Solution
= Solution
For every $n\geq1$, $|x^n\sin(1/x)|\leq|x|^n\to0$, so the <squeeze theorem> gives limit zero. For $n=0$, sequences $x_k=(\pi/2+2k\pi)^{-1}$ and $\widetilde x_k=(3\pi/2+2k\pi)^{-1}$ give function values $1$ and $-1$, so no limit exists. Hence
$$
\boxed{n=1,2,3,4.}
$$
The value assigned at zero when $n=4$ does not affect this punctured <limit>.