Extension: Proving the Division by Infinity Case of L'Hopital

The simplest method to prove the $\frac{\pm\infty}{\pm\infty}$ case is to assume the $\frac00$ case of L'Hopital's rule:

Given continuous differentiable functions $f,g$ such that $f(a)=g(a)=0$, $$\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(a)}{g'(a)}.$$

Now, consider continuous differentiable functions $f$ and $g$ that satisfy $\lim_{x\to a}f(x)=\pm\infty$, $\lim_{x\to a}g(x)=\pm\infty$. It follows that $\frac1{f(a)}=\frac1{g(a)}=0$. By L'Hopital's rule,

$$\lim_{x\to a}\frac{g(x)}{f(x)}=\lim_{x\to a}\frac{\frac{1}{f(x)}}{\frac{1}{g(x)}} = \lim_{x\to a}\frac{-\frac{f'(x)}{f(x)^2}}{-\frac{g'(x)}{g(x)^2}} = \lim_{x\to a}\frac{f'(x)}{g'(x)}\left(\frac{g(x)}{f(x)}\right)^2.$$

If $\lim_{x\to a}\frac{g(x)}{f(x)}$ approaches a finite value, we may split the limits on the right-hand side (note that, due to the equality, $\frac{f'(x)}{g'(x)}$ cannot be infinite if $\frac{g(x)}{f(x)}$ is locally finite): $$\lim_{x\to a}\frac{g(x)}{f(x)}=\left(\lim_{x\to a}\frac{f'(x)}{g'(x)}\right)\left(\lim_{x\to a}\frac{g(x)}{f(x)}\right)^2.$$

Therefore $$\left(\lim_{x\to a}\frac{g(x)}{f(x)}\right)^{-1}=\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)},\;\;\; \lim_{x\to a}\frac{g(x)}{f(x)}\ne\pm\infty.$$ Now we have proven the theorem for all cases except where $\lim_{x\to a}\frac{f(x)}{g(x)}=0.$

In this case, note that $\lim_{x\to a}\frac{g(x)}{f(x)}=\pm\infty\ne0$ and thus we may apply: $$\lim_{x\to a}\frac{g(x)}{f(x)}=\lim_{x\to a}\frac{g'(x)}{f'(x)}.$$ By taking the reciprocal of both sides, we obtain $$\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)},$$ with $\lim_{x\to a}\frac{f(x)}{g(x)}=0.$

QED