Where is the mistake ( in using mean value theorem)? $$   f(x)=
      \begin{cases} 
      x^2\sin \frac1x &  x \ne 0 \\
      0 & x=0\\ 
   \end{cases}
$$
$f$ is differentiable everywhere and 
$$   f'(x)=
      \begin{cases} 
      2x\sin \frac1x-\cos \frac1x &  x \ne 0 \\
      0 & x=0\\ 
   \end{cases}
$$
$f$ satisfies the MVT. Using it on $(0,x)$ we get:
$$\frac{x^2\sin \frac1x-0}{x-0}= 2c\sin \frac1c-\cos \frac1c$$
$c\in(0,x)$
When $x\to0$ then $c\to0$. So we have a contradiction
$$0=\lim \limits_{x \to 0}x\sin \frac1x=\lim \limits_{c \to 0}2c\sin\frac1c-\cos\frac1c$$
Last limit doesn't exist where is the mistake ?
I see that $\lim \limits_{x \to 0}f'(x) $ doesn't exist but MVT still applies?
Using a limit inside of an interval is something i dont understand dont we then get a single point? This process is important it used in the proof of l"Hospital rule.

The mean value theorem says that there exists some $c$ in the interval $(0,x)$ such that [...]. And it is indeed the case that for any $x>0$ you can find such a $c$. That is not to say that any $c$ in $(0,x)$ satisfies the MVT, or even that the other numbers in $(0,x)$ behave nicely.
So what the MVT actually tells you in this case is that for any $x>0$, there is a $c_x\in(0,x)$, and these $c_x$ are such that $$2c_x\sin\frac1{c_x}-\cos \frac1{c_x}\to 0$$
