4.4: L'Hopital's Rule

We now prove a result that allows us to compute various limits by calculating a related limit involving derivatives. All four theorems in this section are known as L'Hopital's Rule.

For this section, we assume \(a,b \in \mathbb{R}\) with \(a < b\).

Theorem \(\PageIndex{1}\)

Suppose \(f\) and \(g\) are continuous on \([a,b]\) and differentiable on \((a,b)\). Suppose \(f(\bar{x})=g(\bar{x})=0\), where \(\bar{x} \in [a,b]\). Suppose further that there exists \(\delta > 0\) such that \(g^{\prime}(x) \neq 0\) for all \(x \in B(\bar{x} ; \delta) \cap[a, b]\), \(x \neq \bar{x}\).

If

\[\displaystyle \lim _{x \rightarrow \bar{x}} \frac{f^{\prime}(x)}{g^{\prime}(x)}=\ell,\]

then

\[\displaystyle \lim _{x \rightarrow \bar{x}} \frac{f(x)}{g(x)}=\ell.\]

Proof

Let \(\left\{x_{k}\right\}\) be a sequence in \([a,b]\) that converges to \(\bar{x}\) and such that \(x_{k} \neq \bar{x}\) for every \(k\). By Theorem 4.2.4, for each \(k\), there exists a sequence \(\left\{c_{k}\right\}\) between \(c_{k}\) between \(x_{k}\) and \(\bar{x}\), such that

\[\left[f\left(x_{k}\right)-f(\bar{x})\right] g^{\prime}\left(c_{k}\right)=\left[g\left(x_{k}\right)-g(\bar{x})\right] f^{\prime}\left(c_{k}\right).\]

Since \(f(\bar{x})=g(\bar{x})=0\), and \(g^{\prime}\left(c_{k}\right) \neq 0\) for sufficiently large \(k\), we have

\[\frac{f\left(x_{k}\right)}{g\left(x_{k}\right)}=\frac{f^{\prime}\left(c_{k}\right)}{g^{\prime}\left(c_{k}\right)}.\]

Under the assumptions that \(g^{\prime}(x) \neq 0\) for \(x\) near \(\bar{x}\) and \(g(\bar{x})=0\), we also have \(g\left(x_{k}\right) \neq 0\) for sufficiently large \(k\). By the squeeze theorem (Theorem 2.1.6), \(\left\{c_{k}\right\}\) converges to \(\bar{x}\). Thus,

\[\displaystyle \lim _{k \rightarrow \infty} \frac{f\left(x_{k}\right)}{g\left(x_{k}\right)}=\displaystyle \lim _{k \rightarrow \infty} \frac{f^{\prime}\left(c_{k}\right)}{g^{\prime}\left(c_{k}\right)}=\displaystyle \lim _{x \rightarrow \bar{x}} \frac{f^{\prime}(x)}{g^{\prime}(x)}=\ell .\]

Therefore, (4.9) follows from Theorem 3.1.2. \(\square\)

Example \(\PageIndex{1}\)

We will use Theorem 4.4.1 to show that \[\displaystyle \lim _{x \rightarrow 0} \frac{2 x+\sin x}{x^{2}+3 x}=1. \nonumber\]

Solution

First we observe that the conditions of Theorem 4.4.1 hold. Here \(f(x)=2 x+\sin x\), \(g(x)=x^{2}+3 x\), and \(\bar{x} = 0\). We may take \([a, b]=[-1,1]\), for example, so that \(f\) and \(g\) are continuous on \([a, b]\) and differentiable on \((a, b)\) and, furthermore, \(\frac{f(x)}{g(x)}\) is well defined on \([a, b] \backslash \{\bar{x}\}\). Moreover, taking \(\delta = 7 / 3\), we get \(g^{\prime}(x)=2 x+3 \neq 0\) for \(x \in B(\bar{x} ; \delta) \cap[a, b]\). Finally we calculate the limit of the quotient of derivatives using Theorem 3.2.1 to get

\[\begin{align*} \displaystyle \lim _{x \rightarrow \bar{x}} \frac{f^{\prime}(x)}{g^{\prime}(x)} & =\displaystyle \lim _{x \rightarrow 0} \frac{2+\cos x}{2 x+3} \\[4pt] &=\frac{\displaystyle \lim _{x \rightarrow 0} 2+\displaystyle \lim _{x \rightarrow 0} \cos x}{\displaystyle \lim _{x \rightarrow 0} 2 x+3} \\[4pt] &=\frac{2+1}{3} \\[4pt] &=1.\end{align*}\]

