Proof of the L'Hôpital Rule for $\frac{\infty}{\infty}$

Let $$\begin{equation} \lim_{x\rightarrow a^{+}}\frac{f^{\prime }(x)}{g^{\prime }(x)}=L. \tag{1} \end{equation}$$ Then for each $\delta >0$ there exists a real $\beta \in \left( a,b\right) $ such that for all $x\in \left( a,\beta \right) $ $$\begin{equation} \left\vert \frac{f^{\prime }(x)}{g^{\prime }(x)}-L\right\vert <\delta . \tag{2} \end{equation}$$ Let $x\in \left( a,\beta \right) ,y\in \left( a,\beta \right) ,x<y$. Since the functions $f,g$ are continuous and differentiable on $\left[ x,y\right] $ we can apply the Cauchy Mean Value Theorem. Consequently, there exists a $c\in \left[ x,y\right] \subset \left( a,\beta \right) $ such that $$\begin{equation} \frac{f(x)-f(y)}{g(x)-g(y)}=\frac{f^{\prime }(c)}{g^{\prime }(c)}. \tag{3} \end{equation}$$ Hence for $x\in \left( a,\beta \right) ,y\in \left( a,\beta \right) ,x<y$ $$\begin{eqnarray} \left\vert \frac{f(x)-f(y)}{g(x)-g(y)}-L\right\vert &<&\delta \\ \\&&\\ \left\vert \frac{f(x)/g(x)-f(y)/g(x)}{1-g(y)/g(x)}-L\right\vert &<&\delta . \tag{4} \end{eqnarray}$$ Assume $\lim_{x\rightarrow a^{+}}f(x)=\lim_{x\rightarrow a^{+}}g(x)=+\infty $ and fix $y$. Then $g(y)/g(x)\rightarrow 0$ and there exists a $\gamma \in \left( a,\beta \right) $ such that for $x\in \left( a,\gamma \right) $, we have $g(x)>0$ and $g(x)/g(y)>1$. Inequality $(4)$ implies $$\begin{equation} \left( 1-\frac{g(y)}{g(x)}\right) \left( L-\delta \right) <\frac{f(x)}{g(x)}-\frac{f(y)}{g(x)}<\left( 1-\frac{g(y)}{g(x)}\right) \left( L+\delta \right) . \tag{5} \end{equation}$$ Letting $x\rightarrow a^{+}$ we conclude that $$\begin{equation} \lim_{x\rightarrow a^{+}}\frac{f(x)}{g(x)}=L. \tag{6} \end{equation}$$

Adapted from J. Campos Ferreira, Introdução à Análise Matemática, Teorema 11, p. 386 and J. Santos Guerreiro, Curso de Análise Matemática, Proposição 5.2.3.2, p. 314.

You're very close to a partial proof. I don't know that the general $\infty / \infty$ rule can be proven from the $0/0$ rule, and it all boils down to exactly the question you ask: does $\lim \dfrac{g^2 f'}{f^2 g'}$ exist? Not always, apriori.

But if we assume it exists, then we know that $\lim \dfrac{g}{f} = \lim \dfrac{1/f}{1/g} = \lim \dfrac{F'}{G'} =\lim \dfrac{F}{G}= \lim \dfrac{g^2 f'}{f^2 g'}$, so that (as we are assuming $\lim f'/g'$ exists and is finite) we may 'cross-multiply' to get that $\lim \dfrac{f}{g} = \lim \dfrac{f'}{g'}$

And so we have a case of the general theorem. (Conceivably, some annoying details may be needed to cover cases where we inadvertently divided by $0$ or whatnot). I do not see how we can use the $0/0$ case to get the complete result. But Américo's answer gives a complete proof independent of the $0/0$ case.