I ask for the proof of the L'Hôpital rule for the indeterminate form $\frac{\infty}{\infty}$ utilizing the rule for the form $\frac{0}{0}$.
Theorem: Let $f,g:(a,b)\to \mathbb{R}$ be two differentiable functions such as that: $\forall x\in(a,b)\ \ g(x)\neq 0\text{ and }g^{\prime}(x)\neq 0$ and $\lim_{x\to a^+}f(x)=\lim_{x\to a^+}g(x)=+\infty$ If the limit $\lim_{x\to a^+}\frac{f^{\prime}(x)}{g^{\prime}(x)}$ exists and is finite, then $\lim_{x\to a^+}\frac{f(x)}{g(x)}=\lim_{x\to a^+}\frac{f^{\prime}(x)}{g^{\prime}(x)}$
My attempt: Since $\lim_{x\to a^+}f(x)=+\infty$, $\exists \delta>0:a
The limit $\lim_{x\to a^+}\frac{f^{\prime}(x)}{g^{\prime}(x)}$ exists by the hypothesis but we don't know if the limit $\displaystyle\lim_{x\to a^+}\frac{g^2(x)}{f^2(x)}$ exists to deduce that the limit $\lim_{x\to a^+}\frac{F^{\prime}(x)}{G^{\prime}(x)}$ exists to use the L'Hôpital Rule for the form $\frac{0}{0}$.
EDIT: After discussing it with other users in the site, we came to the conclusion that this proof is only partial and can't logically be continued to yield the Theorem. As a result, the rule for the $\frac{0}{0}$ form can't be used to proove the rule for the $\frac{\infty}{\infty}$ form. Mr. Tavares and myself have already given two different proofs (with the pretty much the same main idea) of the Theorem in question using Cauchy's Mean Value Theorem. You can read them below. You can also read the proof Rudin gives for a stronger version of the Theorem (that does not suppose that $\lim_{x\to a^+}f(x)=+\infty$) in his book Principle of Meathematical Analysis. If you have any objections in either proofs please let me know. Thank you.