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Course Notes for Calculus I

Section 7.3 L’Hôpital’s Rule

Subsection 7.3.1 Derivatives in Limits

The first application of derivatives is a strange reversal: after using limits to define derivatives, I am going to use derivatives to calculate limits. L’Hôpital’s rule is a method that applies to indeterminante form limits of type \(\frac{\infty}{\infty}\) or type \(\frac{0}{0}\text{.}\) (I defined these types of limits in Section 5.2.) In this case, if \(f\) and \(g\) are differentiable functions, L’Hôpital’s rule states that the limit is preserved if I differentiate both numerator and denominator.
\begin{equation*} \lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \lim_{x \rightarrow a} \frac{f^\prime(x)}{g^\prime(x)} \end{equation*}
This is particularly useful in asymptotic analysis, where many limits (as \(x \rightarrow \infty\)) have type \(\frac{\infty}{\infty}\text{.}\) Indeed, L’Hôpital’s rule can be used to prove many of the ideas in asymptotic analysis.

Subsection 7.3.2 Examples

Example 7.3.1.

Consider the claim that \(\ln x\) is asymptotically slower than \(\sqrt{x}\text{.}\) I will use L’Hôpital’s rule to prove the claim. Both the numerator and the denominator here approach \(\infty\text{,}\) so the rule applies. I differentiate both numerator and denominator, then simplify the nested fraction. The result is a limit I know how to do, where the numerator is constant and the denominator becomes very large.
\begin{equation*} \lim_{x \rightarrow \infty} \frac{\ln x}{\sqrt{x}} = \lim_{x \rightarrow \infty} \frac{\frac{1}{x}}{\frac{1}{2\sqrt{x}}} = \lim_{x \rightarrow \infty} \frac{2 \sqrt{x}}{x} = \lim_{x \rightarrow \infty} \frac{2}{\sqrt{x}} = 0 \end{equation*}
A limit of \(0\) means that the numerator has lower asymptotic order than the denominator; I was justified in Subsection 5.2.4 when I said that the logarithm \(\ln x\) is asymptotically slower that \(\sqrt{x}\text{.}\)