"$f^\prime (c) = \lim_{n \to \infty} f^\prime (x_n)$" does not imply the continuity. Question: suppose $f : [a,b] \rightarrow R$ is a differentiable and $c \in [a,b]$. Then show there exists a sequence $\{x_n\}$ converging to $c$, $x_n \not = c$ for all $n$, such that $f^\prime (c) = \lim_{n \to \infty} f^\prime (x_n)$. Do note this does not imply that $f^\prime$ is continuous (why?).
Attempt: Let $x_n \in [a,b]$ Then, $f(x_n)$ is differentiable.  Then, $f^\prime (c) = f^\prime(\lim_{n \to \infty}x_n)=\lim_{n \to \infty}f^\prime(x_n)$.
I don't understand why this does not imply continuity because if $f(c)=\lim_{n \to \infty}f(x_n)$, $f$ is continuous at $c$. Isn't it the case for the derivative function?  
Thank you in advance. 
Edit: I agree that my attempt above assumes that $f^\prime$ is continuous, which might not be true. So, could you give me some hint to solve this problem without using that assumption? 
Let $x_n \in [a,b]$. Then, $f(x_n)$ is differentiable. Then, $\lim_{x_n \to c} \frac {f(x_n)-f(c)}{x_n-c}=f^\prime (c)$.I don't know how to proceed from here. 
 A: For $f(x)=x^{2}\cos(1/x)$, $x\ne 0$, $f(0)=0$, then $f'(0)=0$ and $f'(x)=2x\cos(1/x)+\sin(1/x)$, and $f'$ is not continuous at $x=0$. Still, $f'(x_{n})\rightarrow 0$ for $x_{n}=\dfrac{1}{2\pi n}$.
A: Your proof is flawed, as it goes by assuming that $f'$ is continuous at $c$ when you write:

$$
f'(c) = f(\lim_{n\to\infty} x_n)
$$
  for (any) sequence $x_n)_n$ converging to $c$.

But $f'$ may not be continuous at $c$, that is the point.
To prove the statement (which is "there exists a sequence $(x_n)_n$," crucially not "for every sequence $(x_n)_n$" as the former would be equivalent to continuity of $f'$), you can use the mean value theorem. (And note that the MVT asks only for differentiability, not that $f$ be continuously differentiable.)
Indeed, by the MVT we have that for every $n\geq 1$, there exists $x_n \in (c, c+\frac{1}{n})$ such that
$$
f'(x_n) = \frac{f(c+\frac{1}{n})-f(c)}{\frac{1}{n}}\,.
$$
But now, by the very definition of $f'(c)$, we have
$$
f'(c) = \lim_{x\to c} \frac{f(x)-f(c)}{x-c}
$$
so in particular 
$$
f'(c) = \lim_{n\to\infty} \frac{f(c+\frac{1}{n})-f(c)}{\frac{1}{n}}\,.
$$
