Determining if function sequence is not uniformly convegent.

Question

We let $f_n(x)=\frac{1}{1+x^{2n}}$
I want to prove that the sequence is not uniformly convergent on $(-1,1)$, but is uniformly convergent on any compact set in $(-1,1)$
One possibility is to use a sequence $x_n$, converging to 1, and show that $|f_n(x_n)-f(x_n)|\geq\epsilon$
If i choose $x_n =(\frac{1}{2})^{1/n}$, will that work?

Answer 1

hint

If the convergence to $f:x\mapsto 1 $ at $ (-1,1) $ was uniform, we would have

$$\lim_{n\to+\infty}\lim_{x\to 1^-}f_n(x)=$$

$$\lim_{x\to 1^-}\lim_{n\to +\infty}f_n(x)$$

Other approach. $ f_n $ is an even function, we study it at $ [0,1)$.

$$g_n(x)=|f_n(x)-1|$$ $$=\frac{x^{2n}}{1+x^{2n}}$$

$$g'_n(x)(1+x^{2n})^2=2nx^{2n-1}$$

$ g_n $ is strictly increasing at $ [0,1) $ and $$\sup_{0\le x<1}g_n(x)=\lim_{1^-}g_n(x)=\frac 12$$ the convergence is not uniform at $ (-1,1)$

but at $ [0,A] $, $$\sup_{[0,A]}g_n=g_n(A)\to 0$$

which proves the uniform convergence at $ [-A,A] $.