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Math Help - need help with limit, just a refresher

  1. #1
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    need help with limit, just a refresher

    Prove that the stated limit is correct by finding a number d (delta) in terms of E (is an element of) such that
    |f(x) - L| < E whenever |x - c|< d:

    lim (4x - 1) = 11
    x -> 3
    Last edited by ffxwolf; September 22nd 2009 at 11:11 AM.
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by ffxwolf View Post
    Prove that the stated limit is correct by finding a number d (delta) in terms of E (is an element of) such that
    |f(x) - L| < E whenever |x - c|< d:

    lim (4x - 1) = 11
    x -> 3
    You want to find \delta such that |x-3|<\delta \implies |(4x-1)-11|<\epsilon.

    Choose \delta=\frac{\epsilon}{4}.

    Thus when |x-3|<\delta, we have that |(4x-1)-11|=|4x-12|=4|x-3|<4\delta=\frac{4\epsilon}{4}=\epsilon
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  3. #3
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    Thank you for your response, I just don't understand why delta = epsilon/4
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  4. #4
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by ffxwolf View Post
    Thank you for your response, I just don't understand why delta = epsilon/4
    If you don't know what \delta is, you can solve for it.

    Whatever \delta is you know that |x-3|<\delta.

    So, work with what you have. You ultimately want |(4x-1)-11|<\epsilon.

    So, similar to before,

    |(4x-1)-11|=|4x-12|=4|x-3|<4\delta.

    Now you want 4\delta to be equal to \epsilon, so you just solve for it.

    4\delta=\epsilon\implies \delta=\frac{\epsilon}{4}
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  5. #5
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    Okay, I see it now. Thanks
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