It now follows from Theorem 4.4.1 that \(\displaystyle \lim _{x \rightarrow 0} \frac{2 x+\sin x}{x^{2}+3 x}=1\) as we wanted to show.

Example \(\PageIndex{2}\)

We will apply L'Hospital's rule to determine the limit \[\displaystyle \lim _{x \rightarrow 1} \frac{3 x^{3}-2 x^{2}+4 x-5}{4 x^{4}-2 x-2}.\nonumber\]

Solution

Here \(f(x)=3 x^{3}-2 x^{2}+4 x-5\) and \(g(x)=4 x^{4}-2 x-2\). Thus \(f(1)=g(1)=0\). Moreover, \(f^{\prime}(x)= 9 x^{2}-4 x+4\) and \(g^{\prime}(x)=16 x^{3}-2\). Since \(g^{\prime}(1)=14 \neq 0\) and \(g^{\prime}\) is continuous we have \(g^{\prime}(x) \neq 0\) for \(x\) near \(1\). Now,

\[\displaystyle \lim _{x \rightarrow 1} \frac{9 x^{2}-4 x+4}{16 x^{3}-2}=\frac{9}{14}.\nonumber\]

Thus, the desired limit is \(\frac{9}{14}\) as well.

Example \(\PageIndex{3}\)

If the derivatives of the functions \(f\) and \(g\) themselves satisfy the assumptions of Theorem 4.4.1 we may apply L'Hospital's rule to determine first the limit of \(f^{\prime}(x) / g^{\prime}(x)\) and then apply the rule again to determine the original limit.

Solution

Consider the limit

\[\displaystyle \lim _{x \rightarrow 0} \frac{x^{2}}{1-\cos x}.\nonumber\]

Here \(f(x)=x^{2}\) and \(g(x)=1-\cos x\) so both functions and all its derivatives are continuous. Now \(g^{\prime}(x)=\sin x\) and, so, \(g^{\prime}(x) \neq 0\) for \(x\) near zero, \(x \neq 0\). Also \(f^{\prime}(0)=0=g^{\prime}(0)\) and \(g^{\prime \prime}(x)=\cos x \neq 0\) for \(x\) near \(0\). Moreover,

\[\displaystyle \lim _{x \rightarrow 0} \frac{f^{\prime \prime}(x)}{g^{\prime \prime}(x)}=\displaystyle \lim _{x \rightarrow 0} \frac{2}{\cos x}=2 .\nonumber\]

By L'Hospital's rule we get

\[\displaystyle \lim _{x \rightarrow 0} \frac{f^{\prime}(x)}{g^{\prime}(x)}=\displaystyle \lim _{x \rightarrow 0} \frac{f^{\prime \prime}(x)}{g^{\prime \prime}(x)}=\displaystyle \lim _{x \rightarrow 0} \frac{2}{\cos x}=2 .\nonumber\]

Applying L'Hospital's rule one more time we get

\[\displaystyle \lim _{x \rightarrow 0} \frac{x^{2}}{1-\cos x}=\displaystyle \lim _{x \rightarrow 0} \frac{f(x)}{g(x)}=\displaystyle \lim _{x \rightarrow 0} \frac{f^{\prime}(x)}{g^{\prime}(x)}=2 .\nonumber\]

Example \(\PageIndex{4}\)

Let \(g(x)=x+3 x^{2}\) and let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be given by

\[f(x)=\left\{\begin{array}{ll}
x^{2} \sin \frac{1}{x}, & \text { if } x \neq 0 \text {;} \\
0, & \text { if } x=0.
\end{array}\right. \nonumber\]

