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Need help with an ε-δ limit proof

  1. Sep 29, 2011 #1
    1. The problem statement, all variables and given/known data

    Use the ε-δ definition of the limit to prove that

    lim as x -> 3 of (2x - 6)/(x-3) = 2

    2. Relevant equations

    3. The attempt at a solution

    I've started a preliminary analysis for the proof:
    For any ε>0, find δ=δ(ε) such that 0 < |x - 3| < δ implies that |(2x - 6)/(x-3) - 2|< ε.
    Simplifying:
    |(2x - 6)/(x-3) - 2| = |(2x - 6 - 2(x - 3))/(x-3)| = |(2x - 6 - 2x + 6)/(x - 3)| = 0

    Where do I go from here? Have I just proven that since the simplified form is 0, which is less than ε, and therefore proven the limit?
     
  2. jcsd
  3. Sep 29, 2011 #2

    SammyS

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    What do you get when you reduce (2x-6)/(x-3) to lowest terms?
     
  4. Sep 29, 2011 #3
    2. Does that mean I've proved the limit? No formal proof required?
     
  5. Sep 29, 2011 #4

    SammyS

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    No. Any δ > 0 will work, but you still need to show that for any x such that 0 < |x - 3| [STRIKE]< δ[/STRIKE] , |(2x - 6)/(x-3) - 2|< ε .

    I crossed out the < δ side, because as long as (x-3) ≠ 0, (2x - 6)/(x-3) - 2 = 0 which, of course is less than ε .
     
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