For $n \in N,$ let $$f_n(x)=\frac{1}{1+x^{2n}} ,\ x\in[-1,1] $$
(i)Find $f(x)=\lim_{n\to \infty}f_n(x)$ for $ x \in [-1,1]$
(ii)Prove that for each $0<r<1 \ f_n\rightarrow f$ uniformly on $[-r,r]$
This is my solution. is it okay?
(i) Consider two cases : (1) $x\in (-1,1) \ $, (2) $ x= \pm1 \ $ $$f(x)=\begin{cases}1\quad, x\in(-1,1)\\0\quad,x\pm 1 \end{cases}$$
(ii)For each $n \in N$, let $f_n(x)=\frac{1}{1+x^{2n}}$, and let $f(x) = 1$ .Then $$|f_n(x)-f(x)|= |\frac{1}{1+x^{2n}} -1| = |\frac{-x^{2n}}{1+x^{2n}}| \le r^{2n}$$
Thus $|f_n -f| \rightarrow 0$ since $\frac{-x^{2n}}{1+x^{2n}} \rightarrow 0$
Therefore $f_n\rightarrow f$ uniformly on $[-r,r]$
$a)$ Taking from your answer, $f(x) = \begin{cases} 1, x \in (-1,1) \\ \frac{1}{2}, x = \pm 1 \end{cases}$.
$b)$ $\displaystyle \text{sup}_{n \in \mathbb{N}}|f_n(x) - f(x)| = \text{sup}_{n \in \mathbb{N}} \left|\dfrac{1}{1+x^{2n}} - 1\right|= \dfrac{x^{2n}}{1+x^{2n}} \leq x^{2n} \to 0$ as $n \to \infty$, proving uniform convergence on $[-r,r]$