## Calculus (3rd Edition)

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Since $$\lim _{x \rightarrow \infty} \frac{\ln x }{x^{1/2}}=\frac{\infty}{\infty}$$ is an intermediate form, then we can apply L’Hôpital’s Rule as follws $$\lim _{x \rightarrow \infty} \frac{1/ x }{1/(2x^{1/2})}=\lim _{x \rightarrow \infty} \frac{2 }{x^{1/2}}=\frac{2}{\infty}=0.$$