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 $[0,1]$.
 A: Every continuous function on $[0,1]$ is a limit of a (bounded) sequence of polynomials (Weierstrass' theorem), so this set is not closed for any $A>0$. However, for $A= 0$ it is closed being a singleton.
Of course, you may pick your favourite example of a sequence of differentiable functions converging uniformly to a non-differentiable one and renormalise it suitably.
A: 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 $x=0$
A: Take $A=2.$ Then the functions
$$f_n(x) = (1/n +(x-1/2)^2)^{1/2}$$
are in $B(2)$ and converge uniformly to $|x-1/2|$ on $[0,1].$ Since $|x-1/2|\not \in C^1,$ $B(2)$ is not closed.
A: $B$ is not closed!
Let
$$f_{n} = \left\{\begin{matrix}
A(x-\frac{1}{2})^{{1+\frac{1}{n}}} &  x \in [\frac{1}{2},1]\\  
- A\left |x-\frac{1}{2}\right |^{{1+\frac{1}{n}}}   &     x \in [0,\frac{1}{2}]
\end{matrix}\right.$$ 
then $\left \| f_{n} \right \|_{\infty} \leqslant A$ and $f_n$ is differentiable  at $x=\frac{1}{2}$ for all $n\in N$. Therefore $f_{n} \in B$ for all $n \in N$ .  It is easy to show that
$$f_{n} \rightarrow f= \left\{\begin{matrix}
A(x-\frac{1}{2})&  x \in [\frac{1}{2},1]\\  
- A\left |x-\frac{1}{2}\right |&     x \in [0,\frac{1}{2}]
\end{matrix}\right.$$ 
but $f \in C[0,1] \setminus C^{1}[0,1]$, so $f \notin B.$  
