Answer
$0$
Work Step by Step
Given
$$ \lim _{x \rightarrow \infty} \sqrt{x} e^{-x / 2}$$
Rewrite
\begin{aligned}
\lim _{x \rightarrow \infty} \sqrt{x} e^{-x / 2}&= \lim _{x\to \infty \:}\left(\frac{1}{e^{\frac{x}{2}}}\sqrt{x}\right)\\
&= \lim _{x\to \infty \:}\left(\sqrt{\frac{x}{e^x}}\right)\\
&= \sqrt{\lim _{x\to \infty \:}\left(\frac{x}{e^x}\right)}
\end{aligned}
Apply L'Hopital's rule for
\begin{aligned}
\lim _{x\to \infty \:}\left(\frac{x}{e^x}\right)&= \lim _{x\to \infty \:}\left(\frac{1}{e^x}\right)\\
&=\frac{1}{\infty}=0
\end{aligned}
Then
\begin{aligned}
\lim _{x\to \infty \:}\left(\sqrt{\frac{x}{e^x}}\right)&= \sqrt{0}=0
\end{aligned}