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Math Help - Lost on Cubic Epsilon Delta proof

  1. #1
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    Lost on Cubic Epsilon Delta proof

    For proving that
    Delta epsilon proof: limit x->2 (x^3 - 3) = 5

    The following proof is given:

    Given e > 0 we need to find d > 0 such that
    |x - 2| < d ==> |(x^3 - 3) - 5| = |x^3 - 8| < e.

    One way to do this is to expand x^3 - 8 into powers of (x - 2).
    x^3 - 8 = [(x - 2) + 2]^3 - 8 = (x - 2)^3 + 6(x - 2)^2 + 12(x - 2).

    Then, by the triangle inequality,
    |x^3 - 8| = |(x - 2)^3 + 6(x - 2)^2 + 12(x - 2)|
    <= |x - 2|^3 + 6|x - 2|^2 + 12|x - 2|.

    Now, we insist on d <= 1. In this manner,
    |x^3 - 8| <= |x - 2| + 6|x - 2| + 12|x - 2|.
    = 19 |x - 2|.

    Finally we can fully specify d. Given e > 0, let d = min{1, e/19}.
    Then, |x - 2| < d ==>
    |x^3 - 8| <= 19 |x - 2|
    < 19 (e/19)
    = e, as required.


    I'm lost as to how the exponents can be dropped as seen in the bolded text. Anyone have an idea?
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  2. #2
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    It's because \displaystyle |x - 2| \leq \delta and you are making \displaystyle \delta \leq 1.

    So \displaystyle |x - 2| \leq 1.


    Therefore \displaystyle |x-2|^3 \leq 1^3 = 1 and \displaystyle |x-2|^2 \leq 1^2 = 1.

    Also note that \displaystyle |x - 2|^3 \leq |x - 2| for \displaystyle |x - 2| \leq 1 and \displaystyle |x - 2|^2 \leq |x - 2| for \displaystyle |x - 2| \leq 1, since exponentiating a fraction between \displaystyle (0, 1) makes the fraction smaller.

    So \displaystyle |x - 2|^3 + 6|x - 2|^2 + 12|x - 2| \leq |x - 2| + 6|x - 2| + 12|x - 2|.
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  3. #3
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    Thanks. I get the whole inequality thing now. All I'm lost on now is the theoretical aspect of it. What allows us to take the 19 and divide E by it such that d = E/19.

    I understand the logic behind doing so for 2nd degree polynomials. For example, when proving that the limit of X^2 +1 = 5 as x approaches 2, I understand why we take X - 2 < d, set d at 1, add to all sides to get 3 < X +2 < 5, and then replace E/|x+2| with E/5.

    I'm not sure what we are doing here. Factoring it all out, I can now get the answer right for any cubic equation because you just take the coefficient. But I'm not sure why exactly we are allowed to divide E by the coefficient like we divide E by 5 above.
    Last edited by Seoulless; December 26th 2010 at 08:01 PM.
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  4. #4
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    You have gotten to the point that

    \displaystyle |x^3 - 8| \leq 19|x - 2| as long as \displaystyle \delta \leq 1.

    Also recall that \displaystyle |x - 2| \leq \delta.

    So surely \displaystyle |x^3 - 8| \leq 19\delta.


    Remember that you were trying to find an \displaystyle \epsilon such that \displaystyle |x^3 - 8| < \epsilon. Surely here you can let \displaystyle \epsilon = 19\delta \implies \delta = \frac{\epsilon}{19}, as long as \displaystyle \delta < 1.

    So we say that \displaystyle \delta := \min \left\{1, \frac{\epsilon}{19} \right\}.
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