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Math Help - Delta Epsilon Proof

  1. #1
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    Delta Epsilon Proof

    Hello, I read the stickied delta epsilon guide at the top of the page and found it very helpful for finding an epsilon for a given delta. The guide represents delta = epsilon/M. If M has an x in it, we need to set a bound on delta in order to get an M that is a number without any x's in it.

    My problem is, for example, proving that the limit of x^3 as x approaches 3 is 27. Here is what I have so far:

    |x-3| < \delta

    |x^3 - 27| < \epsilon

    (|x - 3|)(|x^2 + 3x + 9|) < \epsilon

    (|x -3|) < \epsilon/(|x^2 + 3x + 9|)

    In the stickied article, it says that I need to set a delta (i.e, delta = 1) and then add things to my first equation so that |x - 3| becomes |x^2 + 3x + 9|. If i do this, my M still ends up having an x in it because the only way to turn |x -3| into a quadratic equation is to multiply by x. I can solve delta epsilon proofs, but the cubic powers are throwing me off here. Any suggestions?
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  2. #2
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    If \delta <1 then \left| {x - 3} \right| < \delta \, \Rightarrow \,\left| x \right| < 4.

    \left| {x^2  + 3x + 9} \right| < 16 + 12 + 9
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  3. #3
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    Thanks for the help, Plato! I didn't realize it was that simple.

    I have one more question that I was wondering if someone could look over.

    I am trying to prove e^x is continuous. Can someone please make sure I'm doing this right?

    Scratch work:

     |x - x_0| < \delta

    |e^x - e^x_0| < \epsilon

    ln |e^x - e^x_0| < ln |\epsilon|

    |x - x_0| < ln |\epsilon|

    \delta < (ln |\epsilon|)

    Proof:

    Let \delta = min{(1, ln |\epsilon|)}

     |x - x_0| < \delta

    |e^x - e^x_0| < e^\delta

    |e^x - e^x_0 |< e^(ln |\epsilon|) (ln |epsilon| should be in the exponent)

    |e^x - e^x_0| < \epsilon

    |f(x) - f(x_0)| < \epsilon, therefore e^x is continuous.
    Last edited by Epiphenomenon; October 10th 2009 at 04:43 PM.
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  4. #4
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    Quote Originally Posted by Epiphenomenon View Post

    ln |e^x - e^x_0| < ln |\epsilon|

    |x - x_0| < ln |\epsilon|
    you assumed that the logarithm is linear, so this is step is wrong.
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  5. #5
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    Oh, ok, I will work on it. How would you recommend I approach this?
    Last edited by Epiphenomenon; October 10th 2009 at 06:06 PM.
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  6. #6
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    Ok, I can solve for delta, but I can't create an equation to substitute it into.

     x - x_0 < \delta

    e^x - e^x_0 < \epsilon

    e^x < \epsilon + e^x_0

    ln|e^x| < ln |\epsilon + e^x_0|

     x < ln |\epsilon + e^x_0|

     x - x_0 < ln|\epsilon + e^x_0| - x_0

     \delta < ln|\epsilon + e^x_0| - x_0

    Now that I have my delta written in terms of epsilon and x0, I can't figure out how to substitute it into e^x - e^x_0 to get it  < \epsilon
    Last edited by Epiphenomenon; October 10th 2009 at 09:43 PM.
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  7. #7
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    Is it ok if I start with my delta equation |x - x_0| < \delta, make the necessary substitions into my delta equation, and end up with |f(x) - f(x_0) < \epsilon for my proof? As opposed to starting with |f(x) - f(x_0)| and ending with epsilon? Can I start my proof with the delta equation instead of the epsilon equation?

    Here's what I'm thinking of doing:

     \delta = min(1, ln |\epsilon + e^x_0| - x_0)

    x - x_0 < \delta

    e^x/e^x_0 < e^\delta (I raised the delta equation by the power of e on both sides)

    e^x/e^x_0 < e^(ln |(\epsilon + e^x_0)| - x_0)

    e^x/e^x_0 < (\epsilon + e^x_0)/e^x_0

    e^x - e^x_0 < \epsilon
    Last edited by Epiphenomenon; October 11th 2009 at 03:23 PM.
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