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.