Calculus, function that has fractured derivative

Question

I have a function which has fractured derivative at $x=0$. How can i prove that it really has fractured derivative at $x=0$. Here is my function $$ f(x)=\begin{cases} x^2 \sin(1/x), & x \ne 0, \\ 0, & x=0 \end{cases} $$.

Answer 1

Hint: Computing $$\frac{f(x_0+h)-f(x_0)}{h}=\frac{f(h)}{h}=h\sin(1/h)$$ and $$|h\sin(1/h)|\le |h|$$ and this tends to zero for $h$ tends to zero.