
Uniform Continuity
Qn: Show that the function f(x) = 1/(1+x^2) is uniformly continuous on R.
I tried to show that the function satifies the Lipschitz condition
1/(1+x^2)  1/(1+y^2) = (x^2  y^2)/(1+x^2)(1+y^2) = (x+y)/(1+x^2)(1+y^2)xy  (*)
Then I observe that since (x1)^2 >= 0. Hence 1+x^2 >= 2x and 1+y^2 >= 2y.
Thus (*) <= (x+y)/4xyxy = 1/4 1/x + 1/yxy
However, I am unable to proceed after that. How should i do it? I have also tried to prove it by using the epsilondelta definition of uniform continuity but the argument is very similar to the one above.
Thanks.

Re: Uniform Continuity
You only have to show that $\displaystyle g(x,y):=\frac{x+y}{(1+x^2)(1+y^2)}$ is bounded independently of $\displaystyle x$ and $\displaystyle y$. Write $\displaystyle \leftg(x,y)\right =\left\frac{x}{(1+x^2)(1+y^2)}\right+\left\frac {y}{(1+x^2)(1+y^2)}\right\leq \frac{x}{1+x^2}+\frac{y}{1+y^2}$, and $\displaystyle x\mapsto \frac{x}{1+x^2}$ is bounded by $\displaystyle \frac 12$.