$f: \mathbb{R} \to \mathbb{R} $ by $f(x) =\frac 1{1+x^2}$ is uniformly continuous on $\mathbb{R} $ The definition of uniform continuity states that a function is uniformly continuous if, given any challenge $\epsilon > 0$ that there exists a response $\delta > 0$ for every value $x_1,x_2 \in R$ s.t. $|f(x_1)-f(x_2)| < \epsilon$ implies that $|x_1-x_2| < \delta$.
I found that for the function $f(x)$, in order to satisfy $|x1-x2| < \delta$, 
$$\left|\frac{1}{1+x_1^2}-\frac{1}{1+x_2^2}\right| < \epsilon$$
I simplified that to $\left|\frac{x_2^2-x_1^2}{(1+x_1^2)(1+x_2^2)}\right| < \epsilon$, but I wasn't sure how to simplify or modify that expression to get $\left|x_1-x_2\right|$ in terms of $x_1, x_2$, and $\epsilon$. Any help would be appreciated!
 A: This is one of the times where instead of proving the result directly,  it's often easier to prove that the function is differentiable with a bounded derivative.   That in turn will imply that the function is lipshitz continuous, which implies uniformly continuous.  Your derivative is $f'(x)=\frac {-2x} {(1+x^2)^2}$,  note $|f'(x)|=2|\frac x {(1+x^2)^2}|$,  and since $|x|\le |(1+x^2)^2|$,   we have $|f'(x)|\le 2$,  hence bounded derivative, and we're done
I highly recommend this technique because it's often easier than playing with epsilons and deltas :)
A: $\left|\dfrac{1}{1+x_1^2} - \dfrac{1}{1+x_2^2}\right| \leq  |x_1-x_2|\cdot \dfrac{|x_1|+|x_2|}{(1+x_1^2)(1+x_2^2)} \leq |x_1-x_2|\cdot \left(\dfrac{|x_1|}{1+x_1^2} + \dfrac{|x_2|}{1+x_2^2}\right) = |x_1-x_2|\cdot \left(\dfrac{1}{2} + \dfrac{1}{2}\right) = |x_1-x_2|$.
A: From where you got:
$$\left|\frac{x^2-y^2}{(1+x^2)(1+y^2)}\right|=|x-y|\,\left|\frac{x+y}{(1+x^2)(1+y^2)}\right|\le|x-y|$$
since
$$\frac{|x+y|}{(1+x^2)(1+y^2)}\le1\iff |x+y|\le1+x^2+y^2+x^2y^2\;\iff$$
$$\iff \pm(x+y)\le x^2y^2+x^2+y^2+1\iff x^2y^2+\left(x\pm\frac12\right)^2+\left(y\pm\frac12\right)^2+\frac12\ge0$$
