Let $f(x)=\underset{n→∞}{\lim}\frac{x^{2n}-1}{x^{2n}+1}$, then

$f(x)=\left\{\begin{matrix}1&for\,|x|>1\\-1&for\,|x|<1\end{matrix}\right.$

For $|x|<1,\,x^{2n}→0$ as $n → ∞$ and for $|x|>1,\,\frac{1}{x^{2n}}→0$ as $n → ∞$

For |x| = 1, we have $x^2=1$

Thus $\frac{x^{2n}-1}{x^{2n}+1}=\left\{\begin{matrix}0&;&|x|=1\\\frac{1-x^{-2n}}{1+x^{-2n}}&;&|x|>1\\\frac{x^{2n}-1}{x^{2n}+1}&;&|x|<1\end{matrix}\right.$

Hence $f(x)=\left\{\begin{matrix}0&;&|x|=1\\1&;&|x|>1\\-1&;&|x|<1\end{matrix}\right.$