L'Hospitals Rule on infinite intervals It seems like every proof I see of L'Hospital's rule is on a finite interval, for example my Real Analysis defines L'Hospital's rule as follows:

If $f$ and $g$ are differentiable functions defined on an interval $(a,b)$, both of which tend to $0$ at $b$, and if the ratio of their derivatives $\frac{f'(x)}{g'(x)}$ tends to a finite limit $L$ at $b$ then $\frac{f(x)}{g(x)}$ also tends to $L$ at $b$. It is assumed that $g(x), g'(x) \neq 0$

How would I go about proving the result on the interval $[a,\infty]$? What about if $L = \infty$?
 A: For $L \to \infty$ apply the theorem to $$\lim_{x \to 0^+} \frac{F(x)}{G(x)},$$ where $F(x) = f(1/x)$ and $G(x) = g(1/x)$
A: The proof given below basically mimic the proof of finite interval case that can be found in page 143-144 of Real Mathematical Analysis by Charles Pugh. That proof can be easily extended.

Assume $f(x)$ and $g(x)$ differentiable on $(a,\infty)$
and
$$\lim_{x\to \infty}f(x)=0,\lim_{x\to \infty}g(x)=0,\lim_{x\to >\infty}\frac{f'(x)}{g'(x)}=L, g'(x)\neq 0, g(x)\neq 0$$
We have
$$\lim_{x\to \infty}\frac{f(x)}{g(x)}=L$$

Proof:
Fixed $x$, we try to estimate $|\frac{f(x)}{g(x)}-L|$. For any $t>x$
$$|\frac{f(x)}{g(x)}-L|=|\frac{f(x)}{g(x)}-\frac{f(x)-f(t)}{g(x)-g(t)}+\frac{f(x)-f(t)}{g(x)-g(t)}-L|\\\leq|\frac{f(x)}{g(x)}-\frac{f(x)-f(t)}{g(x)-g(t)}|+|\frac{f(x)-f(t)}{g(x)-g(t)}-L|\\=|\frac{g(x)f(t)-f(x)g(t)}{g(x)(g(x)-g(t))}|+|\frac{f'(\theta)}{g'(\theta)}-L|$$
where $\theta\in (x,t)$ by Cauchy Mean Value Theorem
For given $\epsilon>0$, pick $x$ large enough  $|\frac{f'(\theta)}{g'(\theta)}-L|< \epsilon/2$ can be satisfied because $\lim_{x\to \infty}\frac{f'(x)}{g'(x)}=L$. For that $x$, pick $t$ large enough that safisfies $|f(t)|+|g(t)|<\frac{g(x)^2\epsilon}{4(|f(x)|+|g(x)|)}$, $|g(t)|<\frac{|g(x)|}{2}$ (can be done because $\lim_{t\to \infty}f(t)=0,\lim_{t\to \infty}g(t)=0$), we have $|\frac{g(x)f(t)-f(x)g(t)}{g(x)(g(x)-g(t))}|\leq\frac{(|f(t)|+|g(t)|)(|f(x)|+|g(x)|)}{|g(x)||\frac{g(x)}{2}|}< \epsilon/2$. We conclude the $x$ we pick satisfies $|\frac{f(x)}{g(x)}-L|<\epsilon$ and the claim is proved.
