Uniform Continuity proof, does it look reasonable?

[spenghali]

Homework Statement

Note: I will use 'e' to denote epsilon and 'd' to denote delta.

Using only the e-d definition of continuity, prove that the function f(x) = x/(x+1) is uniformly continuous on [0, infinity).

Homework Equations

The Attempt at a Solution

Proof:

Must show that for each e>0 there is d>0 s.t.

|x/(x+1) - a/(a+1)| < e whenever x,a are elements of [0, infinity) |x-a| < d.

|x/(x+1) - a/(a+1)| = |(-x+a)/[(x+1)(a+1)]| [tex]\leq[/tex] |-x+a| = |x-a|.

Thus, given e>0, if we choose d=e then,

|x/(x+1) - a/(a+1)| < e whenever |x-a| < d.

This implies that f(x) = x/(x+1) is uniformly continuous on [0,infinity). QED

 

[Dick]

Sure. That works. You could clean up few details, like x/(x+1) - a/(a+1)=(x-a)/((x+1)(a+1)), not (-a+x)/((x+1)(a+1)) and you could also explicitly justify why |(x-a)/((x+1)(a+1))|<=|x-a| but the proof works fine.

 

[spenghali]

Cool, thanks for the input.