Answer
$$\lim_{x\to\infty}\sqrt xe^{-x/2}=0$$
Work Step by Step
$$A=\lim_{x\to\infty}\sqrt xe^{-x/2}$$
$$A=\lim_{x\to\infty}\frac{\sqrt x}{e^{x/2}}$$
$$A=\lim_{x\to\infty}\frac{\sqrt x}{\sqrt{e^x}}$$
$$A=\lim_{x\to\infty}\sqrt{\frac{x}{e^x}}$$
$$A=\sqrt{\lim_{x\to\infty}\frac{x}{e^x}}$$
As $x\to\infty$, $e^x\to\infty$. We have an indeterminate form of $\infty/\infty$, which can dealt with by L'Hospital's Rule:
$$A=\sqrt{\lim_{x\to\infty}\frac{x'}{(e^x)'}}$$
$$A=\sqrt{\lim_{x\to\infty}\frac{1}{e^x}}$$
We know from above that $\lim_{x\to\infty}(e^x)=\infty$, therefore, $\lim_{x\to\infty}\frac{1}{e^x}=0$.
So, $$A=\sqrt0=0$$
