(x)}{g'(x)},
$$
thus
$$
\lim_{x\to a^{-}}\frac{f(x)}{g(x)}=\lim_{x\to a^{-}}\frac{f(\beta_x)}{g(\beta_x)}=\lim_{x\to a^{-}}\frac{\frac{f(\beta_x)}{g(\beta_x)}-\frac{f(a-\delta_x)}{g(\beta_x)}}{1-\frac{g(a-\delta_x)}{g(\beta_x)}}=\lim_{x\to a^{-}}\frac{f(\beta_x)-f(a-\delta_x)}{g(\beta_x)-g(a-\delta_x)}=\lim_{x\to a^{-}}\frac{f'(x)}{g'(x)}.
$$
The proof that $\lim_{x\to a^{+}}\frac{f(x)}{g(x)}=\lim_{x\to a^{+}}\frac{f'(x)}{g'(x)}$ is similar.
