I was working on a problem and I just can not figure out how to accomplish what it is asking. Attached is the problem and figure.
$\displaystyle f'(x)\lim\limits_{h\to 0}\frac{f(x+h)-f(x)}{h}\Rightarrow f'(0)=\lim\limits_{h\to 0}\frac{f(0+h)-f(0)}{h}$
$\displaystyle f'(0)=\lim\limits_{h\to 0}\frac{f(h)-f(0)}{h}=\frac{h^2\sin \frac{1}{h}-0}{h}=$
$\displaystyle \lim\limits_{h\to 0}h\sin \frac{1}{h}= \lim\limits_{h\to 0}\frac{\sin \frac{1}{h}}{\frac{1}{h}}=0. $
So if you use that extension of the function f, you can say that at x=0, f'(0)=0 and the interpretation is that the tangent line would be the x-axis.
$\displaystyle f'(x)=\left\{ \begin{array}{rl} \left(x^2\sin\frac{1}{x}\right)', & \forall x \neq 0,\\
0, & \text(if ) x=0. \end{array}\right.$
No you have to prove that h it is continuous.