Closed subset of $C([0,1],\mathbb R$) 
Let $B=\{f\in C^1[0,1]:\Vert f\Vert _\infty \le A\}$. Is $B$ a closed subset of $C([0,1],\mathbb R)$?

Here's what I tried to do:
Let $\{f_n\}$ be a sequences of functions of $B$ such that $f_n \to f$, with $f\in C([0,1],\mathbb R)$.
If I prove that  $f\in B$, then I finish the proof, i.e. $B$ will be closed in  $C([0,1],\mathbb R)$.
But I don't know how could I do that.
Note:  $C([0,1],\mathbb R)$ is the space of continuous functions with domain $[0,1]$.

Ahh, another simple example of sequence of functions is $$ f_{n} = \sqrt{x + \frac{1}{n}} $$
which uniformly convergences to $f(x)= \sqrt x $ which is not differentiable at $x=0$
