# L'Hôpital's Rule/Corollary 2

## Corollary to L'Hôpital's Rule

Let $f$ and $g$ be real functions which are continuous on the closed interval $\left[{a \,.\,.\, b}\right]$ and differentiable on the open interval $\left({a \,.\,.\, b}\right)$.

Suppose that $\forall x \in \left({a \,.\,.\, b}\right): g' \left({x}\right) \ne 0$.

Suppose that $f \left({x}\right) \to \infty$ and $g \left({x}\right) \to \infty$ as $x \to a^+$.

Then:

- $\displaystyle \lim_{x \mathop \to a^+} \frac {f \left({x}\right)} {g \left({x}\right)} = \lim_{x \mathop \to a^+} \frac {f' \left({x}\right)} {g' \left({x}\right)}$

provided that the second limit exists.

## Proof

Let $\displaystyle \lim_{x \mathop \to a^+} \frac {f' \left({x}\right)} {g' \left({x}\right)} = L$.

Let $\left({x_n}\right)$ be a sequence such that:

- $\quad x_n \in \left({a \,.\,.\, b}\right)$ for all $n \in \N$ and $\displaystyle \lim_{n \to \infty}x_n = a$

From Intermediate Value Theorem for Derivatives and the definition of limit of real function follows that:

- $\displaystyle \lim_{n \to \infty}g \left({x_n}\right) = \infty$

and $\left({g(x_n)}\right)$ is strictly increasing.

Consider the range $\left[{{x_{n-1}}\,.\,.\,{x_n}}\right] \subset \left({a \,.\,.\, b}\right)$ where $n \geq 2$.

By Cauchy Mean Value Theorem, there exists $c_n \in \left({x_{n-1} \,.\,.\, x_n}\right)$ such that:

- $\displaystyle \frac{f \left({x_n}\right) - f \left({x_{n-1}}\right)}{g \left({x_n}\right) - g \left({x_{n-1}}\right)} = \frac{f' \left({c_n}\right)}{g' \left({c_n}\right)}$

From the above and the Squeeze Theorem for Real Sequences:

- $\displaystyle \lim_{n \to \infty} c_n = a$

and:

- $\displaystyle \lim_{n \to \infty} \frac{f \left({x_n}\right) - f \left({x_{n-1}}\right)}{g \left({x_n}\right) - g \left({x_{n-1}}\right)} = \lim_{n \to \infty}\frac{f' \left({c_n}\right)}{g' \left({c_n}\right)} = L$

So, by Stolz-Cesàro Theorem:

- $\displaystyle \lim_{n \mathop \to \infty} \frac {f \left({x_n}\right)} {g \left({x_n}\right)} = L$

From the definition of limit of a function deduce that:

- $\displaystyle \lim_{x \mathop \to a^+} \frac {f \left({x}\right)} {g \left({x}\right)} = L = \lim_{x \mathop \to a^+} \frac {f' \left({x}\right)} {g' \left({x}\right)}$

$\blacksquare$

## Remarks

- The proof does not actually use the assumption $\displaystyle \lim_{x \mathop \to a^+} f \left({x}\right) = \infty$.

- Cases $x \to b^-$, $x \to \pm \infty$ and $g \left({x}\right) \to -\infty$ can be proven similarly.

## Source of Name

This entry was named for Guillaume de l'Hôpital.

However, this result was in fact discovered by Johann Bernoulli.

Because of variants in the rendition of his name, this result is often seen written as **L'Hospital's Rule**.