limits

Note that in the limit $x \ne 0$ thus $\left| {\sin \left( {\frac{1}{{x^2 }}} \right)} \right| \leqslant 1$
So $\left| {x\sin \left( {\frac{1}{{x^2 }}} \right)} \right| \leqslant \left| x \right|$.
Just let $\varepsilon > 0 \Rightarrow \quad \delta = \varepsilon$.