5.3: L'Hôpital's Rule

We shall now prove a useful rule for resolving indeterminate limits. Below, $G_{\neg p}$ denotes a deleted globe $G_{\neg p}(\delta)$ in $E^{1},$ or one about $\pm \infty$ of the form $(a,+\infty)$ or $(-\infty, a).$ For one-sided limits, replace $G_{\neg p}$ by its appropriate "half."

Theorem $\PageIndex{1}$ (L'Hôpital's rule)

Let $f, g : E^{1} \rightarrow E^{*}$ be differentiable on $G_{\neg p}$, with $g^{\prime} \neq 0$ there. If $|f(x)|$ and $|g(x)|$ tend both to $+\infty,^{1}$ or both to $0,$ as $x \rightarrow p$ and if

$$\lim _{x \rightarrow p} \frac{f^{\prime}(x)}{g^{\prime}(x)}=r \text { exists in } E^{*},$$

then also

$$\lim _{x \rightarrow p} \frac{f(x)}{g(x)}=r;$$

similarly for $x \rightarrow p^{+}$ or $x \rightarrow p^{-}$.

Proof

It suffices to consider left and right limits. Both combined then yield the two-sided limit.

First, let $-\infty \leq p<+\infty$,

$$\lim _{x \rightarrow p^{+}}|f(x)|=\lim _{x \rightarrow p^{+}}|g(x)|=+\infty \text { and } \lim _{x \rightarrow p^{+}} \frac{f^{\prime}(x)}{g^{\prime}(x)}=r\text { (finite)}.$$

Then given $\varepsilon>0,$ we can fix $a>p\left(a \in G_{\neg p}\right)$ such that

$$\left|\frac{f^{\prime}(x)}{g^{\prime}(x)}-r\right|<\varepsilon, \text { for all } x \text { in the interval }(p, a).$$

Now apply Cauchy's law of the mean (§2, Theorem 2) to each interval $[x, a],$ $p<x<a.$ This yields, for each such $x,$ some $q \in(x, a)$ with

$$g^{\prime}(q)[f(x)-f(a)]=f^{\prime}(q)[g(x)-g(a)].$$

As $g^{\prime} \neq 0$ (by assumption), $g(x) \neq g(a) \neq g(a)$ by Theorem 1, §2, so we may divide to obtain

$$\frac{f(x)-f(a)}{g(x)-g(a)}=\frac{f^{\prime}(q)}{g^{\prime}(q)}, \quad \text { where } p<x<q<a.$$

This combined with (1) yields

$$\left|\frac{f(x)-f(a)}{g(x)-g(a)}-r\right|<\varepsilon,$$

or, setting

$$F(x)=\frac{1-f(a) / f(x)}{1-g(a) / g(x)},$$

we have

$$\left|\frac{f(x)}{g(x)} \cdot F(x)-r\right|<\varepsilon \text { for all } x \text { inside }(p, a).$$

As $|f(x)|$ and $|g(x)| \rightarrow+\infty$ (by assumption), we have $F(x) \rightarrow 1$ as $x \rightarrow p^{+}$. Hence by rules for right limits, there is $b \in(p, a)$ such that for all $x \in(p, b)$, both $|F(x)-1|<\varepsilon$ and $F(x)>\frac{1}{2}$. (Why?) For such $x$, formula (2) holds as well. Also,

$$\frac{1}{|F(x)|}<2 \text { and }|r-r F(x)|=|r||1-F(x)|<|r| \varepsilon.$$

Combining this with (2), we have for $x \in(p, b)$

$$\begin{aligned}\left|\frac{f(x)}{g(x)}-r\right| &=\frac{1}{|F(x)|}\left|\frac{f(x)}{g(x)} F(x)-r F(x)\right| \\ &<2\left|\frac{f(x)}{g(x)} \cdot F(x)-r+r-r F(x)\right| \\ &<2 \varepsilon(1+|r|). \end{aligned}$$

Thus, given $\varepsilon>0,$ we found $b>p$ such that

$$\left|\frac{f(x)}{g(x)}-r\right|<2 \varepsilon(1+|r|), \quad x \in(p, b).$$

As $\varepsilon$ is arbitrary, we have $\lim _{x \rightarrow p^{+}} \frac{f(x)}{g(x)}=r,$ as claimed.

