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What is wrong with the following argument?

Let $f$ be a real continuous function which is differentiable at every point $x\not=a$. By mean value theorem, $$ \frac{f(a+h)-f(a)}{h} = f'(\xi) \qquad (h\not=0)$$ where $\xi$ is between $a$ and $a+h$. Letting $h\rightarrow0$, we obtain $$ f'(a) = \lim\limits_{\xi\rightarrow a}f'(\xi) \mbox{.} $$ Therefore, f'(a) exists if and only if $\lim\limits_{\xi\rightarrow a}f'(\xi)$ exists.

That the only if statement is false can be seen from $f(x)=x^2\sin\frac{1}{x}$ for $x\not=0$, $f(0)=0$, which shows that $f'(0)$ exists but $\lim\limits_{x\rightarrow0}f'(x)$ does not. Why?

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  • $\begingroup$ Mean value theorem uses existence of derivative on interval. $\endgroup$
    – zkutch
    May 21, 2021 at 1:42
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    $\begingroup$ Mean value theorem requires that the function be continuous on a closed interval and the differentiability on the interior of that interval, and it does apply to my function. $\endgroup$
    – user912011
    May 21, 2021 at 1:44
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    $\begingroup$ You switched a $\exists$ with a $\forall$. There exists a net $\xi_h\to a$ (indexed by $h\downarrow 0$) such that $f'(\xi)\to D^+f(a)$, but it doesn't mean for all nets $\xi\to a$ we have $f'(\xi)\to D^+f(a)$. Also there is no reason why we would have $D^+f(a)=D^-f(a)$. $\endgroup$
    – user10354138
    May 21, 2021 at 1:44
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    $\begingroup$ For $x\neq0$ you have that $f'(x)=2x\sin(1/x)-\cos(1/x)$, which doesn't have a limit when $x\to0$. This is because $2x\sin(1/x)\to0$, but $\cos(1/x)$ keeps oscillating back and forth from $-1$ to $1$. The function $f$ is not only differentiable at $x\neq0$, but also at $x=0$, since $f'(0)=\lim_{x\to0}\frac{f(x)-f(0)}{x-0}=\lim_{x\to0}x\sin(1/x)=0$. $\endgroup$
    – plop
    May 21, 2021 at 1:58
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    $\begingroup$ If $f$ is differentiable at $a$, then, since for each interval $(a, a+h)$ there exists some $c$ with $f'(c)=(f(a+h)-f(a))/h$, we know that there is a sequence of points $c_n$ which converges to $a$, for which $f'(c_n) \to f'(a)$. But that doesn't mean that $\lim_{c\to a}f'(c)$ exists. $\endgroup$
    – Joe
    May 21, 2021 at 2:02

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