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