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