Answer
$f'(0)$ does not exist.
Work Step by Step
$f(x) = x~sin(\frac{1}{x})~~~$ if $x\neq 0$
$f(x) = 0~~~~~~~$ if $x = 0$
We can try to evaluate $f'(0)$ using Definition 4:
$\lim\limits_{h \to 0} \frac{f(0+h)-f(0)}{h}$
$=\lim\limits_{h \to 0} \frac{(0+h)~sin(\frac{1}{0+h})-0}{h}$
$=\lim\limits_{h \to 0} \frac{h~sin(\frac{1}{h})}{h}$
$=\lim\limits_{h \to 0} sin(\frac{1}{h})$
As $h$ approaches $0$, the value of $sin(\frac{1}{h})$ moves back and forth continuously between $1$ and $-1$. The function does not converge on one single point. Therefore, the limit does not exist.
$f'(0)$ does not exist.
