Answer
Apply the squeeze theorem to prove that $\lim\limits_{x\to0^+}\sqrt xe^{\sin(\frac{\pi}{x})}=0$.
Work Step by Step
1) We know that $$-1\leq \sin(\frac{\pi}{x})\leq1$$
2) For $\geq1$, function $e^x$ increases as $x$ increases.
So, $$e^{-1}\leq e^{\sin(\frac{\pi}{x})}\leq e^1$$
3) Multiply by $\sqrt x$ throughout.
The inequality direction would not change, since $\sqrt x\geq0$ for $\forall x\in R$
$$\sqrt xe^{-1}\leq\sqrt xe^{\sin(\frac{\pi}{x})}\leq\sqrt xe^1$$
4) Now consider the following limits
$\lim\limits_{x\to0^+}\sqrt xe^{-1}=\sqrt0\times e^{-1}=0$
$\lim\limits_{x\to0^+}\sqrt xe^1=\sqrt0\times e^1=0$
Therefore, $\lim\limits_{x\to0^+}\sqrt xe^{-1}=\lim\limits_{x\to0^+}\sqrt xe^1=0$
5) Applying the sqeeze theorem to the findings in 3) and 4), we find that $$\lim\limits_{x\to0^+}\sqrt xe^{\sin(\frac{\pi}{x})}=0$$
