Answer
$0$
Work Step by Step
Given: $\lim\limits_{x\to\infty}\frac{\ln\sqrt x}{x^2}$
Here, $x\to\infty$, $\ln\sqrt x\to\infty$ and $x^2\to\infty$.
This shows an indeterminate form of type $\dfrac{\infty}{-\infty}$, and so we will apply L'Hospital Rule.
$\lim\limits_{x\to\infty}\dfrac{(\ln\sqrt x)'}{(x^2)'}=\lim\limits_{x\to\infty}\dfrac{\dfrac{1}{\sqrt x}(\sqrt x)'}{2x}=\lim\limits_{x\to\infty}\dfrac{\frac{1}{\sqrt x}\frac{1}{2\sqrt x}}{2x}$
or, $=\lim\limits_{x\to\infty}\dfrac{1}{4x^2}$
or, $=\dfrac{1}{4}\lim_{x\to\infty}\dfrac{1}{x^2}$
Because $\lim\limits_{x\to\infty}\dfrac{1}{x^2}=0$
or, $=\dfrac{1}{4}(0)$
Thus, we have $=0$