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Limit Proof

  1. Jun 29, 2010 #1
    The question is:

    Suppose f(x) <= g(x). Prove that lim [x->a] f(x) <= lim[x->a] g(x).

    I've been able to prove it by contradiction. I let lim [x->a] f(x) = L and lim[x->a] g(x) = M and I suppose that L > M. I then went on to choose epsilon = (L - M)/2 and a contradiction easily follows. But my problem is how do you arrive/choose this choice of epsilon? I just immediately thought of using this epsilon, but if I hadn't I don't know how/why I would have arrived at this epsilon. Can someone please help me out with this?

  2. jcsd
  3. Jun 29, 2010 #2
    You chose this [itex]\varepsilon[/itex] because it worked for the problem. The mental processes leading to this choice are probably extremely difficult. Experience certainly plays a big part.
  4. Jun 29, 2010 #3
    Not the answer I was hoping for. There is no way I could lets say draw a diagram and decide which epsilon would work and which one wouldn't?

    For example, when you prove that f(x) can't approach two different limits I drew a diagram:


    And I can "see" that epsilon = |L-M|/2 would make it impossible for both |f(x) - L| < epsilon and |f(x) - M| < epsilon

    So I was hoping I could do something similar to understand my choice of epsilon.

    Okay I'm going to try to explain this and I'd appreciate if you could tell me if this all makes sense.

    I have:


    Why is it that I can't choose epsilon = |l-m|? Is it because then the intervals |f(x) - L| < |l-m| and |g(x) - M| < |l-m| overlap and hence I can't really tell whether g(x) < or > f(x)? However, when I do choose |l-m|/2 then |f(x) - L| < |l-m|/2 and |g(x) - M| < |l-m|/2 don't overlap anymore and so I will be guaranteed that g(x) < f(x) and hence a contradiction will follow somehow...

    Do you see what I mean? I appreciate the help.
  5. Jun 29, 2010 #4
    I think the explanation/visualization with the overlapping intervals is appropriate.
  6. Jun 29, 2010 #5
    Thanks a lot for you help :)
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