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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

    HallsofIvy

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    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

    HallsofIvy

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    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.
     
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