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:
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.
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.