Solution

Now consider the limit

\[\displaystyle \lim _{x \rightarrow 0} \frac{f(x)}{g(x)}=\displaystyle \lim _{x \rightarrow 0} \frac{x^{2} \sin \frac{1}{x}}{x+3 x^{2}} . \nonumber\]

Using the derivative rules at \(x \neq 0\) and the definition of derivative at \(x = 0\) we can see that \(f\) is differentiable and \[f^{\prime}(x)=\left\{\begin{array}{ll}
2 x \sin \frac{1}{x}-\cos \frac{1}{x}, & \text { if } x \neq 0 \text {;}\\
0, & \text { if } x=0 \text {,}
\end{array}\right. \nonumber\]

However, \(f^{\prime}\) is not coniitnuous at \(0\) (since \(\displaystyle \lim _{x \rightarrow 0} f^{\prime}(x)\) does not exists) and, hence, L'Hospital's rule cannot be applied in this case.

On the other hand \(\displaystyle \lim _{x \rightarrow 0} \frac{x^{2} \sin \frac{1}{x}}{x+3 x^{2}}\) does exist as we can see from

\[\displaystyle \lim _{x \rightarrow 0} \frac{x^{2} \sin \frac{1}{x}}{x+3 x^{2}}=\displaystyle \lim _{x \rightarrow 0} \frac{x \sin \frac{1}{x}}{1+3 x}=\frac{\displaystyle \lim _{x \rightarrow 0} x \sin \frac{1}{x}}{\displaystyle \lim _{x \rightarrow 0}(1+3 x)}=0 . \nonumber\]

Theorem \(\PageIndex{2}\)

Let \(a, b \in \mathbb{R}\), \(a < b\), and \(\bar{x} (a, b)\). Suppose \(f, g:(a, b) \backslash\{\bar{x}\} \rightarrow \mathbb{R}\) are differentiable on \((a, b) \backslash\{\bar{x}\}\) and assume \(\displaystyle \lim _{x \rightarrow \bar{x}} f(x)=\displaystyle \lim _{x \rightarrow \bar{x}} g(x)=\infty\). Suppose further that there exists \(\delta > 0\) such that \(g^{\prime}(x) \neq 0\) for all \(x \in B(\bar{x} ; \delta) \cap(a, b)\), \(x \neq \bar{x}\).

If \(\ell \in \mathbb{R}\) and

\[\displaystyle \lim _{x \rightarrow \bar{x}} \frac{f^{\prime}(x)}{g^{\prime}(x)}=\ell ,\]

then

\[\displaystyle \lim _{x \rightarrow \bar{x}} \frac{f(x)}{g(x)}=\ell .\]

Proof

Since \(\displaystyle \lim _{x \rightarrow \bar{x}} f(x)=\displaystyle \lim _{x \rightarrow \bar{x}} g(x)=\infty\), choosing a smaller positive \(\delta\) if necessary, we can assume that \(f(x) \neq 0\) and \(g(x) \neq 0\) for all \(x \in B(\bar{x} ; \delta) \cap(a, b)\).

We will show that \(\displaystyle \lim _{x \rightarrow \bar{x}^{+}} \frac{f(x)}{g(x)}=\ell\). The proof that \(\displaystyle \lim _{x \rightarrow \bar{x}^{-}} \frac{f(x)}{g(x)}=\ell\) is completely analogous.

Fix any \(\varepsilon > 0\). We need to find \(\delta_{0}>0\) such that \(|f(x) / g(x)-\ell|<\varepsilon\) whenever \(x \in B_{+}\left(\bar{x} ; \delta_{0}\right) \cap (a, b)\).

From \(\PageIndex{18}\), one cn choose \(K > 0\) and a positive \(\delta_{1}<\delta\) such that \[\left|\frac{f^{\prime}(x)}{g^{\prime}(x)}\right| \leq K \text { and }\left|\frac{f^{\prime}(x)}{g^{\prime}(x)}-\ell\right|<\frac{\varepsilon}{2}\] whenever \(x \in B\left(\bar{x} ; \delta_{1}\right) \cap(a, b)\), \(x \neq \bar{x}\).

