analysis - prove this problem - MVT Let f be a differentiable function on (a,b) and let c $\in(a,b)$
Then there exist a sequence{$x_n$} in (a,b) with $x_n\neq c$ and $x_n \to c$ such that
$f'(x_n)\to f'(c)$
 A: I'll quickly give a proof of Darboux's theorem which you can find here.

Theorem: If $f:I\to\mathbb R$ is differentiable then $f'$ has the intermediate value property.

Proof: Take $\alpha,\beta\in I$ and assume w.l.o.g. that $f'(\alpha)<f'(\beta)$. Take $y$ with $f'(\alpha)<y<f'(\beta)$. Define $\phi:[\alpha,\beta]\to\mathbb R$ by
$$\phi(x)=f(x)-yx.$$
since $\phi$ is continuous with compact domain, it achieves a maximum, and since it's differentiable, it can be found at a critical point. Note $\phi'(\alpha)=f'(\alpha)-y<y-y=0$ so the maximum cannot be achieved at $\alpha$, or by a similar argument at $\beta$. Thus there exists an $x\in(\alpha,\beta)$ that is a maximum of $\phi$, ie. $f'(x)-y=0$. This proves the intermediate value property. QED
Now the proof of your question is very easy. We'll construct the sequence $(x_n)$ inductively. Take $x_0\in(a,c)$. By the Darboux theorem, given $x_n<c$, there exists $x_{n+1}\in(x_n,c)$ such that $f'(x_{n+1})=(f'(x_n)+f'(c))/2$. Thus
$$|f'(x_{n+1})-f'(c)|=\frac{1}{2}|f'(x_n)-f'(c)|$$
so
$$|f'(x_n)-f'(c)|=\frac{1}{2^n}|f'(x_0)-f'(c)|\to0\text{ as }n\to\infty$$
which proves the claim.
