Is there a form $\lim_{x\to\infty}\left[\frac{\infty}{\infty}\right] $ of l'Hôpital's rule? In course of learning L'Hospital rule I've learned both $\displaystyle\lim_{x\to\infty}\left[\dfrac{0}{0}\right]$ and $\displaystyle\lim_{x\to0}\left[\dfrac{\infty}{\infty}\right]$ form. But is there a form $\displaystyle\lim_{x\to\infty}\left[\dfrac{\infty}{\infty}\right]?$
For example can I evaluate $\displaystyle\lim_{x\to\infty}\dfrac{x}{e^x}$ as follows:
$$\displaystyle\lim_{x\to\infty}\dfrac{x}{e^x}=\displaystyle\lim_{x\to\infty}\dfrac{1}{e^x}=0?$$
 A: Yes. Probably the simplest way to see this is to apply the transformation $y = 1/x$:
$$\lim_{x\to\infty} \frac{f(x)}{g(x)} = \lim_{y\to0^+} \frac{f(1/y)}{g(1/y)}$$
after which you can apply De l'Hôpital's rule in its familiar form, since by the chain rule:
$${\partial\over\partial y}f(1/y) = \frac{-1}{y^2}f'(1/y)$$
and both the $\dfrac{-1}{y^2}$ factors cancel:
$$\lim_{x\to\infty} \frac{f(x)}{g(x)} = \lim_{y\to0^+} \frac{f(1/y)}{g(1/y)} = \lim_{y\to0^+} \frac{\frac{-1}{y^2}f'(1/y)}{\frac{-1}{y^2}g'(1/y)} = \lim_{y\to0^+} \frac{f'(1/y)}{g'(1/y)} = \lim_{x\to\infty} \frac{f'(x)}{g'(x)}$$
A: Yes, you can do this.  In general, L'hopital's theorem says that for any point $a \in \mathbb{R}$ and under certain conditions you have
$$
\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}
$$
There is nothing special about $a = 0$ or $a = \infty$.
Specifically, the required conditions are that $\lim_{x \to a} f(x)$ and $\lim_{x \to a} g(x)$ are either both $\pm \infty$ or both zero, and that $\lim_{x \to a} \frac{f'(x)}{g'(x)}$ exists (including if it is $\pm \infty$).
