Question: Let $\{f\_n\}$ be sequence of continously differentiable functions converging uniformly to $f$ such that $f'\_n(1)=0$ for all $n$. Is it necessary that $f'(1)=0$?

I am trying to prove that this is not true as the linear functional from $C[0,1]$ to the set of complex numbers as f goes to $f'(1)$ is not continuous so Kernel of this linear functional should not be closed

Answer: 
If we let
$$g\_n(x)=\frac1n\sin nx$$
then $g\_n'(0)=1$ and $g\_n\to0$ uniformly.

Set $f\_n(x)=g\_n(1-x)+x$. Then $f\_n(x)\to f(x)=x$ uniformly, and
$f\_n'(1)=0$. But $f'(1)=1$.
Try 
$$
f\_n(x)=\frac{1}{\sqrt{n}}\sin\bigg(\frac{\pi (2n^2+1)}{2}x\bigg)
$$
Then we have 
$$
\sup\_{x\in \mathbb{R}}|f\_n(x)|<\frac{1}{\sqrt{n}}\to0
$$
and 
$$
f\_n'(1)=\pi n\cos\bigg(\frac{\pi }{2}(2n^2+1)\bigg)=0
$$
What about $f\_n'(x)$ in general? Does it converge to something nice?