L’Hôpital’s Rule is named for the French mathematician Guillaume François Antoine Marquis de L’Hôpital (1661–1704), who was initially credited as the author of the first textbook on calculus in 1696. The name L’Hôpital is pronounced “Lo-pee-tal.” In fact, the real author was the Swiss mathematician Johann Bernoulli. L'Hôpital merely translated it from German into French.

THEOREM 1 L’Hôpital’s Rule

Assume that \(f(x)\) and \(g(x)\) are differentiable on an open interval containing \(a\) and that

Also assume that \(g'(x) \neq 0\) (except possibly at \(a\)). Then

if the limit on the right exists or is infinite (\(\infty\) or \(-\infty\)). This conclusion also holds if \(f(x)\) and \(g(x)\) are differentiable for \(x\) near (but not equal to) \(a\) and

Furthermore, this rule is valid for one-sided limits.

Example 1

Use L’Hôpital’s Rule to evaluate \(\lim\limits_{x\rightarrow 2}\dfrac{x^3-8}{x^4+2x-20}.\)

Solution Let \(f(x) = x^{3} -8\) and \(g(x) = x^{4} + 2x - 20\). Both \(f\) and \(g\) are differentiable and \(\frac{f(x)}{g(x)}\) is indeterminate of type \(\frac{0}{0}\) at \(a = 2\) because \(f(2) = g(2) = 0\):

• Numerator: \(f(2) = 2^{3} - 8 = 0\)
• Denominator: \(g(2) = 2^{4} + 2(2) - 20 = 0\)

Furthermore, \(g'(x) = 4x^{3} + 2\) is nonzero near \(x = 2\), so L’Hôpital’s Rule applies. We may replace the numerator and denominator by their derivatives to obtain

Example 2

Evaluate \(\lim\limits_{x\rightarrow 2}\dfrac{4-x^2}{\sin\pi x}\).

Solution The quotient is indeterminate of type \(\frac{0}{0}\) at \(x = 2\):

• Numerator: \(4 - x^{2} = 4 - 2^{2} = 0\)
• Denominator: \(\sin \pi x = \sin 2\pi = 0\)

The other hypotheses (that \(f\) and \(g\) are differentiable and \(g'(x) \neq 0\) for \(x\) near \(a = 2\)) are also satisfied, so we may apply L’Hôpital’s Rule:

Example 3

Evaluate \(\lim\limits_{x\rightarrow \tfrac{\pi}{2}}\dfrac{\cos^2x}{1-\sin x}\).

Solution Again, the quotient is indeterminate of type \(\frac{0}{0}\) at \(x=\frac{\pi}{2}\):

The other hypotheses are satisfied, so we may apply L’Hôpital’s Rule:

Note that the quotient \(\frac{-2\cos x\sin x}{-\cos x}\) is still indeterminate at \(x = \frac{\pi}{2}\). We removed this indeterminacy by cancelling the factor \(-\cos x\).

Example 4 The Form \(0\cdot\infty\)

Evaluate \(\lim\limits_{x\rightarrow 0^+}x\ln x\).

Solution This limit is one-sided because \(f(x) = x \ln x\) is not defined for \(x \leq 0\). Furthermore, as \(x \rightarrow 0+\),

• \(x\) approaches 0,
• \(\ln x\) approaches \(-\infty\).

So \(f(x)\) presents an indeterminate form of type \(0 \cdot \infty\) . To apply L’Hôpital’s Rule we rewrite our function as \(f(x) = \frac{\ln x}{x^{-1}}\) so that \(f(x)\) presents an indeterminate form of type \(-\frac{\infty}{\infty}\). Then L’Hôpital’s Rule applies:

Example 5 Using L’Hôpital’s Rule Twice

Evaluate \(\lim\limits_{x\rightarrow 0}\dfrac{e^x -x -1}{\cos x -1}\).

Solution For \(x = 0\), we have

A first application of L’Hôpital’s Rule gives

This limit is again indeterminate of type 0/0, so we apply L’Hôpital’s Rule again:

Example 6 Assumptions Matter

Can L’Hôpital’s Rule be applied to \(\lim\limits_{x\rightarrow 1}\dfrac{x^2 + 1}{2x+1}\)?

