Uniform convergence of functions and intervals We define $f_n:\mathbb{R}\to\mathbb{R}$ by $f_n(x)=\dfrac{x}{1+nx^2}$ for each $n\ge 1$.
I compute that $f(x):= \displaystyle\lim_{n\to \infty}f_n(x) = 0$ for each $x\in\mathbb{R}$.
Now, I want to know in which intervals $I\subseteq \mathbb{R}$ the convergence is uniform.
Any hint? Thanks.
 A: If you use the technique, then the max over $x\in(-\infty,\infty)$ of the function is achieved at $x=\frac{1}{\sqrt{n}}$ and it equals $\frac{1}{2\sqrt{n}}$. So, we have
$$ \sup| f_n(x)-f(x) |= \sup \Big| \dfrac{x}{1+nx^2} \Big|= \frac{1}{2\sqrt{n}} < \epsilon.$$
This shows the sequence converges uniformly over $\mathbb{R}$.
Added: Here is the plot of the function for $n=1,2,3$.

A: The sequence converges uniformly on $I$ if for every $\epsilon >0, \exists N$ such that $\forall x \in I $ and $\forall n \geq N$, we have $|f_n(x) - f(x)| < \epsilon$
For the interval $|x| \leq 1$, we see that $x(1+nx) = x +nx^2 \leq 1 + nx^2$, so that $\frac{x}{1+nx^2} \leq \frac{x}{x(1+nx)} = \frac{1}{1+nx}$.  So, to make this smaller than $\epsilon$, choose $N_1 = (\frac{1}{\epsilon}-1)$.  
For the interval $|x| >1$, we have that $|\frac{x}{1+nx^2}| \leq |\frac{x}{nx^2}| =|\frac{1}{nx}| < \frac{1}{n}$.  So, choose $N_2 > \frac{1}{\epsilon}$.
Comparing $N_1, N_2$, we see that we can choose $N = N_2$.  Convergence is uniform on $\mathbb{R}$. 
