Can someone help me prove this?
sqrt(x)e^sin(pi/x) = 0 , x--> 0+ (from the right)
$\displaystyle sin(\pi/x)$ lies between -1 and 1 for all x so $\displaystyle e^{sin(\pi/x)}$ lies between $\displaystyle e^{-1}$ and $\displaystyle e^1$ for all x. The important point is that it is bounded. Since $\displaystyle \sqrt{x}$ goes to 0, the limit of this function is 0.