Equicontinuity of $x^n$

I'm stuck in an analysis problem, it states:

Let $\lbrace f_n:[0,1)\rightarrow \mathbb{R}\rbrace_{n\in \mathbb{N}}$, with $f_n(x)=x^n$. Prove that:

$i)$ The familiy is pointwise equicontinuous (i.e. for every $x_0\in[0,1)$.

$ii)$ The familiy is NOT uniformly equicontinuous.

My attempt:

Por $i)$, we need to prove that for every $x_0\in[0,1)$, and $\forall \varepsilon>0 \quad \exists \delta=\delta(\varepsilon;x_0)>0$ such that if $|x-x_0|<\delta$, then $|f_n(x)-f_n(x_0)|<\varepsilon$, for all $n\in \mathbb{N}$

So working with the last inequality, and using the mean value theorem, it exists $c\in (0,1)$, such that $f_n(x)-f_n(x_0)=f_n^{'}(c)(x-x_0)$

$|f_n(x)-f_n(x_0)|=|f_n^{'}(c)(x-x_0)|=|nc^{n-1}||x-x_0|<n|c|^{n-1}\delta<n\delta<\varepsilon$

So, by this kind of estimation, $\delta=\varepsilon/n$. But this choice of $\delta$ clearly depends on $n$. And it shouldn't!

And for part 2 I don't have clue. Any help or advise will be appreciated.

For part ii): Suppose $\{f_n\}$ is uniformly equicontinuous on $[0,1).$ Let $\epsilon= (e^{-1/2} - e^{-1})/2.$ Then there exists $\delta > 0$ such that $x,y \in [0,1),$ $|y-x| < \delta,$ implies $|f_n(y)-f_n(x)| < \epsilon.$ Now for large $n,$ $1/2n< \delta.$ So for large $n,$

$$\tag 1 |f_n(1-1/n + 1/2n) -f_n(1-1/n)| < \epsilon.$$

But the left side of $(1)$ converges to $e^{-1/2} - e^{-1}.$ Thus for large $n,$ $(1)$ fails. We have a contradiction, proving $\{f_n\}$ is not uniformly equicontinuous on $[0,1).$

Let $\varepsilon>0$ be given, and let $K$ be a nonempty compact subset of $[0, 1)$. Set $s = \max K $. Either $s=0 $ or $s>0$. The former case is trivial ($K=\{0\}$), and so we assume there is a number $r>0$ such that $s=\frac{1}{1+r}$. By induction we obtain \begin{align} \displaystyle\frac{1}{(1+r)^n} \leq \frac{1}{1+nr} \end{align} for all positive integers $n$ (Bernoulli's Inequality). We may find a positive integer $N$ such that $\varepsilon N > \frac{1}{r}$ (Archimedean Principle). Thus \begin{align} \displaystyle \left| \,f_n (s) \right| < \varepsilon \, \text{ whenever } \, n\geq N \,. \end{align}

Since $K$ was given as an arbitrary compact subset of $[0,1)$, we have shown that the sequence of functions $\{\, f_n\}_{n=1}^\infty$ converges uniformly on every compact subset of $[0, 1)$. In other words, the sequence of functions $\{\,f_n\}_{n=1}^\infty$ is normal in $[0,1)$. I think knowing some Italians would be a good thing to have to your avail at this point.

If you really want to hold my feet to a fire, I can explain how to further show that the family is equicontinuous.