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$$ \lim_{x \to a} f(x) = \pm \infty \,, \quad \lim_{x \to a} g(x) = \pm \infty\,, \quad \lim_{x \to a} \frac{f'(x)}{g'(x)} = m\,,$$ 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$$ \absW{ \frac{f'(\xi)}{g'(\xi)} - m} < \epsilon$$ whenever $0 < \abs{\xi - a} < \delta$ .

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 $0 < \abs{\xi_x -a} < \delta$ ) such that$$ \frac{f(x) - f(c)}{g(x) - g(c)} = \frac{f'(\xi_x)}{g'(\xi_x)}\,.$$ 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 $0 < \abs{x - a} < \delta'$ for some $0 < \delta' < \delta$ . (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

This proves$$ \lim_{x \to a} \frac{f(x)}{g(x)} = m = \lim_{x \to a} \frac{f'(x)}{g'(x)}\,.$$

Bibliography

1

Michael Spivak, Calculus, 3rd ed. Publish or Perish, 1994.