[Math] Show that $f_n(x)=\frac{x^{2n}}{1+x^{2n}}$ is not uniformly convergent

Question

Show that $f_n(x)=\frac{x^{2n}}{1+x^{2n}}$ is not uniformly convergent. For $f_n$ to be uniform convergent, $\sup_{x\in\mathbb{R}}\{|f_n(x)-f(x)|\}\lt\epsilon$. I know that $f_n$ converges pointwise to $0$ if $ |x|\leq1$, pointwise to $1/2$ if $x=\pm 1$ and to $1$ otherwise. How can i formally prove, using above definition, that $f_n$ doest not converge uniformly? I know that the limit function is discontinuous and hence uniform convergence isn't possible but I would like a formal proof.

Thanks!

Answer 1

Let $x_n=1+\frac{1}{n}$. Then $x_n >1$, hence $|f_n(x_n)-f(x_n)|=f_n(x_n) \to \frac{e^2}{1+e^2}$. Now consider $\epsilon >0$ with $\epsilon < \frac{e^2}{1+e^2}$.