Fix \(\alpha \in B_{+}\left(\bar{x} ; \delta_{1}\right) \cap(a, b)\) (in particular, \(\alpha > \bar{x}\). Since \(\displaystyle \lim _{x \rightarrow \bar{x}} f(x)=\infty\), we can find \(\delta_{2}>0\) such that \(\delta_{2}<\min \left\{\delta_{1}, \alpha-\bar{x}\right\}\) and \(f(x) \neq f(\alpha)\) for \(x \in B_{+}\left(\bar{x} ; \delta_{2}\right) \cap(a, b)=B_{+}\left(\bar{x} ; \delta_{2}\right)\). Moreover, for such \(x\), since \(g^{\prime}(z) \neq 0\) if \(x < z < \alpha\), Rolle's theorem (Theorem 4.2.2) guarantees that \(g(x) \neq g(\alpha)\). Therefore, for all \(x \in B_{+}\left(\bar{x} ; \delta_{2}\right)\) we can write,

\[\frac{f(x)}{g(x)}=\frac{f(x)-f(\alpha)}{g(x)-g(\alpha)} \frac{1-\frac{g(\alpha)}{g(x)}}{1-\frac{f(\alpha)}{f(x)}} .\]

Now, define

\[H_{\alpha}(x)=\frac{1-\frac{g(\alpha)}{g(x)}}{1-\frac{f(\alpha)}{f(x)}} \text { for } x \in B_{+}\left(\bar{x} ; \delta_{2}\right) .\] Since \(\displaystyle \lim _{x \rightarrow \bar{x}} f(x)=\displaystyle \lim _{x \rightarrow \bar{x}} g(x)=\infty\), we have that \(\displaystyle \lim _{x \rightarrow \bar{x}^{+}} H_{\alpha}(x)=1\). Thus, there exists a positive \(\gamma<\delta_{2}\) such that

\[\left|H_{\alpha}(x)-1\right|<\frac{\varepsilon}{2 K} \text { whenever } x \in B_{+}(\bar{x} ; \gamma) .\]

For any \(x \in B_{+}(\bar{x} ; \gamma)\), applying Theorem 4.2.4 on the interval \([x, \alpha]\), we can write \([f(x)-f(\alpha)] g^{\prime}(c)= [g(x)-g(\alpha)] f^{\prime}(c)\) for some \(c \in (x, \alpha)\) (note that, in particular, \(c \in B\left(\bar{x}: \delta_{1}\right) \cap(a, b)\)). For such \(c\) we get \[\frac{f(x)}{g(x)}=\frac{f^{\prime}(c)}{g^{\prime}(c)} H_{\alpha}(x) .\] Since \(c \in B\left(\bar{x}: \delta_{1}\right) \cap(a, b)\), applying \(\PageIndex{20}\) we get that, for \(x \in B_{+}(\bar{x} ; \gamma)=B_{+}(\bar{x} ; \gamma) \cap(a, b)\),

\[\begin{aligned}
\left|\frac{f(x)}{g(x)}-\ell\right| &=\left|\frac{f^{\prime}(c)}{g^{\prime}(c)} H_{\alpha}(x)-\ell\right| \\
&=\left|\frac{f^{\prime}(c)}{g^{\prime}(c)}\left(H_{\alpha}(x)-1\right)+\frac{f^{\prime}(c)}{g^{\prime}(c)}-\ell\right| \\
& \leq\left|\frac{f^{\prime}(c)}{g^{\prime}(c)}\right|\left|H_{\alpha}(x)-1\right|+\left|\frac{f^{\prime}(c)}{g^{\prime}(c)}-\ell\right| \\
&<K \frac{\varepsilon}{2 K}+\frac{\varepsilon}{2}=\varepsilon \text {.}
\end{aligned}\]

Setting \(\delta_{0}=\gamma\) completes the proof. \(\square\)

Example \(\PageIndex{5}\)

