Answer
See graph and explanations.
Work Step by Step
See graph. It can be seen that the function will oscillate within $[-1,1]$ as x approaches 0 from the left or the right. This is because, as $x\to0^-$ or $x\to0^+$, we have $\frac{1}{x}\to-\infty$ or $\frac{1}{x}\to\infty$. In each case $y=sin(\frac{1}{x})$ will not converge, but will oscillate within $[-1,1]$.
