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?
The problem is that $\xi\rightarrow a$ doesn't imply $h\rightarrow0$.
Observe that if $a&lt;\xi&lt;a+h$ or $a&gt;\xi&gt;a+h$, then $|h|&gt;|\xi-a|$. Now examine the definition of limit, one can see at once that $|h|$ may very well be large given that $|\xi-a|$ is arbitrarily small.
The problem is that $\xi\rightarrow a$ doesn't imply $h\rightarrow0$.
Observe that if $a&lt;\xi&lt;a+h$ or $a&gt;\xi&gt;a+h$, then $|h|&gt;|\xi-a|$. Now examine the definition of limit, one can see at once that $|h|$ may very well be large given that $|\xi-a|$ is arbitrarily small.