Prove that $f(x)=\frac{1}{x^2+1}$ is uniformly continuous in $\mathbb{R}$

Question

I'm trying to prove this from the definition, but i don't now how to proceed.

Answer 1

Hints :

1. $\lim_{x\to \pm \infty} f(x) = 0$. Hence, you can find an interval $[-M,M]$ so that it is smaller than $\epsilon/2$ outside that interval.
2. Any continuous function on $[-M,M]$ is uniformly continuous.

Answer 2

Your function is differentiable and has a bounded derivative in the whole of $\mathbb R$. You can use the mean value theorem to prove it is in fact Lipschitz.