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How is the epsilon delta proof an actual proof?

  1. Sep 16, 2010 #1
    I am a first year freshman at UC Berkeley, in Math 1A. We learned the delta-epsilon proof for proving the limit of functions. I won't go through a whole proof or anything, but the general idea is you have a delta that is less than |x-a| (and greater than zero) and an epsilon less than |f(x)-L|. The way we have learned to prove this is to find a delta in terms of epsilon, and then solve for epsilon.

    Here is my problem.
    This proof really just does simple algebra in one direction (to find delta in terms of epsilon), and then the reverse of that algebra to isolate epsilon. So, what goes wrong if you prove a limit that is not true?

    My graduate student instructor said you wouldn't know unless you plugged in epsilon values. My professor said it was a good question, and it would be hard to prove that you were doing an incorrect proof. The way he said to prove that that limit is incorrect would be to do the proof for the correct limit. A function can have only one limit, and so the wrong one would be supplanted.

    But I still think that there is a real problem with a proof that proves things that are incorrect. How is that a proof at all?
     
  2. jcsd
  3. Sep 17, 2010 #2

    jgens

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    You can't prove a limit that isn't true. Mathematics wouldn't be much use if we could prove things that were false. If you mean, what happens if you try to prove a limit that is not true, then that's a different question.

    This works in some cases, but it fails if a function fails to approach a limit at all. Another (more general) way to prove that a function doesn't approach a specific limit l, is to show that there is some ε > 0 such that for every δ > 0 there is some x which satisfies 0 < |x-a| < δ but not |f(x)-l| < ε.

    Why is that a problem? If I prove that the square of every real number is non-negative, that means I prove that it's impossible for the square of a real number to be less than zero. If you don't object to this last example (I hope you don't), then why would you object to the example with limits?
     
  4. Sep 17, 2010 #3
    Right, so if you try and prove a limit that is not true, where exactly in the proof is there a contradiction? There isn't. There should be. That is my point.

    The square root example I'll try and make analogous to this problem, but I don't think it really works. You prove that you square any real number and therefore you conclude that the square of any number must be greater than zero. However, say for kicks you set out to prove that you can find a negative number by squaring a real number...and nothing goes wrong with the proof.

    I'll write out a proof.
    Prove lim x->4 of 4x is 20.

    this is the scratchwork to find delta (E= epsilon, d=delta)

    -E<|f(x)-L|<E
    -E<|4x-20|<E
    -E<|4(x-5)|<E
    -E/4<|x-5|<E/4
    -E/4+1<|x-4|<E/4+1

    0<|x-a|<d
    0<|x-4|<d
    |x-4|<E/4+1
    d=E/4+1

    okay, now the proof.
    d=E/4+1
    -E/4+1<|x-4|<E/4+1
    (skipping a few steps...it's just the same thing backwards)
    -E<|4x-20|<E

    where's the problem?
     
  5. Sep 17, 2010 #4

    jgens

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    There's a contradiction in that it can't be done. What you'll eventually end up showing is that there exists an [itex]\varepsilon > 0[/itex] such that for all [itex]\delta > 0[/itex] there is an [itex]x[/itex] which satisfies [itex]0 < |x-a| < \delta[/itex] but not [itex]|f(x)-l| < \varepsilon[/itex].

    That wouldn't be possible. What good is math if you can prove things that aren't true?


    This step isn't justified.

    No. Be careful with your signs when you find [itex]-\delta[/itex].
     
  6. Sep 17, 2010 #5

    jgens

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    Sorry! In my last reply, I said that one of your steps wasn't quite valid when it more or less was (you should omit absolute value signs when you don't need them; granted, that doesn't pardon my mistake). The problem with your 'proof' is that you should have

    [tex]0 < |x-4| < \frac{\varepsilon}{4}+1[/tex]

    [tex]-\frac{\varepsilon}{4}-1<x-4<\frac{\varepsilon}{4}+1[/tex]

    [tex]-\frac{\varepsilon}{4}-2<x-5<\frac{\varepsilon}{4}[/tex]

    [tex]|x-5| < \frac{\varepsilon}{4}+2[/tex]

    [tex]|4x-20| < \varepsilon+8[/tex]

    So you should see that your choice of [itex]\delta[/itex] doesn't allow you to make [itex]|4x-20|[/itex] small enough.
     
    Last edited: Sep 17, 2010
  7. Sep 17, 2010 #6
    I don't understand why your instructors would explain it like that. You can't find a delta for every epsilon if the limit is incorrect or DNE.
     
  8. Sep 19, 2010 #7
    I see, that's interesting! I've never seen it solved like this, on both sides of the expression...

    0<|x-4|<E/4+1
    -E/4-1<|x-4|<E/4+1

    But that does lead to a contradiction in this problem, and works if you plug in the limit as 16.
    Well that is wonderful...but it is still strange that we don't actually do it this way in class. We only solve from this step

    0<|x-4|<E/4+1

    Without plugging in -E on the left of the inequality.

    Thank you for your help...I'm going to discuss this with my GSI.

    and deluks-

    Given a function f we we say that
    lim f(x) = L
    x->a

    if given any  E> 0 we can fi nd some delta such that for every x satisfying 0 < |x-a|< delta,
    |f(x)-L|< E

    That was a direct quote from a handout in discussion...albeit the symbols didn't carry over from copy and paste, so I typed out delta where there was a delta.
     
  9. Sep 19, 2010 #8

    HallsofIvy

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    Well, there doesn't have to be a -E on the left because any negative number is "< 0".
    You may be confusing 0< |x- 4|< E/4+ 1 with (-E/4+ 1)< x- 4< E/4+ 1 where I have removed the absolute value signs.

    And that is exactly what everyone else has been saying! that is precisely the usual definition of "limit". I suspect you are "actually doing it that way" in class and do not realize it.
     
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