
# 6.5: L'Hopital's Rule


The following result is one case of $$l^{\prime}$$ l'Hópital's rule.

## Theorem $$\PageIndex{1}$$

Suppose $$a, b \in \mathbb{R}, f$$ and $$g$$ are differentiable on $$(a, b), g^{\prime}(x) \neq 0$$ for all $$x \in(a, b),$$ and

$\lim _{x \rightarrow a^{+}} \frac{f^{\prime}(x)}{g^{\prime}(x)}=\lambda .$

If $$\lim _{x \rightarrow a^{+}} f(x)=0$$ and $$\lim _{x \rightarrow a^{+}} g(x)=0,$$ then

$\lim _{x \rightarrow a^{+}} \frac{f(x)}{g(x)}=\lambda .$

Proof

Given $$\epsilon>0,$$ there exists $$\delta>0$$ such that

$\lambda-\frac{\epsilon}{2}<\frac{f^{\prime}(x)}{g^{\prime}(x)}<\lambda+\frac{\epsilon}{2}$

whenever $$x \in(a, a+\delta) .$$ Now, by the Generalized Mean Value Theorem, for any $$x$$ and $$y$$ with $$a<x<y<a+\delta,$$ there exists a point $$c \in(x, y)$$ such that

$\frac{f(y)-f(x)}{g(y)-g(x)}=\frac{f^{\prime}(c)}{g^{\prime}(c)}.$

Hence

$\lambda-\frac{\epsilon}{2}<\frac{f(y)-f(x)}{g(y)-g(x)}<\lambda+\frac{\epsilon}{2}.$

Now

$\lim _{x \rightarrow a^{+}} \frac{f(y)-f(x)}{g(y)-g(x)}=\frac{f(y)}{g(y)}$

and so we have

$\lambda-\epsilon<\lambda-\frac{\epsilon}{2} \leq \frac{f(y)}{g(y)} \leq \lambda+\frac{\epsilon}{2}<\lambda+\epsilon$

for any $$y \in(a, a+\delta) .$$ Hence

$\lim _{x \rightarrow a^{+}} \frac{f(x)}{g(x)}=\lambda .$

Q.E.D.

## Exercise $$\PageIndex{1}$$

Use l'Hôpital's rule to compute

$\lim _{x \rightarrow 0^{+}} \frac{\sqrt{1+x}-1}{x}. \nonumber$

## Exercise $$\PageIndex{2}$$

Suppose $$a, b \in \mathbb{R}, f$$ and $$g$$ are differentiable on $$(a, b), g^{\prime}(x) \neq 0$$ for all $$x \in(a, b),$$ and

$\lim _{x \rightarrow b^{-}} \frac{f^{\prime}(x)}{g^{\prime}(x)}=\lambda . \nonumber$

Show that if $$\lim _{x \rightarrow b^{-}} f(x)=0$$ and $$\lim _{x \rightarrow b^{-}} g(x)=0,$$ then

$\lim _{x \rightarrow b^{-}} \frac{f(x)}{g(x)}=\lambda . \nonumber$