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

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