Any help on this question would be appreciated as I'm stuck. I would love to think some more but I'm not getting anywhere no matter how much I read about pointwise/continuous convergence and I would really like to know how this is done.. Thanks
$f_n(x) = \frac{x^n}{1 + x^n}$ for $x\ge0$
Show that${f_n}$ converges uniformly on every interval of the form [0,a] with a<1, and of the form $[b,\infty)$ with b>1
Since $x^n\ge 0$, we have $|f_n(x)|\le x^n\le a^n$ for $x\in[0,a]$. Since $a<1$, the result folows, i.e. for $n$ big enough we have $a^n<\epsilon$ no matter how small $\epsilon>0$ we are given.
And we have $|1-f_n(x)|=|\frac1{1+x^n}|\le \frac 1{1+b^n}$ for $x\ge b$. Again, this is $<\epsilon$ for $n$large enough. More specifically, it suffices to have $b^n>\frac1\epsilon$.
Hints:
i) For $\,x\in [0,a]\;,\;a<1\,$ , we have that
$$\frac{x^n}{1+x^n}=1-\frac{1}{1+x^n}\xrightarrow[n\to\infty]{}1-\frac1{1-0}=0$$
ii) For $\,x\in[b,\infty)\;,\;b>1\,$ , we get
$$1-\frac{1}{1+x^n}\xrightarrow[n\to\infty]{}1-0=1$$
In both cases we have (assume $\,n\ge m\,$ for simplicity)
$$\left|\,\frac{x^n}{1+x^n}-\frac{x^m}{1+x^m}\,\right|=\left|\,\frac1{1+x^m}-\frac1{1+x^n}\,\right|=$$
$$\frac{|x^n-x^m|}{(1+x^n)(1+x^m)}\begin{cases}\le |x^n-x^m|=x^m(1-x^{n-m})\xrightarrow[n,m\to\infty]{}0\cdot 1=0\;\; (\;\text{case (i)}\;)\\{}\\\frac{x^n-x^m}{x^{n+m}}\xrightarrow[n,m\to\infty]{}0\;(\;\text{case (ii)}\;)\end{cases}$$
