Answer
$$0$$
Work Step by Step
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
$$
\lim _{x \rightarrow \infty} \frac{1/ x }{1/(2x^{1/2})}=\lim _{x \rightarrow \infty} \frac{2 }{x^{1/2}}=\frac{2}{\infty}=0.
$$
