Answer
$$\lim_{x\to\infty}\frac{\ln x}{\sqrt x}=0$$
Work Step by Step
$$A=\lim_{x\to\infty}\frac{\ln x}{\sqrt x}$$
As $x\to\infty$, $\ln x$ approaches $\infty$ and $\sqrt x$ also approaches $\infty$.
So this limit is in an indeterminate form of $\infty/\infty$. We can use L'Hospital's Rule here:
$$A=\lim_{x\to\infty}\frac{\frac{d}{dx}(\ln x)}{\frac{d}{dx}(\sqrt x)}$$
$$A=\lim_{x\to\infty}\frac{\frac{1}{x}}{\frac{1}{2\sqrt x}}$$
$$A=\lim_{x\to\infty}\frac{2\sqrt x}{x}$$
Divide both numerator and denominator by $x$, we have
$$A=\lim_{x\to\infty}\frac{\frac{2\sqrt x}{x}}{\frac{x}{x}}$$
$$A=\lim_{x\to\infty}\frac{2\sqrt{\frac{x}{x^2}}}{1}$$
$$A=\lim_{x\to\infty}2\sqrt{\frac{1}{x}}$$
We remember that $\lim_{x\to\infty}(\frac{1}{x})=0$. So,
$$A=2\sqrt0$$
$$A=0$$
