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Delta-epsilon method

  1. Sep 27, 2010 #1

    haushofer

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    Hi,

    I had the following question of a student this day about proving the following limit:

    [tex]
    \lim_{x \rightarrow 3} x^2 = 9
    [/tex]

    So this means that I should prove that

    [tex]
    |x-3| < \delta \ \rightarrow \ \ |x^2 - 9| < \epsilon(\delta)
    [/tex]

    So I had the following idea:

    [tex]
    |x^2 - 9| = |x-3||x+3|
    [/tex]

    The first term on the RHS is smaller than delta. For the second term I write

    [tex]
    |x+3| = |x-3+6| < |x-3| + 6
    [/tex]

    So I get in total

    [tex]
    |x-3| < \delta \ \rightarrow \ |x^2 - 9| = |x-3||x+3| < |x-3|(|x-3| + 6 ) < \delta(\delta + 6)
    [/tex]

    So choosing

    [tex]
    \epsilon = \delta(\delta + 6)
    [/tex]

    should prove the statement, right? But if I see the book they're using (Apostol) I see that they "choose" |x-3|<C, try to make an inequality then for |x+3| etc. But for me that seems making things more difficult then they are. So my question is: is my answer described above right?
     
  2. jcsd
  3. Sep 27, 2010 #2
    Hi,

    I'm new to this forum so please excuse me if I mess up the latex. For the most part your method is sound. However, delta should be a function of epsilon since delta must exist for all epsilon. With the way you've done things getting delta as a function of epsilon is a bit messy. I would suggest limiting yourself to an interval around 3. Doing this with the interval (0,6) allows you to simply say |x+3| < 9 and you get [tex]|x^2 - 9| = |x-3||x+3|<9|x-3|[/tex]. This way you can choose [tex]\delta = \epsilon/9[/tex].
     
  4. Sep 28, 2010 #3

    haushofer

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    I would get

    [tex]
    \delta = \frac{-6 \pm \sqrt{36 + 4 \epsilon}}{2}
    [/tex]

    If I take the positive sign I get

    [tex]
    \delta = \frac{-6 + \sqrt{36 + 4 \epsilon}}{2}
    [/tex]

    For this particular delta it's true that it exists for all epsilon>0, right?

    However, aren't you getting then an extra condition on |x-3|? So |x-3| is not only smaller than delta anymore in your proof; you also get |x-3|<3, right? So you get

    [tex]
    |x-3| < \delta \ \ \ \ AND \ \ \ \ |x-3|<3
    [/tex]
     
    Last edited: Sep 28, 2010
  5. Sep 28, 2010 #4

    haushofer

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    Maybe it's a matter of taste, but I find these extra conditions (limiting first to a certain interval) more messy than using the Cauchy-Schwarz inequality.
     
  6. Sep 28, 2010 #5

    HallsofIvy

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    You are NOT allowed to choose [itex]\epsilon[/itex]. You are given [itex]\epsilon[/itex] and want to choose [itex]\delta[/itex] to give that [itex]\epsilon[/itex].

    No, it does not "prove the statement". You are going the wrong way.
     
  7. Sep 28, 2010 #6

    haushofer

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    Why I'm going the wrong way? I've now found that
    [tex]
    |x-3| < \delta \ \rightarrow \ \ |x^2 - 9| < \epsilon(\delta)
    [/tex]
    where the arrow is read as "implies", for the delta
    [tex]
    \delta = \frac{-6 + \sqrt{36 + 4 \epsilon}}{2}
    [/tex]
     
    Last edited: Sep 28, 2010
  8. Sep 28, 2010 #7
    You have written ε and a function of δ. You have it reversed. δ must be written as a function of ε.

    "For every ε, there exists a δ such that...."

    Because ε is in the outermost scope, δ may be written in terms of ε. However, since ε is a constant with respect to the rest of the expression, what you have above is incorrect.

    But you're pretty close. In particular, if you haven't made a mistake in your algebra (I haven't checked), then you can easily turn:

    [tex]

    \delta = \frac{-6 + \sqrt{36 + 4 \epsilon}}{2}

    [/tex]

    into an equation of the form "ε = ..."
     
  9. Sep 28, 2010 #8
    Whether [tex]|x-3|<3[/tex] is imposed or not does not change the fact that [tex]|x-3|<\delta[/tex]. These statements do not contradict eachother, you still end up with [tex]|x-3|<min(\delta,3)\leq\delta[/tex].
     
    Last edited: Sep 28, 2010
  10. Sep 29, 2010 #9

    haushofer

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    In my openingpost I had

    [tex]

    \epsilon = \delta(\delta + 6)

    [/tex]

    directly. I have to think about this more, but at this point I really don't see why I cannot use this epsilon to complete the proof.
     
  11. Sep 29, 2010 #10

    haushofer

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    Ok, that's what I understand, but not why my approach in my openingspost is wrong. Sorry if I'm being a pain in the ***.
     
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