this is my attempt to prove that $ \ \lim_{x\rightarrow 0}(x^2-x)\sin\left ( \frac{1}{x} \right )=0 \ \ $ using $\epsilon -\delta$ definition
if $\left | x \right |< \delta \leq 1$
$\left | (x^2-x)\sin\left ( \frac{1}{x} \right ) \right |\leq \left | x^2-x \right |=\left | x \right |\left | x-1 \right |\leq \left | x \right |(\left | x \right |+1)=x^2+\left | x \right |\leq 2\left | x \right |$
it is enough to take $\delta =\min\left \{ 1,\frac{\epsilon }{2} \right \}$
please tell me if this solution is true
is there better way to do that?