1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Advanced calc

  1. Dec 6, 2008 #1
    1. The problem statement, all variables and given/known data

    let f(x)= (x^2)/(1+x) for all x in [ifinity, 0) proof that f(x) is uniformly continuous. can anyone help me with this problem

    2. Relevant equations

    using the definition of a uniform continuous function

    3. The attempt at a solution

    i did long division to simplify the problem and got (x-1) - 1/(x+1)
  2. jcsd
  3. Dec 6, 2008 #2


    User Avatar
    Science Advisor

    What is the definition of "uniformly continuous" and how have you tried to use it?
  4. Dec 6, 2008 #3
    Find a delta that doesn't depend on any specific point in the domain of f.
  5. Dec 6, 2008 #4
    for u and v in D,

    |f(u)-f(v)|<epsilon if |u-v|<delta
  6. Dec 7, 2008 #5


    User Avatar
    Science Advisor

    Do you mean the domain to be [0, infinity) or (0, infinity)? Since "infinity" isn't a real number, "[infinity, 0)" doesn't make sense. I suspect it was supposed to be [0, infinity).

    "for u and v in D,

    |f(u)-f(v)|<epsilon if |u-v|<delta"

    is much too "shorthand" to be useful here. f is uniformly continuous in D if f(x) is defined for all x in d and given epsilon> 0, there exist delta> 0 such that if |u-v|< delta, then |f(u)- f(v)|< epsilon.

    The crucial point is, as JG89 said, the delta depends only on epsilon, not u or v.

    |f(u)- f(v)|= |u^2/(1+u)- v^2/(1+v)|= |(u^2(1+v)- v^2(1+u))/(1+ u+ v+ uv)|
    = |u^2- v^2+ uv(u- v))/(1+ u+ v+ uv)|= |(u+v- uv)/(1+u+v+uv)||u-v|.

    If you can find an upper bound for that first fraction, you are home free.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook