Why does not differentiability at a point imply the existence of limit of derivative tending to the point

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?

• Mean value theorem uses existence of derivative on interval. May 21, 2021 at 1:42
• 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. May 21, 2021 at 1:44
• 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 $D^+f(a)=D^-f(a)$. May 21, 2021 at 1:44
• 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 $f'(0)=\lim_{x\to0}\frac{f(x)-f(0)}{x-0}=\lim_{x\to0}x\sin(1/x)=0$.
– plop
May 21, 2021 at 1:58
• 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.
– Joe
May 21, 2021 at 2:02