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