Let $f_{n}(x)=\frac{x^{n}}{n}$. Show that $f_{n}$ converges uniformly to a differentiable function on $[0,1]$.
I am confident this function converges pointwise to $f(x)=0$, which is differentiable. It is easy to see that by the squeeze theorem. For that, since $0\leq x\leq 1$ we have
$$ 0\leq \frac{x^{n}}{n}\leq \frac{1}{n}\Longrightarrow \lim_{n\to\infty} 0 \leq \lim_{n\to\infty}\frac{x^{n}}{n}\leq \lim_{n\to\infty}\frac{1}{n}\Longrightarrow 0\leq \lim_{n\to\infty}\frac{x^{n}}{n}\leq 0 $$
So, $\lim_{n\to\infty}x^{n}/n=0$. So I am attempting to show that $f_{n}$ converges uniformly to $f(x)=0$ as well.
My attempt at the proof:
Proof. Let $\epsilon >0$ be given and let $x\in[0,1]$. Then
$$ |f_n(x)-f(x)|=\Big|\frac{x^n}{n}-0\Big|=\Big|\frac{x^n}{n}\Big|\leq \frac{1}{n}$$ Since $|f_n(x)-f(x)|<\epsilon$ is equivalent to $1/n<\epsilon$, we choose $N=\frac{1}{\epsilon}$. Then
$$\forall n>N \implies |f_n(x)-f(x)|<\epsilon \text{ for all } x\in[0,1]$$
Therefore, $f_n$ converges uniformly to $f$.
I believe this is correct, should I make changes?
The second problem I have is that I am unsure how I can show that $f'(1)\ne \lim_{n\to\infty}f_n'(1)$. Is that as simple as saying that $f'(1)=0$ since $f(x)=0$ for all $x\in[0,1]$, but $f_n'(x)=x^{n-1}$?