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Limit Proof

  1. Jun 28, 2010 #1
    1. The problem statement, all variables and given/known data

    If f(x) <= g(x) then lim[x->a] f(x) <= lim[x->a] g(x) provided that both of these limits exist.


    2. The attempt at a solution

    I've been able to prove it by contradiction. So I assumed that l = lim[x->a] f(x) > lim[x->a] g(x) = m. Therefore, l - m > 0 and I could choose epsilon = (l - m)/2 and the contradiction follows. However, what I need someone to help me with is how can I "see" that (l - m)/2 will actually work (i.e. yield a contradiction)? I just immediately thought of using this choice of epsilon and a contradiction followed. But going back to the problem now, I have no clue why I choose this epsilon. Can someone be able to help me as to why this one actually works?

    Thanks
     
  2. jcsd
  3. Jun 28, 2010 #2

    Mark44

    Staff: Mentor

    (l - m)/2 is half the difference of l and m.
     
  4. Jun 29, 2010 #3
    LOL! Obviously...
     
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