This is the proof of L’Hôpital’s Rule in the case of the indeterminate form $\pm\infty/\infty$. Compared to the proof for the $0/0$ case, more complicated estimates are needed.

Assume that

where $a$ and $m$ are real numbers. The case when $a$ or $m$ is infinite only involves slight modifications to the arguments below.

Given $\epsilon>0$. there is a $\delta>0$ such that

whenever $0<\lvert\xi-a\rvert<\delta$.

Let $c$ and $x$ be points such that $a-\delta<c<x<a$ or $a<x<c<a+\delta$. (That is, both $c$ and $x$ are within distance $\delta$ of $a$, but $x$ is always closer.) By Cauchy’s mean value theorem, there exists some $\xi_{x}$ in between $c$ and $x$ (and hence $0<\lvert\xi_{x}-a\rvert<\delta$) such that

We can assume the values $f(x)$, $g(x)$, $f(x)-f(c)$, $g(x)-g(c)$ are all non-zero when $x$ is close enough to $a$, say, when $0<\lvert x-a\rvert<\delta^{{\prime}}$ for some $0<\delta^{{\prime}}<\delta$. (So there is no division by zero in our equations.) This is because $f(x)$ and $g(x)$ were assumed to approach $\pm\infty$, so when $x$ is close enough to $a$, they will exceed the fixed values $f(c)$, $g(c)$, and $0$.

We write

Note that

but $\xi_{x}$ is not guaranteed to approach $a$ as $x$ approaches $a$, so we cannot just take the limit $x\to a$ directly. However: there exists $0<\delta^{{\prime\prime}}<\delta^{{\prime}}$ so that

whenever $0<\lvert x-a\rvert<\delta^{{\prime\prime}}$. Then

for $0<\lvert x-a\rvert<\delta^{{\prime\prime}}$.

This proves