1. ## limits

how can we evaluate

limit [x sin (1/x)]
x->0

2. The sandwich theorem should work. Recall that $\displaystyle -1 \leq \sin{\left(\frac{1}{x}\right)} \leq 1$.

This would mean
\displaystyle \begin{align*} -x \leq x\sin{\left(\frac{1}{x}\right)} &\leq x \\ \lim_{x \to 0}(-x) \leq \lim_{x \to 0}{x\sin{\left(\frac{1}{x}\right)}} &\leq \lim_{x \to 0}x \end{align*}