Let $\{f_n\}$ be sequence of continously diff. func. converging uniformly to $f$ such that $f_n'(1)=0$ for all $n$. Is it necessary that $f'(1)=0$?

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 1

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 $f'(1)=1$.

Answer 2

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?