Prove that $\displaystyle f(x) = \frac {1}{x^2+1} $ is either uniform continuous or not.

Proof:

I claim that this function is uniform continuous, since its derivative is bounded, here is my proof.

Let $\displaystyle \epsilon > 0 $ be given, and pick $\displaystyle 0 < \delta < \epsilon $, then for $\displaystyle x,y \in \mathbb {R} , |x-y|< \delta $, we have $\displaystyle | f(x) - f(y) | = | \frac {1}{x^2+1} - \frac {1}{y^2+1} | = | \frac {y^2+1-x^2-1}{(x^2+1)(y^2+1)} | $ $\displaystyle \frac { | x^2 - y^2 | }{|(x^2+1)(y^2+1)|} = \frac {|x-y||x+y|}{|(x^2+1)(y^2+1)|} < \frac { \delta |x+y|}{|(x^2+1)(y^2+1)|} < \delta < \epsilon $

Is this right? Thanks.