Uniform convergence of $\frac{1+x^{n+1}}{1+x^n}$ on $[0,\infty)$ - better (easier) solution?

Question

I needed to prove uniform convergence of the sequence $f_n(x)=\frac{1+x^{n+1}}{1+x^n}$ on $[0,\infty)$. Clearly, $$f_n\rightarrow\begin{cases}1,& x\in[0,1],\\x, & x\in[1,\infty)\end{cases}$$

I proved it by considering three cases.

1. For $x\in[0,1/2]$, it holds that $\left|\frac{1+x^{n+1}}{1+x^n}-1\right|=x^n\frac{1-x}{1+x^n}\leq \frac{x^n}{1+x^n}$.
2. For $x\in[1/2,1]$, it holds that $x^n\frac{1-x}{1+x^n}\leq x^n (1-x)$.
3. For $x\geq1$, it holds that $\left|\frac{1+x^{n+1}}{1+x^n}-x\right|=\frac{x-1}{1+x^n}\leq\frac{x-1}{x^n}$.

By analysing the functions on the RHSs on the corresponding intervals, we obtain that $$\|f_n-f\|_\infty\rightarrow0$$

If you try analysing the function $f_n-f$ itself, things become nasty. I was wondering if somebody sees a better (easier) proof of the uniform convergence than mine.

Thank you!

Answer 1

Not sure if simpler, but you may use that for $x\neq0$ $$f_n(x)=xf_n\left(\frac1x\right)$$ So, if you show that $f_n(x)$ uniformly to $x$ for $x\ge 1$ as you have it, then for $0<x<1$, the result follows immediately from the above.