Pointwise convergence of $ f_n : \mathbb{R} \rightarrow \mathbb{R} $

Question

I'm studying Real Analysis and I'm trying to solve the following problem

Problem: Show that this sequence $f_n:\Bbb R\to\Bbb R$ given by $$ f_n = \frac{(x+1)^{2n+1}+(x-1)^{2n+1}}{(x+1)^{2n+1}-(x-1)^{2n+1}} $$ converges to a function $f$.

I need some hint to deal with that sequence. I've tried many kind of algebraic manipulation but none gave me a good expression to evaluated this limit when $n\mapsto \infty$ at different domains of $x$.

My last idea was wrrite $f_n$ as

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

But it simply turns into a mess at the moment to know which occours in different domains of $x$ because we have to deal with more than one inequality simultaniously.

Answer 1

hint

If $ x= 1 $ then $$f_n(1)=1\implies \lim_{n\to+\infty}f_n(1)=1$$

for $ x\ne 1 $,

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

$$x>1\implies 1+\frac{2}{x-1}>1$$ $$\lim_{n\to+\infty}f_n(x)=1$$

$$-1<1+\frac{2}{x-1}<1\iff$$ $$-1<\frac{1}{x-1}<0\iff$$ $$x-1<-1\iff x<0$$ $$\implies \lim_{n\to +\infty}f_n(x)=-1$$

$$-1=1+\frac{2}{x-1}\implies x=0$$ $$\implies \lim_{n\to+\infty}f_n(0)=0$$

finally

$$0<x<1\implies -1<x-1<0$$ $$\implies 1+\frac{2}{x-1}<-1$$ $$\lim_{n\to+\infty}f_n(x)=1$$

$$------------$$ The pointwise function limit $ f $ is defined by

$$f(x)=-1 \text{ if } x<0$$ $$f(0)=0$$ and $$f(x)=1 \text{ if } x>0$$