Prove that $f'(0)$ does not exist for the given function $f(x)$ If we want to show this we must show that $f$ is not differentiable at $x=0$.  
The function is defined as follows:  
$$f(x)=
\begin{cases}
x\sin{\frac{1}{x}}, & \text{if $x\ne0$} \\
0,  &\text{if $x=0$} 
\end{cases}
$$  which is a piecewise function.  
I say $f'(0)$ is not defined because $\lim_{x \to 0} f(x) = 1$ but $f(0) = 0$ is not the same value, so $f'(0)$ DNE, so $f$ is not differentiable at $0$, but my professor say this is completely bad!  
I say $\lim_{x \to 0} x\sin(\frac{1}{x}) = \lim_{x \to 0} \dfrac{\sin(\frac{1}{x})}{\frac{1}{x}} = 1$.  But I think this is not right and I don't know how to show this does not exist!  

Short answer:
$$\lim_{h\to0}\frac{h\sin\dfrac1h-0}h=\lim_{h\to0}\sin\dfrac1h=\lim_{t\to\infty}\sin t$$
doesn't exist (because for any $L$, $\max(|\sin t-L|)\ge 1$).
