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Epsilon delta proof.

  1. Dec 3, 2011 #1
    1. The problem statement, all variables and given/known data

    if |x-3| < ε/7 and 0 < x ≤ 7 prove that |x^2 - 9| < ε

    2. Relevant equations



    3. The attempt at a solution

    So ths is what I did so far.

    |x+3|*|x-3| < ε (factored out the |x^2 - 9|)

    |x+3|*|x-3| < |x+3|* ε/7 < ε (used the fact that |x-3| < ε/7)

    |x+3|* ε/7 *7 < ε*7|x-3| < ε/7*7 (multiplied both sides of the inequality by 7)

    I suck at epsilon delta proofs and have no idea where to go from here.
     
  2. jcsd
  3. Dec 3, 2011 #2

    SammyS

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    It looks to me like the problem should read something like:
    if |x-3| < ε/7 and 0 < x-3 ≤ 1, prove that |x2 - 9| < ε​
    Otherwise you can only show that |x2 - 9| < (10/7) ε .

    More importantly, you are to prove that |x2 - 9| < ε. The way you are trying to go about this is to assume the fact which you are trying to prove. That is not a valid way to go about this.

    Added in Edit:

    I had a typo originally. See the correction in Red.
     
    Last edited: Dec 3, 2011
  4. Dec 3, 2011 #3
    How do i get started on this proof then/?
     
  5. Dec 3, 2011 #4

    SammyS

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    If x-3 ≤ 1, what does that say about x+3 ?
     
  6. Dec 3, 2011 #5
    x+3<7
     
  7. Dec 3, 2011 #6

    SammyS

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    Correct.

    I suppose I should have asked "What if -1 ≤ x-3 ≤ 1" or any suitable left hand value.

    So, if 5 ≤ x+3 ≤ 7, then you can definitely say that |x+3| ≤ 7 . Correct?
     
    Last edited: Dec 3, 2011
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