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Math Help - Limit Proof

  1. #1
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    Limit Proof

    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?

    Thanks
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  2. #2
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    It takes a bit of intuition to choose things like that. Sometimes, similar to what happened to you, you can just tell what epsilon you need. But sometimes it's not as obvious. For me, I'll sometime leave my choice of epsilon blank on my paper and continue working out the proof as if I hadn't until I get to the important usage of epsilon and see what I need there to complete the proof, and go back and fill it in. Also sometimes you might pick the wrong epsilon, but that's okay, you just go back and adjust it till it works out.
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  3. #3
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    Quote Originally Posted by MattMan View Post
    It takes a bit of intuition to choose things like that. Sometimes, similar to what happened to you, you can just tell what epsilon you need. But sometimes it's not as obvious. For me, I'll sometime leave my choice of epsilon blank on my paper and continue working out the proof as if I hadn't until I get to the important usage of epsilon and see what I need there to complete the proof, and go back and fill it in. Also sometimes you might pick the wrong epsilon, but that's okay, you just go back and adjust it till it works out.
    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 you can draw a diagram:

    L________M

    And "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:

    M__________L

    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.
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  4. #4
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    Drawing a mental diagram like that, I kind of consider it intuition from experience. What you did is perfectly acceptable for the problem. However that thought process will change as the problem changes. I think I misunderstood your first question, as in asking the best way to choose epsilon depending on the problem.
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  5. #5
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    Yes of course. Thanks a lot for your help!
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  6. #6
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    I would probably use a sequence \{x_n\} approaching a to get at this. Presumably it should be easier to prove that if

    \displaystyle f(x_n) \le g(x_n)

    then

    \displaystyle \lim_{n \to \infty} f(x_n) \le \lim_{n \to \infty} g(x_n).

    It can make your life easier proving things if you can figure out whether it's easier to work with sequences or the \epsilon - \delta definition.
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