Describe the intervals where $f_n(x)$ converges uniformly 
Problem: $f_n(x) = \frac{6}{(1+x^{2n})}$  and $x\in\mathbb{R}$.
Find all real numbers $x$ where $f_n(x)$ converges and describe the limit function. 

I found the limit function to be
$$f(x)=\begin{cases}0&\text{ if }|x|>1,\\
6&\text{ if }0 \leq |x| < 1,\\
3&\text{ if }|x| = 1.\end{cases}$$
So $f_n(x)$ converges for all $x\in\mathbb{R}$.
Now I have to find the intervals for which $f_n$ converges uniformly to $f$. I am considering the intervals $[0,1)$ and $(1,\infty)$. But  I do not know where to go from here. I am thinking the M-test? But I am not sure how to apply it. Any suggestions or hints would be helpful.
 A: Enough to consider $x\ge0$. 
Next, instead of comparing with limit function, we look at increments: 
$$|f_n (x)- f_m(x)|  = \left|\frac{6}{1+x^{2n} } - \frac{6}{1+x^{2m}}\right| =6  \frac{|x^{2m} - x^{2n}|}{(1+x^{2m})(1+x^{2n})}. $$
Recall that the convergence is uniform on a set $A$ if and only if $$\lim_{m\to\infty} \sup_{n \ge m} \sup_{x\in A} |f_n (x)-f_m(x)|=0.$$
Without loss of generality, assume $n>m$. 
1a.  Let $a>1$. On $[a,\infty)$ we have uniform convergence because RHS is bounded above by $6\frac{ x^{2n}}{1+x^{2n}} \frac{6}{a^{2m}}\le 6a^{-2m}$. 
1b. On $(1,\infty)$ we don't have uniform convergence because for example, looking at  $x=2^{1/m^2}$, and $n=m^2$, the righthand side becomes 
$$ 6 \frac{|2^{2/m} - 2^{1}|}{(1+2^{2/m})(1+2^{1})}\underset{m\to\infty} {\to} 6 \frac{ 2-1}{2(1+2)}=1.$$
2a. Let $a<1$. On the interval $[0,a]$ we again have uniform convergence. This is easy to see because righthand side is bounded above by $6a^{2m}$. 
2b. On the interval $[0,1)$ there is no uniform convergence: similarly to 1b,  take  $x=2^{-1/m^2}$, $n=m^2$ to obtain 
$$6 \frac{|2^{-2/m}-2^{-1}|}{(1+2^{-2/m})(1+2^{-1})}\underset{m\to\infty} {\to}6\frac{\frac12}{2 \times \frac 32}=1.$$
Also the sequence trivially converges uniformly on the degenerate interval $[1,1]$. 