Solution The answer is no. The function does not have an indeterminate form because

However, the limit can be evaluated directly by substitution: \(\lim\limits_{x\rightarrow 1}\frac{x^2 + 1}{2x+1} = \frac{2}{3}\). An incorrect application of L’Hôpital’s Rule gives the wrong answer:

Example 7 The Form \(\infty-\infty\)

Evaluate \(\lim\limits_{x\rightarrow 0}\left(\dfrac{1}{\sin x} - \dfrac{1}{x}\right)\).

Solution Both \(\frac{1}{\sin x}\) and \(\frac{1}{x}\) become infinite at \(x = 0\), so we have an indeterminate form of type \(\infty - \infty\) . We must rewrite the function as

to obtain an indeterminate form of type \(\frac{0}{0}\). L’Hôpital’s Rule yields (see Figure 4.48):

Figure 4.48: The graph confirms that \(y=\frac{1}{\sin x} - \frac{1}{x}\) approaches \(0\) as \(x \rightarrow 0\).

Example 8 The Form \(0^0\)

Evaluate \(\lim\limits_{x\rightarrow 0^+}x^x\).

It is a fascinating question: what happens to \(x^x\) as \(x > 0 \) gets closer to 0. It is difficult enough to understand \(x^x\) and \(\lim\limits_{x\rightarrow 0^+}x^x\) is a limit that is hard to guess. However, using algebra and L'Hôpital's rule, we can calculate this limit.

Solution First, compute the limit of the logarithm \(\ln x^{x} = x \ln x\):

Since \(f(x) = e^{x}\) is continuous, we can exponentiate to obtain the desired limit (see Figure 4.49):

Figure 4.49: The function \(y = x^{x}\) approaches \(1\) as \(x \rightarrow 0^+\).

THEOREM 2 L’Hôpital’s Rule for Limits at Infinity

Assume that \(f(x)\) and \(g(x)\) are differentiable in an interval \((b, \infty )\) and that \(g'(x) \neq 0\) for \(x > b\). If \(\lim\limits_{x\rightarrow \infty}f(x)\) and \(\lim\limits_{x\rightarrow \infty}g(x)\) exist and either both are zero or both are infinite, then

provided that the limit on the right exists. A similar result holds for limits as \(x \rightarrow -\infty\) .

Which of \(f(x) = x^{2}\) and \(g(x) = x \ln x\) grows faster as \(x \rightarrow \infty\) ?

Solution Both\( f(x)\) and \(g(x)\) approach infinity as \(x \rightarrow \infty\) , so L’Hôpital’s Rule applies to the quotient:

We conclude that \(x \ln x \ll x^{2}\) (Figure 4.50).

Figure 4.50: ---

Example 10

Jonathan is interested in comparing two computer algorithms whose average run times are approximately \((\ln n)^{2}\) and \(\sqrt{n}\). Which algorithm takes less time for large values of \(n\)?

Solution Replace \(n\) by the continuous variable \(x\) and apply L’Hôpital’s Rule twice:

This shows that \((\ln x)^2\ll\sqrt x\). We conclude that the algorithm whose average time is proportional to \((\ln n)^{2}\) takes less time for large \(n\).

In other words, \(\lim\limits_{x\rightarrow \infty}\frac{e^x}{x^n}=\infty\) for all \(n\).

Proof

The theorem is true for \(n = 0\) since \(\lim\limits_{x\rightarrow \infty}e^x = \infty\). We use L’Hôpital’s Rule repeatedly to prove that \(\frac{e^x}{x^n}\) tends to \(\infty \) for \(n = 1, 2, 3,\ldots\). For example,

Then, having proved that \(\frac{e^x}{x} \rightarrow \infty\), we use L’Hôpital’s Rule again

Proceeding in this way, we prove the result for all whole numbers \(n\). A more formal proof would use the principle of induction. Finally, if \(k\) is any exponent, choose any whole number \(n\) such that \(n > k\). Then \(\frac{e^x}{x^n} < \frac{e^x}{x^k}\) for \(x > 1\), so \(\frac{e^x}{x^k}\) must also tend to infinity as \(x \rightarrow \infty\).

A full proof of L’Hôpital’s Rule, without simplifying assumptions, is presented in a supplement on the text’s Companion Web Site.