1. ## uniformly continuous

Hello, I need to show that h(x) = 1/(x^2 +1) is uniformly continuous on R

This is what I have so far:
Let x,y be in R
I know |x-y|< delta
I want to show |f(x)-f(y)|< epsilon taking some delta
so, |1/(x^2 +1) - 1/(y^2 +1)|
= |(y^2-x^2)/(x^2 +1)(y^2 +1)|
= |(x-y)(x+y)/(x^2 +1)(y^2 +1)|
< |delta(x+y)/(x^2 +1)(y^2 +1)|
so i could chose an epsilon = (x^2 +1)(y^2 +1)/(x+y)
so that = epsilon.

However, to be uniformly continuous, my choice of delta shouldn't depend on x to be uniformly continuous.
please provide me with some feedback

thanks

2. There's a theorem about differentiable functions with bounded derivative. Have you covered it yet?

3. No, we haven't covered that

4. The proof is pretty straightforward, and is probably the best way to tackle this. Intuitively, if the graph can only get *this* steep and no steeper, for every x in the reals, can you see why that lets you choose an epsilon that depends only on delta and not on x?