TITLE: Proof of uniform continuity
QUESTION [5 upvotes]: I seem to have hit a dead-end in the following proof.

Define $f:\mathbb{R}\to\mathbb{R}$ by:
$f(x)=\frac{1}{1+x^2}$
Show that $f$ is uniformly continuous.

My proof:

Let $x_{0}\in \mathbb{R}$.
Also let $\epsilon >0$
Choose $\delta = ?$
Then, for $x\in \mathbb{R}$, such that $|x-x_{0}|<\delta$, we have:
|$f(x)-f(x_{0})|=|\frac{1}{1+x^2}-\frac{1}{1+x_{0}^2}|=|\frac{x_{0}^2-x^2}{(1+x^2)(1+x_{0}^2)}|=\frac{|x_{0}-x||x_{0}+x|}{(1+x^2)(1+x_{0}^2)}\le \delta|x+x_{0}|$

The last line uses the fact that $x^2, x_{0}^2\ge 0$.  How can I finish the proof?

REPLY [3 votes]: Hint:
Prove this theorem and use it,
THEOREM
If $f$ continuous on $[a,\infty)$ and $\lim_{x\to\infty}f(x)$ exist, then $f$ uniformly continuous on $[a,\infty)$