I am looking for an adapted version to the case where $\lim_{a^+} f = \infty$ and $\lim_{a^+} g = \infty$ of this proof.
It goes like this:
Suppose $f'(x), g'(x)$ exist and $g'(x) \neq 0$ for all $x$ in an interval $(a,b]$. If $$ \lim_{x\to a^+} f(x) = 0 = \lim_{x\to a^+} g(x)$$ and $$\lim \frac{f'(x)}{g'(x)}$$ exists, then $$\lim_{x\to a^+} \frac{f(x)}{g(x)} = \lim_{x\to a^+} \frac{f'(x)}{g'(x)}$$
Proof: We may assume that $f(a)=0=g(a)$ (since the limit is not affected by the value of the function at $a$). Also $g(b) \neq 0$, else $g'(x)=0$ at some $x\in (a,b)$ by Rolle's theorem. Define $$h(x)= f(x) - \frac{f(b)}{g(b)} g(x)$$, then $$h(a)=0=h(b)$$, and $h$ is continuous on $[a,b]$ and $$h'(x)=f'(x) - \frac{f(b)}{g(b)} g'(x)$$ exists on $(a,b)$. By Rolle's Theorem there exists $x\in (a,b)$ such that $h'(x)=0$, hence $$\frac{f'(x)}{g'(x)} = \frac{f(b)}{g(b)}$$. Since $a<x<b$, it follows that $$\lim_{b\to a^+} \frac{f(b)}{g(b)} = \lim_{x\to a^+} \frac{f'(x)}{g'(x)}$$
Then, at the bottom of the proof, the author writes "the theorem can be adapted for $\lim f(x) = \infty = \lim g(x)$."
Hence my question.
I'm self-answering my post, please do tell me if something is wrong with what's written below.
Suppose all the condition of the question above are met, with the exception that $\lim_{a^+} f(x) = \pm \infty = \lim_{a^+} g(x)$ (instead of $\lim_{a^+} f = 0 = \lim_{a^+} g$).
Then, letting $c\in (a,b]$, the application of the Cauchy MVT between $c$ and $b$ gives: $$\exists x\in (c,b), \frac{f'(x)}{g'(x)}= \frac{f(b)-f(c)}{g(b)-g(c)}$$
Then, noticing $a<c<x<b$ and taking the limit, gives: $$\lim_{x\to a^+}\frac{f'(x)}{g'(x)} = \lim_{c\to a^+} \frac{f(c)}{g(c)}$$
Indeed, the terms $f(b)$ and $g(b)$ vanish since $f$ is defined on $b$ and $\lim_{a^+} f= \infty = \lim _{a^+} g$.
EDIT: One should add the extra assumption that $f(x)\ne 0$ $\forall (a,b]$.