The case $\lim _{x \rightarrow p^{+}} f(x)=\lim _{x \rightarrow p^{+}} g(x)=0$ is simpler. As before, we obtain

$$\left|\frac{f(x)-f(a)}{g(x)-g(a)}-r\right|<\varepsilon.$$

Here we may as well replace $" a^{\prime \prime}$ by any $y \in(p, a).$ Keeping $y$ fixed, let $x \rightarrow p^{+}$. Then $f(x) \rightarrow 0$ and $g(x) \rightarrow 0,$ so we get

$$\left|\frac{f(y)}{g(y)}-r\right| \leq \varepsilon \text { for any } y \in(p, a).$$

As $\varepsilon$ is arbitrary, this implies $\lim _{y \rightarrow p^{+}} \frac{f(y)}{g(y)}=r.$ Thus the case $x \rightarrow p^{+}$ is settled for a finite $r.$

The cases $r=\pm \infty$ and $x \rightarrow p^{-}$ are analogous, and we leave them to the reader. $\quad \square$

Note 1. $\lim \frac{f(x)}{g(x)}$ may exist even if $\lim \frac{f^{\prime}(x)}{g^{\prime}(x)}$ does not. For example, take

$$f(x)=x+\sin x \text { and } g(x)=x.$$

Then

$$\lim _{x \rightarrow+\infty} \frac{f(x)}{g(x)}=\lim _{x \rightarrow+\infty}\left(1+\frac{\sin x}{x}\right)=1 \quad(\text { why? }),$$

but

$$\frac{f^{\prime}(x)}{g^{\prime}(x)}=1+\cos x$$

does not tend to any limit as $x \rightarrow+\infty$.

Note 2. The rule fails if the required assumptions are not satisfied, e.g., if $g^{\prime}$ has zero values in each $G_{\neg p};$ see Problem 4 below.

Often it is useful to combine L'Hôpital's rule with some known limit formulas, such as

$$\lim _{z \rightarrow 0}(1+z)^{1 / z}=e \text { or } \lim _{x \rightarrow 0} \frac{x}{\sin x}=1 \text { (see §1, Problem 8).}$$

Examples

(a) $\lim _{x \rightarrow+\infty} \frac{\ln x}{x}=\lim _{x \rightarrow+\infty} \frac{(\ln x)^{\prime}}{1}=\lim _{x \rightarrow+\infty} \frac{1}{x}=0.$

(b) $\lim _{x \rightarrow 0} \frac{\ln (1+x)}{x}=\lim _{x \rightarrow 0} \frac{1 /(1+x)}{1}=1.$

(c) $\lim _{x \rightarrow 0} \frac{x-\sin x}{x^{3}}=\lim _{x \rightarrow 0} \frac{1-\cos x}{3 x^{2}}=\lim _{x \rightarrow 0} \frac{\sin x}{6 x}=\frac{1}{6} \lim _{x \rightarrow 0} \frac{\sin x}{x}=\frac{1}{6}.$

(Here we had to apply L'Hôpital's rule repeatedly.)

(d) Consider

$$\lim _{x \rightarrow 0^{+}} \frac{e^{-1 / x}}{x}.$$

Taking derivatives (even $n$ times), one gets

$$\lim _{x \rightarrow 0^{+}} \frac{e^{-1 / x}}{n ! x^{n+1}}, \quad n=1,2,3, \ldots \text { (always indeterminate!).}$$

Thus the rule gives no results. In this case, however, a simple device helps (see Problem 5 below).

(e) $\lim _{n \rightarrow \infty} n^{1 / n}$ does not have the form $\frac{0}{0}$ or $\frac{\infty}{\infty},$ so the rule does not apply directly. Instead we compute

$$\lim _{n \rightarrow \infty} \ln n^{1 / n}=\lim _{n \rightarrow \infty} \frac{\ln n}{n}=0 \text { (Example (a))}.$$

Hence

$$n^{1 / n}=\exp \left(\ln n^{1 / n}\right) \rightarrow \exp (0)=e^{0}=1$$

by the continuity of exponential functions. The answer is then 1.