Uniform Continuity of function $\frac{1}{1+x^2}$ How to show $f(x)$=$\frac{1}{1+x^2}$ is uniform continuous on $\Bbb R$. 
Although, of course for any interval $[a,b]$, this function is continuous and bounded, therefore also uniformly continuous. Following Continuous Extension Theorem  it is uniformly continuous on any $(a,b)$. Therefore proceeding this way, we can show it is uniformly continuous on $ \Bbb R$.  
I wish to prove the same analytically. I assumed there exists $x,u \in \Bbb R$, such that $ |x-u|< \delta$.  
Now,
$|f(x)-f(u)|$=$\frac {|x^2-u^2|}{|(1+x^2)(1+u^2)|}$ $\le$ $\frac{|x-u||x+u|}{x^2u^2}$    $\le$ $\delta$$\frac{|x+u|}{x^2u^2}$.
Here I stuck. I wish to find an $\epsilon$ so that the $|f(x)-f(y)|\lt \epsilon$, where $\delta$ depends only on $\epsilon$, not on $x$. But unable to do that. Tried to apply A.M-G.M inequality but could not find a fruitful result. What to do? 
 A: Note that the function is Lipschitz, as for any $x,y$ we have
$$
|f(x)-f(y)|\leq \sup_{x\in \mathbb{R}}|f'(x)||x-y|=\frac{3\sqrt{13}}{16}|x-y|
$$
Allowing you to choose 
$$
\delta=\frac{16\epsilon}{3\sqrt{13}}
$$
A: If $f'$ is bounded in $\mathbb{R}$, then $f$ is uniform continuous. Indeed you know that
$$
\left|f\left(x\right)-f\left(u\right)\right| \leq \underset{c \in \mathbb{R}}{\text{sup}} \ f'\left(c\right) \left|x-u\right|
$$
Hence the result. ( Lipschitz continuous imples uniformly continous )
A: You can try to prove the following general result which is interesting by itself.
A continuous real function defined on $\mathbb R$ which has finite limits on $\pm \infty$ is uniform continuous.
Then you just have to apply this result to your function $f$.
A: To complement the other answers; if you do wish to continue on your path:

$|f(x)-f(u)|$=$\color{blue}{\frac {|x^2-u^2|}{|(1+x^2)(1+u^2)|}}$ $\le$ $\frac{|x-u||x+u|}{x^2u^2}$    $\le$ $\delta$$\frac{|x+u|}{x^2u^2}$.

Picking up at the blue step:
$$\begin{align}
\frac {|x^2-u^2|}{|(1+x^2)(1+u^2)|} & \le |x-u|\frac{|x|+|u|}{(1+x^2)(1+u^2)} \\[6pt]
& \le|x-u|\left(\frac{|x|}{1+x^2}+\frac{|u|}{1+u^2}\right) \\[6pt]
& \le 2|x-u|
\end{align}$$
A: $f(x)=\frac{1}{1+x^2}\implies f'(x)=-\frac{2x}{(1+x^2)^2} $
For $|x|\le1$ we have 
$$\frac{|x|}{(1+x^2)^2} \le\frac{1}{(1+x^2)^2}\le  1$$
For $|x|\ge1$ we have 
$$|x|\le x^2 \le (1+x^2)^2\implies \frac{|x|}{(1+x^2)^2} \le 1$$
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 therefore, uniformly continuous.
