Uniform convergence of $\frac{x}{1+n^2x^2}$ I am interested in the pointwise limit of $f_n = \frac{x}{1+n^2x^2}$ and whether the sequence of functions converges uniformly on the interval $A=[1,\infty)$, $B=[0,1]$ and $C=[-1,1]$ respectively.
I want to use the fact that $(f_n)$ converges uniformly on a interval $I$ to a limit function $F$
\begin{align}
&\iff\\
\lim_{n\to\infty}\sup_{x \in I} |F(&x)- f_n(x)|=0.\qquad (*)
\end{align} 
In $A,B,C$, we presume of course that $F(x)=0$. In order to check the criterium in $(*)$, we calculate 
\begin{align}
f'_n(x)=\frac{1-n^2x^2}{(1+n^2x^2)^2}=0,
\end{align}
which gives $x=\frac{-1}{n}$ and $x=\frac{1}{n}$. Therefore, $f'_n(x)> 0$ if $\frac{-1}{n}<x<\frac{1}{n}$ and $f'_n(x)< 0$ if $x<\frac{-1}{n},x>\frac{1}{n}$.
Since $\frac{1}{n}\leq 1$, we know that $f_n$ decreases on $A$. So for $A$,
\begin{align}
\lim_{n\to\infty}\sup_{x \in I} |F(&x)- f_n(x)| \leq \lim_{n\to\infty}f_n(1)=\lim_{n\to\infty}\frac{1}{1+n^2}=0.
\end{align}
Hence, $(f_n)$ converges uniformly on $A$. How to deal with the maximum $x=\frac{1}{n}$ for interval $B$ and $C$? 
 A: Why complicate things?! Given $\epsilon >0$ you want to show that there exists an $N$, such that if $n>N$,
$$|f_n(x)-0|=\frac{|x|}{1+n^2x^2}<\epsilon, \forall x.$$
Now, a couple observations:
1) since the function is odd, let's focus on positive $x$.
1) for $x \geq \delta >0$,
$$\frac{x}{1+n^2x^2}<\frac{x}{n^2x^2}=\frac{1}{n^2x}<\frac{1}{n^2\delta}.$$
2) for $0\leq x < \delta$,
$$\frac{x}{1+n^2x^2} \leq \frac{x}{1}< \delta.$$
Now, starting with a given $\epsilon$, first pick the right $\delta$, which here can just be the epsilon, then pick $N$ large enough so that $\frac{1}{N^2\delta}$  is also less that $\epsilon$. This satisfies the conditions of uniform close-ness to the limit function by $\epsilon.$
A: Since $f_n'(x)=\frac{1-n^2x^2}{(1+n^2x^2)^2}$, then $f_n'(x)=0$ at $x=\pm 1/n$.  
At the local extrema, $f_n(1/n)=\frac{1}{2n}$ and $f_n(-1/n)=-\frac{1}{2n}$.  Inasmuch as $f_n'(x)<0$ for $|x|>1/n$, $|f_n(x)|\le \frac1{2n}$ for all $x\in \mathbb{R}$.
Hence, given $\epsilon>0$
$$\left|\frac{x}{1+n^2x^2}-0\right|\le \frac{1}{2n}<\epsilon$$
whenever $n>N(\epsilon)=1+\lfloor \frac{1}{2\epsilon}\rfloor$.  And we are done!
