How does the L.Hopital rule work when numerator $\neq$ $\infty$ L.Hopital rule can be used to find the limits of the form
$\lim_{x \to a} \dfrac{f(x)}{g(x)}$ when $\lim_{x \to a}f(x)=\lim_{x \to a}g(x)=\infty$
Today I saw page claiming that $\lim_{x \to a} f(x)=\infty$ is not necessary (When $\lim _{x \to a}g(x)=\infty$) . Is it true? Can it be proved that this condition isn't necessary.
Edit: To clarify, I am solving a limit problem with L.Hopital where $f(x)\neq \infty$[This doesn't prove it, I am trying to clear what I am trying to ask]
$\lim_{x \to \infty}\dfrac{1}{x^2}$ (as $\lim_{x \to \infty} x^2 =\infty$, limit of numerator need not ne $\infty$) 
Applying L.Hopital once:
$\implies\dfrac{0}{2x}$ (as $\lim_{x \to \infty}2x=\infty$,limit of numerator need not ne $\infty$) 
Applying L.Hopital again:
$\implies \dfrac{0}{2}=0$
Edit:

 A: I think what you are looking for is the proof found in here using the Cesaro-Stolz Theorem. http://www.imomath.com/index.php?options=686
One of the cases is when $\lim_{x\to a}g(x)=\infty$ but makes no hypothesis on $f$.
Another link that may help you is  this one:
In this link a proof is provided that seems to match what your picture wants.
A: They can also both have limits of 0. You can prove this by taking the reciprocal of each function and of the entire ratio. The resulting expression is equivalent to the original but now the numerator and denominator approach zero
A: if you need to find $\lim_{x\rightarrow a} \frac{f(x)}{g(x)}$ and you have $\lim_{x\rightarrow a} f(x) \neq \infty$ and $\lim_{x\rightarrow a} g(x) = \infty$, then l'hopital's rule doesn't apply: limiting to (something finite)/$\infty$ gives 0, as opposed to something indeterminate.
Even worse, trying to use L'Hopital's rule when it's inappropriate can get you into trouble: $$\lim_{x\rightarrow 0}\frac{\sqrt[3]{x}}{\csc(x)}=\frac{0}{\infty}=0$$
But if you apply l'hopital's rule this gives $$\frac{1}{3}\frac{x^{-\frac{2}{3}}}{\csc x \cot x} = \frac{\infty}{\infty}$$
and repeated applications of the rule don't get rid of this indeterminate form.
