If $c$ is critical point and $x_{n}\to c$ then $f''(c)=0.$ Let $f:I \to \mathbb{R}$ be a function. Let $c\in \mathrm{Crit}(f)$ with $x_n\neq c$ and $(x_n)\in \mathrm{Crit}(f)$ where $\mathrm{Crit}(f)=\{x \in I:f'(x)=0\}$
Asumme that $x_{n}\to c$  then $f''(c)=0.$
Any suggestions .Thanks 
Is there counterexample for the case that $f\not \in C^2$?
 A: Consider 
$$ f(x)=\begin{cases}x^2\cos x^{-3}&x\ne0\\0&x=0\end{cases}$$
This function is differentiable with
$$ f'(x)=\begin{cases}3x\cos x^{-3}+3\sin x^{-3}&x\ne 0\\0&x=0\end{cases}$$
Near $x=0$, $f'$ is dominated by the sine term, hence oscillates between positive and negative. Hence $c=0$ is critical for $f$ and is the limit of a sequence of critical points.
However, $f''(0)$ does not exist because the numerator of $\frac{f'(0+h)-f'(0)}{h}$ is $\approx \pm 3$ infinitely often for $h\approx 0$.
A: I assume that $f$ is $C^2$ MVT applied to $f'$ implies that there exists $c_n$ in the interval $[x_n,c]$ or $[c,x_n]$ such that $f'(c)-f(x_n)=f"(c_n)(x_n-c)$. This implies that $f"(c_n)=0$, since $c_n$ converges towards $c$, we deduce that $f"(c)=0$.
A: Let $g(t) = t\sin (1/t), t\ne 0,$ with $g(0) = 0.$ Then $g$ is continuous everywhere. Define
$$f(x) = \int_0^x g(t)\,dt, x\in \mathbb R.$$
By the FTC, $f'(x) = g(x)$ everywhere, hence $f'(1/n\pi) = 0, n = 1,2,\dots $ But for $x\ne 0,  (f'(x) - f'(0))/x = \sin (1/x),$ which has no limit as $x\to 0.$ Therefore $f''(0)$ does not exist.
