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 \(\sqrt{x}\text{.}\)