Recursive L'Hospital's rule 
Is it possible to prove that if:
  $$\lim_{x\rightarrow\infty} \frac{f(x)}{g(x)} =\lim_{x\rightarrow\infty} \frac{f(x)}{g(x)}\frac{1}{l(x)}, \quad\text{and}\quad \lim_{x\rightarrow\infty}\frac{1}{l(x)} = 0$$
  then:
  $$\lim_{x\rightarrow\infty} \frac{f(x)}{g(x)}=0.$$

Note: This question came to my mind after noting that in this problem, L'Hospital's rule fails. But I think it might be interpreted as some sort of recursion.
The problem is to show:
$$\lim_{x\rightarrow\infty} \frac{f(x)}{g(x)} = \lim_{x\rightarrow\infty}\frac{2^{\sqrt{\log(x)}}}{\sqrt{x}} = 0$$
Using  L'Hospital's rule we may write:
$$\lim_{x\rightarrow\infty} \frac{f(x)}{g(x)} =\lim_{x\rightarrow\infty} \frac{f'(x)}{g'(x)}= \log(2)\lim_{x\rightarrow\infty}\frac{2^{\sqrt{\log(x)}}}{\sqrt{x}} \frac{1}{\sqrt{log(x)}} $$
Hence in this particular example we have:
$$\lim_{x\rightarrow\infty} \frac{f(x)}{g(x)} =\log(2)\lim_{x\rightarrow\infty} \frac{f(x)}{g(x)}\frac{1}{\sqrt{log(x)}}$$
 A: IF we know that $\displaystyle\lim_{x\to a}\frac{f(x)}{g(x)}$ exists (where $a$ can be a number or infinity, as that doesn't matter), then yes, we can make this conclusion — simply because we would be looking at an equation $\color{blue}{A=A\cdot0}$, which implies $A=0$. The reason we even have this equation in blue is that by properties of limits $\displaystyle\lim_{x\to a}\left[F(x)\cdot G(x)\right]=\lim_{x\to a}F(x)\cdot\lim_{x\to a}G(x)$ provided both limits on the right exist.
But this "IF" above is a really big deal! What you're missing here is one of the conditions of L'Hospital's Rule:

If $\displaystyle\lim_{x\to a}\frac{f(x)}{g(x)}$ is an indeterminate form and if $\displaystyle\lim_{x\to a}\frac{f'(x)}{g'(x)}$ exists and if  … (one more technical condition), then $\displaystyle\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}$.

That's what's really going on in L'Hospital's Rule, even though (sadly) this important point usually gets glossed over when teaching it in the classroom, thus leading to an unfortunate misconception. Let's look at a typical classroom presentation of an example on L'Hospital's Rule:

Since $\displaystyle\lim_{x\to2}\frac{2x-4}{x^2-4}=\frac{0}{0}$ is an indeterminate form, then $\displaystyle\lim_{x\to2}\frac{2x-4}{x^2-4}\color{red}{=}\lim_{x\to2}\frac{2}{2x}=\frac{2}{4}=\frac{1}{2}$.

That red equals sign is, technically speaking, wrong! The correct logic is:

Since $\displaystyle\lim_{x\to2}\frac{2x-4}{x^2-4}=\frac{0}{0}$ is an indeterminate form, and since $\displaystyle\lim_{x\to2}\frac{2}{2x}=\frac{2}{4}=\frac{1}{2}$ exists, then $\displaystyle\lim_{x\to2}\frac{2x-4}{x^2-4}=\lim_{x\to2}\frac{2}{2x}=\frac{1}{2}$.

That's the logical flaw in your proposed approach. No, we can NOT write
$$\lim_{x\to\infty}\frac{f(x)}{g(x)}\color{red}{=}\log(2)\lim_{x\to\infty}\frac{f(x)}{g(x)}\frac{1}{\sqrt{\log(x)}}$$
unless we know that the latter limit exists. But we don't know that yet. Nor do we even know that $\displaystyle\lim_{x\to\infty}\frac{f(x)}{g(x)}$ exists — which would have helped, because it would have implied that both exist and are equal to zero.
