Show that the function f(x) = 1/(1 + x^2) for all x in the Reals is uniformly continuous on the Reals.
I'm trying to prove this using the definition of uniform continuity, but I'm having some trouble.
In that case, you'll have to use algebra (if that's allowed ):
$\displaystyle \left|\frac1{1+x^2} - \frac1{1+y^2}\right| =\frac{|y^2-x^2|}{(1+x^2)(1+y^2)} \leqslant \frac{|x|+|y|}{(1+x^2)(1+y^2)}|y-x|\leqslant |y-x|$, because $\displaystyle \frac{|x|}{1+x^2}\leqslant\tfrac12$.