1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Uniform Continuity proof, does it look reasonable?

  1. Mar 3, 2010 #1
    1. The problem statement, all variables and given/known data
    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).

    2. Relevant equations



    3. 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
     
  2. jcsd
  3. Mar 3, 2010 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
  4. Mar 3, 2010 #3
    Cool, thanks for the input.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Uniform Continuity proof, does it look reasonable?
Loading...