how can we evaluate
limit [x sin (1/x)]
x->0
The sandwich theorem should work. Recall that $\displaystyle \displaystyle -1 \leq \sin{\left(\frac{1}{x}\right)} \leq 1$.
This would mean
$\displaystyle \displaystyle \begin{align*} -x \leq x\sin{\left(\frac{1}{x}\right)} &\leq x \\ \lim_{x \to 0}(-x) \leq \lim_{x \to 0}{x\sin{\left(\frac{1}{x}\right)}} &\leq \lim_{x \to 0}x \end{align*}$