Differentiable Sequences Suppose that $f:[a, b]\rightarrow\mathbb{R}$ is differentiable and $c\in[a,b]$. Then show that there exists a sequence $\{x_n\}$, $x_n\neq c$, such that 
$$f'(c) = \lim_{n\rightarrow\infty} f'(x_n).$$
I am lost in knowing how to use this and where to go from here.
 A: Let us assume that $c\in (a,b)$; I'll leave you to consider the cases $c=a$ and $c=b$ which are identical in nature to the one that I'm going to consider. Choose $x_1\in (c,b)$ such that:
(1) $f'(x_1)=\frac{f(b)-f(c)}{b-c}$
using the mean value theorem applied to $f$ on the interval $[c,b]$. (Check that you understand this!) Now repeat: choose $x_2\in (c,x_1)$ such that:
(2) $f'(x_2)=\frac{f(x_1)-f(c)}{x_1-c}$
using the mean value theorem applied to $f$ on the interval $[c,x_1]$. Again: choose $x_3\in (c,x_2)$ such that:
(3) $f'(x_3)=\frac{f(x_2)-f(c)}{x_2-c}$
using the mean value theorem applied to $f$ on the interval $[c,x_2]$. Continue this process inductively to obtain a sequence $\{x_n\}_{n\in \mathbb{N}}$. 
We've obtained a sequence $\{x_n\}_{n\in \mathbb{N}}$ such that $c<\cdots<x_{n+1}<x_n<x_{n-1}<\cdots<x_2<x_1$ and:
(n) $f'(x_n)=\frac{f(x_{n-1})-f(c)}{x_{n-1}-c}$. 
Exercise 1: Prove that $\{x_n\}_{n\in \mathbb{N}}$ converges and that $f'(\lim_{n\to \infty} x_n)=\lim_{n\to\infty} f'(x_n)$. (Hint: Use (1), (2), ..., (n), ... above. You can assume $\{x_n\}$ converges if you can't prove this and solve the second part of this Exercise. Also, you need to use the differentiability of $f$ at $\lim_{n\to \infty} x_n$ here!)
Now here's the subtlety: we don't actually know that $\lim_{n\to \infty} x_n=c$. For example, it could just so have turned out that $c=\frac{1}{2}\in [0,1]$ and $x_n=\frac{3}{4}+\frac{1}{n+4}$ for all $n\in \mathbb{N}$. Of course, in this case, $\lim_{n\to \infty} x_n = \frac{3}{4}\neq c=\frac{1}{2}$. So:
Exercise 2: How can we modify the construction of the $x_n$'s to ensure that also $\lim_{n\to \infty} x_n = c$. Once we have this, Exercise 1 completes the proof! (Hint: In the construction of $x_n$ we applied the mean value theorem of $f$ to the interval $[c,x_{n-1}]$. Why not apply the mean value theorem to a smaller closed interval with $c$ still as the left endpoint to ensure that the $x_n$'s get closer and closer to $c$?)
I hope this helps! Notice I haven't given a complete answer if you don't do the Exercises. If you do Exercise 2, then you'll have a complete answer in addition to a solution of Exercise 1. Let me know if you have any difficulties with the two Exercises and I'm happy to help!
