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How to do a delta-epsilon proof by contradiction?

  1. Dec 11, 2011 #1
    1. The problem statement, all variables and given/known data

    Prove that lim x→1 of x2 does not equal 1+10-10. You could use a proof by contradiction.

    (It is question 2.b here)

    2. Relevant equations

    δ-ε proofs!

    3. The attempt at a solution

    Given ε > 0, there is some number δ > 0 such that if:

    |x - a | < δ
    |x - 0 | < δ
    |x| < δ

    Then:

    | f(x) - L | < ε
    | x2 - 1+10-10 | < ε


    ...and here's where I get stuck. Delta-epsilon proofs always seemed a bit circular to me, and what confuses me about proving "by contradiction" here is the fact that I should be able to choose some δ and the limit WOULD approach 1+10-10 :s... I'm a bit lost on where to go from here!
     
    Last edited: Dec 11, 2011
  2. jcsd
  3. Dec 11, 2011 #2

    SammyS

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    So the (sketch of a) proof would go something like this ...
    Assume that the limit, L = 1+10-10.

    Then pick ε small enough so that no matter how small δ is, it's impossible to find an x for which both of the following are true:
    0 < |x - 1| < δ​
    and
    | x2 - (1+10-10) | < ε​
    How small do you suppose ε must be?
     
  4. Dec 11, 2011 #3
    Is it a matter of picking something that is smaller than 10-10?
     
  5. Dec 11, 2011 #4

    SammyS

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    How about 1/2 of that?
     
  6. Dec 11, 2011 #5
    One half of that... so 0.5-10 (or 5-11)?
    Did you choose this for some specific reason or just arbitrarily?

    Another question -- If I'm choosing a specific value of ε here that isn't based on δ (so far as I can tell!), then do I still have to go through those initial steps of simplifying |f(x) - L | < ε to make the left-hand side look like the expression for delta (i.e., | x - 1| < δ )? Or am I missing the point of something here completely... :frown:

    Hmm... Because to show that something isn't true, I just need one example that works to disprove it. So I really can just choose a small ε (like, an actual number that isn't in terms of δ) and show that it doesn't work once and that's it?
     
  7. Dec 11, 2011 #6

    SammyS

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    That's the idea.
     
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