x times sin(1/x)

Idea. As $x\to0$, the factor $\sin\frac1x$ oscillates faster and faster between $-1$ and $1$, but the factor $x$ shrinks to $0$ and wins.

Computation. For every $x\neq0$, $\left|\sin\frac1x\right|\le1$, so
$$0\le\left|x\sin\frac1x\right|\le|x|.$$
Since $|x|\to0$ as $x\to0$, the squeeze theorem gives $\left|x\sin\frac1x\right|\to0$, hence $x\sin\frac1x\to0$.