Consider the limit \[\displaystyle \lim _{x \rightarrow 0} \frac{\ln x^{2}}{1+\frac{1}{\sqrt[3]{x^{2}}}} . \nonumber\]

Solution

Here \(f(x)=\ln x^{2}\), \(g(x)=1+\frac{1}{\sqrt[3]{x^{2}}}\), \(\bar{x} = 0\), and we may take as \((a, b)\) any open interval containing \(0\). Clearly \(f\) and \(g\) satisfy the differentiability assumptions and \(g^{\prime}(x) \neq 0\) for all \(x \neq 0\). Moreover, \(\displaystyle \lim _{x \rightarrow \bar{x}} f(x)=\displaystyle \lim _{x \rightarrow \bar{x}} g(x)=\infty\). We analyze the quotient of the derivatives. We have

\[\displaystyle \lim _{x \rightarrow 0} \frac{2 / x}{-\frac{2}{3} \frac{1}{\sqrt[3]{x^{5}}}}=\displaystyle \lim _{x \rightarrow 0}-3 \frac{\sqrt[3]{x^{5}}}{x}=\displaystyle \lim _{x \rightarrow 0}-3 \sqrt[3]{x^{2}}=0 . \nonumber\]

It now follows from Theorem 4.4.2 that

\[\displaystyle \lim _{x \rightarrow 0} \frac{\ln x^{2}}{1+\frac{1}{\sqrt[3]{x^{2}}}}=0 . \nonumber\]

Remark \(\PageIndex{3}\)

The proofs of Theorem 4.4.1 and Theorem 4.4.2 show that the results in these theorems can be applied for left-hand and right-hand limits. Moreover, the results can also be modified to include the case when \(\bar{x}\) is an endpoint of the domain of the functions \(f\) and \(g\).

The following theorem can be proved following the method in the proof of Theorem 4.4.1.

Theorem \(\PageIndex{4}\)

Let \(f\) and \(g\) be differentiable on \((a, \infty\). Suppose \(g^{\prime}(x) \neq 0\) for all \(x \in (a, \infty\) and

\[\displaystyle \lim _{x \rightarrow \infty} f(x)=\displaystyle \lim _{x \rightarrow \infty} g(x)=0 .\]

If \(\ell \in \mathbb{R}\) and

\[\displaystyle \lim _{x \rightarrow \infty} \frac{f^{\prime}(x)}{g^{\prime}(x)}=\ell ,\] then \[\displaystyle \lim _{x \rightarrow \infty} \frac{f(x)}{g(x)}=\ell .\]

Example \(\PageIndex{6}\)

Consider the limit \[\displaystyle \lim _{x \rightarrow \infty} \frac{1}{x\left(\frac{\pi}{2}-\arctan x\right)} . \nonumber\]

Solution

Writing the quotient in the form \(\frac{1 / x}{\frac{\pi}{2}-\arctan x}\) we can apply Theorem 4.4.4. We now compute the limit of the quotient of the derivatives

\[\displaystyle \lim _{x \rightarrow \infty} \frac{-1 / x^{2}}{-\frac{1}{x^{2}+1}}=\displaystyle \lim _{x \rightarrow \infty} \frac{x^{2}+1}{x^{2}}=1 . \nonumber\]

In view of Theorem 4.4.4 the desired limit is also \(1\).

The following theorem can be proved following the method in the proof of THeorem 4.4.2.

Theorem \(\PageIndex{5}\)

Let \(f\) and \(g\) be differentiable on \((a, \infty)\). Suppose \(g^{\prime}(x) \neq 0\) for all \(x \in (a, \infty)\) and \[\displaystyle \lim _{x \rightarrow \infty} f(x)=\displaystyle \lim _{x \rightarrow \infty} g(x)=\infty . \] If \(\ell \in \mathbb{R}\) and \[\displaystyle \lim _{x \rightarrow \infty} \frac{f^{\prime}(x)}{g^{\prime}(x)}=\ell ,\] then \[\displaystyle \lim _{x \rightarrow \infty} \frac{f(x)}{g(x)}=\ell .\]

