## Calculus 8th Edition

$0$
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 $=0$