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Thread: 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; Sep 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 $\displaystyle \delta$ such that $\displaystyle |x-3|<\delta \implies |(4x-1)-11|<\epsilon$.

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

    Thus when $\displaystyle |x-3|<\delta$, we have that $\displaystyle |(4x-1)-11|=|4x-12|=4|x-3|<4\delta=\frac{4\epsilon}{4}=\epsilon$
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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 $\displaystyle \delta$ is, you can solve for it.

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

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

    So, similar to before,

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

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

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