Example \(\PageIndex{7}\)

Consider the limit \[\displaystyle \lim _{x \rightarrow \infty} \frac{\ln x}{x} .\]

Solution

Clearly the functions \(f(x)=\ln x\) and \(g(x)=x\) satisfy the conditions of Theorem 4.4.5. We have

\[\displaystyle \lim _{x \rightarrow \infty} \frac{f^{\prime}(x)}{g^{\prime}(x)}=\displaystyle \lim _{x \rightarrow \infty} \frac{1 / x}{1}=0 \nonumber\]

It follows from Theorem 4.4.5 that \(\displaystyle \lim _{x \rightarrow \infty} \frac{\ln x}{x}=0\)

Exercise \(\PageIndex{1}\)

Use L'Hospital's Rule to find the following limits (you may assume known all the relevant derivatives from calculus):

1. \(\displaystyle \lim _{x \rightarrow-2} \frac{x^{3}-4 x}{3 x^{2}+5 x-2}\).
2. \(\displaystyle \lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}}{\sin x \cos x}\).
3. \(\displaystyle \lim _{x \rightarrow 1} \frac{x-1}{\sqrt{x+1}-\sqrt{2}}\).
4. \(\displaystyle \lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}}{\ln (1+x)}\).
5. \(\displaystyle \lim _{x \rightarrow 1} \frac{\ln x}{\sin (\pi x)}\).

Exercise \(\PageIndex{2}\)

For the problems below use L'Hospital's rule as many times as appropriate to determine the limts.

1. \(\displaystyle \lim _{x \rightarrow 0} \frac{1-\cos 2 x}{x \sin x}\).
2. \(\displaystyle \lim _{x \rightarrow 0} \frac{\left(x-\frac{\pi}{2}\right)^{2}}{1-\sin x}\).
3. \(\displaystyle \lim _{x \rightarrow 0} \frac{x-\arctan x}{x^{3}}\).
4. \(\displaystyle \lim _{x \rightarrow 0} \frac{x-\sin x}{x-\tan x}\).

Exercise \(\PageIndex{3}\)

Use the relevant version of L'Hospital's rule to compute each of the following limits.

1. \(\displaystyle \lim _{x \rightarrow \infty} \frac{3 x^{2}+2 x+7}{4 x^{2}-6 x+1}\).
2. \(\displaystyle \lim _{x \rightarrow 0^{+}} \frac{-\ln x}{\cot x}\).
3. \(\displaystyle \lim _{x \rightarrow \infty} \frac{\frac{\pi}{2}-\arctan x}{\ln \left(1+\frac{1}{x}\right)}\).
4. \(\displaystyle \lim _{x \rightarrow \infty} \sqrt{x} e^{-x}\). (Hint: first rewrite as a quotient.)

Exercise \(\PageIndex{4}\)

Prove that the following functions are differentiable at \(1\) and \(-1\).

1. \(f(x)=\left\{\begin{array}{ll}
   x^{2} e^{-x^{2}}, & \text { if }|x| \leq 1 \text {;}\\
   \frac{1}{e}, & \text { if }|x|>1 \text {.}
   \end{array}\right.\)
2. \(f(x)=\left\{\begin{array}{ll}
   \arctan x, & \text { if }|x| \leq 1 \text {;}\\
   \frac{\pi}{4} \operatorname{sign} x+\frac{x-1}{2}, & \text { if }|x|>1 \text {.}
   \end{array}\right.\)

Exercise \(\PageIndex{5}\)

Let \(P(x)\) be a polynomial. Prove that \[\displaystyle \lim _{x \rightarrow \infty} P(x) e^{-x}=0 . \nonumber\]

Exercise \(\PageIndex{6}\)

Consider the function

\[f(x)=\left\{\begin{array}{ll}
e^{-\frac{1}{x^{2}}}, & \text { if } x \neq 0 \text {;}\\
0, & \text { if } x=0 \text {.}
\end{array}\right.\]

Prove that \(f \in C^{n}(\mathbb{R})\) for every \(n \in \mathbb{N}\).