Let $f : (a,b)\to\mathbb{R}$ be continuous and differentiable on $(a,b)$ and differentiable at a point k. Does $\lim_{x\to k} f'(x)$ exist?

Question

I'd like to prove that this proposition holds. Is it enough to say that by supposition, $\displaystyle{f'(k)=\lim_{h\to0} \frac{f(k+h)-f(k)}{h}}$ exists? Then, how can we prove that $\displaystyle{\lim_{x\to k} f'(x)=f'(k)}$ from this proposition?

Answer 1

If f is not differentiable, no... If $f$ is not differentiable at $k$, then no. For instance, take $x\mapsto|x|$.

Its derivative coincides with$\frac{|x|}{x}$ for every $x\in(a,b)\setminus \{0\}$, but is not defined in $0$. You can easily see that $$ \lim_{x\to 0^{+}}\frac{|x+h|-|x|}{h}=\frac{x+h-x}{h}=1 \\ \lim_{x\to 0^{-}}\frac{|x+h|-|x|}{h}=\frac{-x-h+x}{h}=-1 $$ so $\lim{x\to 0} \frac{\mathrm d}{\mathrm dx}|x|$ does not exist.

... but even if it is.

The hypothesis that also $f'(k)$ exists it still does not suffice.

The classic evil example is $$ f(x)=x^2\sin \left (\frac{1}{x} \right) \qquad \textrm {for } x\neq 0 $$ and $f(0)=0$. If you differentiate this function you obtain $$ f'(x)=2x\sin \left (\frac{1}{x} \right )-\cos \left ( \frac{1}{x} \right) \qquad \textrm {for } x\neq 0 $$ and $f'(0)=0$. The bad news is that the limit $$ \lim_{x\to 0} f'(x) $$ does not exist.