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.

According to the mean value theorem,
$$
\left|\frac1{1 + x^2} - \frac1{1 + a^2}\right| = f'(c)|x-a|
$$
where $f(x)=\dfrac 1 {1+x^2}$ and $c$ is somewhere between $x$ and $a$.  But $|f'(c)|\le\max |f'|$, the absolute maximum value of $|f'|$.  In order for this to make sense, you need to show that $|f'|$ does have an absolute maximum value.  But that is not hard.  So you have
$$
|f(x)-f(a)|\le M|x-a| \text{ for ALL values of $x$ and $a$},
$$
(where $M$ is the absolute maximum of $|f'|$).  So $f$ is Lipschitz-continuous and therefore uniformly continuous.
