$$ f(x)=
\begin{cases}
x^2\sin \frac1x & x \ne 0 \\
0 & x=0\\
\end{cases}
$$
$f$ 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.
Best Answer
In the last equation, the "constant" $c$ actually depends on $x$, so it can really be viewed as a function $c(x)$:
$$0=\lim \limits_{x \to 0}x\sin \frac1x=\lim \limits_{c \to 0}2c(x)\sin\frac1{c(x)}-\cos\frac1{c(x)}$$
All you know about $c(x)$ is that $0<c(x)<x$ (which implies $c(x)\to 0$ when $x\to 0$), but it may have nice additional properties that "smoothen" the expression on the right so that the limit would exist. (Imagine, for example, if $\frac{1}{c(x)}$ is always an odd multiple of $\frac{\pi}{2}$ so the "offending" $\cos$ is always $0$...)