Evaluation of Derivative Using $\epsilon−\delta$ Definition Consider the function $f \colon\mathbb R \to\mathbb R$ defined by
$f(x)=
\begin{cases}
  x^2\sin(1/x); & \text{if }x\ne 0, \\
  0 & \text{if }x=0.
\end{cases}$
Use $\varepsilon$-$\delta$ definition to prove that the limit $f'(0)=0$.
Now I see that h should equals to delta; and delta should equal to epsilon in this case. Thanks for everyone contributed!

$$\left|{\dfrac{f(h)-f(0)}{h}}\right|=\left|{\dfrac{2h^2 \sin{\dfrac{1}{h}}}{h}}\right|=2 \left|{h \sin{\dfrac{1}{h}}}\right|<2\left|h\right|<\varepsilon.$$
Choose $\delta<\dfrac{\varepsilon}{2}.$
