Show $1/(1+ x^2)$ is uniformly continuous on $\Bbb R$. 
Prove that the function $x \mapsto \dfrac 1{1+ x^2}$ is uniformly continuous on $\mathbb{R}$.

Attempt: By definition a function $f: E →\Bbb R$ is uniformly continuous iff for every $ε > 0$, there is a $δ > 0$ such that $|x-a| < δ$ and $x,a$ are elements of $E$ implies $|f(x) - f(a)| < ε.$
Then suppose $x, a$ are elements of $\Bbb R. $
Now
\begin{align}
|f(x) - f(a)| 
&= \left|\frac1{1 + x^2} - \frac1{1 + a^2}\right|
\\&= \left|  \frac{a^2 - x^2}{(1 + x^2)(1 + a^2)}\right|
\\&= |x - a| \frac{|x + a|}{(1 + x^2)(1 + a^2)}
\\&≤ |x - a| \frac{|x| + |a|}{(1 + x^2)(1 + a^2)}
\\&= |x - a| \left[\frac{|x|}{(1 + x^2)(1 + a^2)} + \frac{|a|}{(1 + x^2)(1 + a^2)}\right]
\end{align}
I don't know how to simplify more. Can someone please help me finish? Thank very much.

First:$f(x)=\frac{1}{1+x^2}\implies f'(x)=-\frac{2x}{(1+x^2)^2} $
Then observes that; For $|x|\le1$ 
$$\frac{|x|}{(1+x^2)^2} \le\frac{1}{(1+x^2)^2}\le  1$$
and for $|x|\ge1$ 
$$|x|\le x^2 \le (1+x^2)^2\implies \frac{|x|}{(1+x^2)^2} \le 1$$
Hence,
$$|f'(x)|=\frac{2|x|}{(1+x^2)^2} \le 2\implies |f(x)-f(y)|\le 2|x-y|$$

This shows that $f$ is Lipschitz which implies, that $f$ is uniformly continuous.

