I'm trying to prove l'Hospital rule for $\lim_{a^+} f(x) = \pm \infty =\lim_{a^+} g(x) $, inferring it from the case $\lim_{a^+} f(x) = 0 =\lim_{a^+} g(x) $.
There are three cases: $\lim_{x\to a^+} \frac{f(x)}{g(x)}$ infinite, $\lim_{x\to a^+} \frac{f(x)}{g(x)}$ finite but not zero, and $\lim_{x\to a^+} \frac{f(x)}{g(x)}$ equal to zero.
If $\lim_{x\to a^+} \frac{f(x)}{g(x)}$ finite but not zero, we can easily find l'Hospital rule, considering the functions $F = \frac{1}{f}$ and $G = \frac{1}{g}$
If $\lim_{x\to a^+} \frac{f(x)}{g(x)} = 0$ we can easily find l'Hospital rule, considering the limit of the ratio $\frac{f+g}{g} = 1 + \frac{f}{g}$
$\lim_{x\to a^+} \frac{f(x)}{g(x)} = \infty$ then either $\lim_{x\to a^+} \frac{f(x)}{g(x)} = +\infty $ or either $\lim_{x\to a^+} \frac{f(x)}{g(x)} = -\infty $, considering the limit of the ratio $\frac{g}{f}$, the previous case 2. gives $ \lim_{a^+}\frac{g'}{f'} = 0$. Hence, $\lim_{a^+} \frac{f'}{g'} = \pm \infty$. However, I cannot see how we can conclude whether $\lim_{a^+} \frac{f'}{g'}$ is $+\infty$ or $-\infty$. This website says "By the Souped up MVT, f/g has the same sign as f′/g′ ". I tried to follow the website's hint, here is what I got:
Let $c\in (a,b]$. By the MVT, we have: $\exists x\in (c,b)$, $$ \frac{f'(x)}{g'(x)} = \frac{f(b)-f(c)}{g(b)- g(c)}$$
Then, since $a<c<x<b$, and since $f(b)$, $g(b)$ are finite, taking the limit yields to: $$\lim _{x\to a^+} \frac{f'(x)}{g'(x)} = \lim _{c\to a^+} \frac{f(c)}{g(c)} \tag{1}$$
Formula $(1)$ allows me to say $f/g$ and $f'/g'$ have the same sign, but in fact it also proves l'Hospital's rule when $\lim_{x\to a^+} \frac{f(x)}{g(x)} = \infty$, by itself.
Do you think I am correctly understanding (and developing) the hint from the website ?