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!  
 A: 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 $L$, $\max(|\sin t-L|)\ge 1$).
A: First, $\lim_{x\to 0}x\sin(1/x)=0\ne 1$.  To show this, simply note that that given $\epsilon>0$
$$|x\sin(1/x)|\le |x|<\epsilon$$
whenever $|x|<\delta=\epsilon$.  And we are done with that.

In the OP, from L'Hospital's Rule, the limit,
$$\lim_{x\to 0}\frac{\sin(1/x)}{1/x}=0$$
since the numerator is bounded by $1$ in absolute value and the denominator approaches infinity.


Next, if $f'(0)$ exists, then $f'(0)=\lim_{h\to 0}\frac{f(h)-f(0)}{h}=\lim_{h\to 0}\sin(1/h)$ exists.
But, let $\epsilon=1$.  Then, for all $\delta>0$ we can take $x=\frac{1}{2n\pi +t}$, $t\in (0,2\pi)$,and by choosing $n$ sufficiently large, we force $|x|<\delta$.
Then, for this $|x|<\delta$, we see that
$|\sin(1/x)|=|\sin(t)|$
By choosing $t\in (0,2\pi)$, we can make $|\sin(1/x)-L|=|\sin(t)-L|\ge 1=\epsilon$ for any $L\in [-1,1]$.

Thus the limit fails to exist and $f$ is not differentiable at $0$.

A: we have $-1< sin(1/x)<1$ implies that $-x< x*sin(1/x) <x$
finally $lim x→0 (x*sin(1/x))$ = $limx→0$  $x$ = $lim$ x→0 $-x$ = $0$
sandwich theorem 
