Prove that the sequence $[f_n]_{n \in N}$, $f_n(x)=\frac{x^{2n}}{1+x^{2n}}$ converges uniformly on $x \in [1+\delta,\infty[$

Question

I've been asked to prove that the sequence $[f_n]_{n \in N}$ with $f_{n}(x)=\frac{x^{2n}}{1+x^{2n}}$ converges uniformly on $x \in [1+\delta,\infty[$ where $\delta > 0 $.

So far I've found that $f(x)=0$ for $x \in ]-1,1[$, $f(x)=\frac{1}{2}$ for $x=\pm 1$ and $f(x)=1$ for $x \notin [-1,1]$. I have calculated the distance between each of them and found it to be $\frac{1}{2}$. It was 0 for $x=\pm 1$. I am told, that I need to prove it using some sort of epsilon/delta proof, however I've been unsuccesful so far. Any help would be greatly appreciated.

Answer 1

Hint

Note that $\frac{x^{2n}}{1+x^{2n}} = 1 - \frac{1}{1+x^{2n}} \in 1-\left[0,\frac{1}{1+(1+δ)^{2n}}\right]$ for any $x \in [1+δ,\infty)$.

So it suffices to show that $\frac{1}{1+(1+δ)^{2n}} \to 0$ as $n \to \infty$.