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.

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 $n=1,2,3$.