Convergence of the sequence, $f_{n}(x)$. Correct me if I am missing something or show me the better way.

Let $0<a<b$ and consider the sequence of functions
  $$f_{n}(x)=\frac{1-(x/b)^{n}}{1+(a/x)^{n}}$$ for $n\in \mathbb{N}$.
i) First consider $x\in [a,b]$. Show that $f_{n}(x)$ does not
  converge uniformly on this set.
ii) Now reduce the domain into $x\in [a+\epsilon, b-\epsilon]$ where
  $\epsilon>0$, and it is assumed that $\epsilon<(b-a)/2$. Prove that
  $f_{n}(x)$ is uniformly convergent.

My answer to i) I have shown that $f_{n}(x)$ converges pointwise to 
$$
f(x)=\begin{cases}
\frac{1}{2} & \text{ if } a=x \\
1 & \text{ if } a<x<b \\ 
0 & \text{ if } x=b.
\end{cases}
$$
Since, for example, $\lim_{x\to a^{-}}f(x)=1\neq f(a)$, so $f$ is not continous on $[a,b]$. Thus $f$ does not converge uniformly on $[a,b]$. Is it enough to answer like that?
My answer to ii) I believe it says proving that
$$
\lim_{n\to\infty}\sup_{x\in [a+\epsilon, b-\epsilon]}\left | f(x)-f_{n}(x) \right |=0.
$$
First, we know that $1-\left ( \frac{x}{b} \right )^{n}\geq 1-\left ( \frac{b-\epsilon}{b} \right )^{n}$ and $1+\left ( \frac{a}{x} \right )^{n}\leq 1+\left ( \frac{a}{a+\epsilon} \right )^{n}$ for all $x\in [a+\epsilon, b-\epsilon]$. Since $[a+\epsilon, b-\epsilon]\subseteq (a,b)$, so
\begin{align*}\left | f(x)-f_{n}(x) \right |&\leq \left | 1-\frac{1-\left ( \frac{b-\epsilon}{b} \right )^{n}}{1+\left ( \frac{a}{a+\epsilon} \right )^{n}} \right |=\left | \frac{\left ( \frac{a}{a+\epsilon} \right )^{n}-\left ( \frac{b-\epsilon}{b} \right )^{n}}{1+\left ( \frac{a}{a+\epsilon} \right )^{n}} \right |\\
&\leq \left | \left ( \frac{a}{a+\epsilon} \right )^{n}-\left ( \frac{b-\epsilon}{b} \right )^{n} \right |\leq\left ( \frac{a}{a+\epsilon} \right )^{n}+\left ( \frac{b-\epsilon}{b} \right )^{n}.\end{align*}
Since $a+\epsilon>a>0$ and $0<b-\epsilon<b$, we finally get $\left | f(x)-f_{n}(x) \right |\to 0$ as $n\to\infty$. 
I do not know if I should use something like that for all $\epsilon>0$ there exist a $N\in\mathbb{N}$ such that $\sup_{x\in [a-\epsilon,b+\epsilon]}\left | f(x)-f_{n}(x) \right |<\epsilon$ for all $n\geq N$.
 A: Your answer to (i) is fine.
Your proof for (ii), in order to make use of the $\varepsilon$-property which in my case is a $\delta$-property as $\varepsilon$ already has been used, could be extended as follows:
Note that for $f_n$ to converge uniformly on $[a+\varepsilon, b-\varepsilon ]$ to $f$ you have to show, that for each $\delta > 0 $ there exists an $n_0 \in \mathbb{N}$ such that for every $n \ge n_0$ and every $x \in [a+\varepsilon, b-\varepsilon]$ it holds that $\lvert f(x) - f_n(x) \rvert \lt \delta$.
Proof:
Let $\delta > 0$. For every $\varepsilon > 0$ the sequences defined as
$$a_n := \left( \frac{a}{a + \varepsilon}\right)^n 
\quad \text{ and } \quad
b_n := \left( \frac{b - \varepsilon}{b}\right)^n $$
are null sequences, i.e. there exists $n_a \in \mathbb{N}$ such that for every $n \ge n_a$
$$ \lvert a_n \rvert < \frac{\delta}{2}.$$
Analogously there exists $n_b \in \mathbb{N}$ such that for every $n \ge n_b$
$$ \lvert b_n \rvert < \frac{\delta}{2}.$$
Now let $n_0 := \max(n_a,n_b)$. Performing the estimates in your proof for all $n \ge n_0$ and all $x \in [a+\varepsilon, b-\varepsilon ]$ the following holds
$$ \lvert f(x) - f_n(x) \rvert 
\leq 
\left( \frac{a}{a + \varepsilon}\right)^n  + 
\left( \frac{b - \varepsilon}{b}\right)^n
\lt 
 \frac{\delta}{2} +  \frac{\delta}{2} = \delta.
$$
