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Delta/Epsilon (Need opinions on my solution, I am new to these)

  1. Sep 30, 2011 #1
    1. The problem statement, all variables and given/known data

    Suppose that δ = min{1,e/7}. Show that
    if |x - 3| < δ, then |x^2 - 9| < e

    2. Relevant equations

    Don't know.


    3. The attempt at a solution

    |x - 3| |x + 3| < e

    Given : |x - 3| < δ => |x - 3| < e/7

    |x + 3| |x-3| < e/7
    |x + 3| 7|x - 3| < e

    Assume δ < 1, => |x - 3| < 1 => -4 < x < -2

    Sub (-4)

    | (-4) + 3| 7|x-3| < e
    |x-3| < e/7


    Is this circular? Please give some feedback!
    Thank you!
     
  2. jcsd
  3. Sep 30, 2011 #2

    SammyS

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    Have an ε (epsilon).
    It's not so much circular as it is helter-skelter.

    Start with 0 < |x-3| < δ , where δ = min(1, ε/7), for an arbitrary ε > 0 .

    Show that |x^2 - 9| < ε .

    Here's a start:
    You know that 0 < |x-3| < 1
    So you know that -1 < x-3 < 1
    From this you can eventually conclude that |x + 3| < 7
    You also know that |x-3| < ε/7.
    so |x-3||x + 3| < (ε/7) |x + 3| < ...

    Maybe I showed you too much ???
     
  4. Oct 1, 2011 #3
    I never liked the standard version of the proof. I prefer to think about it like this:

    say x = 3 + s. Were s is small. Then |3-x| = |s|. So we need to show:

    if |s| < δ then |(3+s)^2 - 9| < ε.

    |(3+s)^2 - 9| = |s2 + 6s| <= |s2| +6|s|

    Okay so s is small so s2 is going to be very small, much smaller then 6s in fact but to make this rigorous lets say |s| < 1.Then we have.
    |
    |(3+s)^2 - 9| = |s2 + 6s| <= |s2| +|6s| <= |s| + 6|s| =7|s|

    so we just need 7|s| < ε. So clearly just make |s| < ε/7. but we still need the condition |s|<1 so we say |s| < min(1,ε/7). but |3-x| = |s|. So let δ = min(1,ε/7). Which is the same answer you got.
     
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