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Math Help - Epsilon - Delta Proof Question

  1. #1
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    Epsilon - Delta Proof Question

    lim ((1/(x+4)) = 1/2
    x-> -2

    or

    a=-2
    L=1/2
    f(x)= (1/(x+4))

    The question says we will need a constraint on |x+2|, so it says to assume |x+2|<1.
    This is the part that is confusing me because I can solve any other limit question.

    The options for answers are:
    a)[0,min{1,4epsilon})
    b)(0,min{1,6epsilon}]
    c)[0,min{1,2epsilon})
    d)(0,min{1,2epsilon}]
    e)(0,min{1,4epsilon})
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  2. #2
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    Re: Epsilon - Delta Proof Question

    Quote Originally Posted by poipoipoi10 View Post
    lim ((1/(x+4)) = 1/2
    x-> -2
    The question says we will need a constraint on |x+2|, so it says to assume |x+2|<1.
    This is the part that is confusing me because I can solve any other limit question.
    The options for answers are:
    a)[0,min{1,4epsilon})
    b)(0,min{1,6epsilon}]
    c)[0,min{1,2epsilon})
    d)(0,min{1,2epsilon}]
    e)(0,min{1,4epsilon})
    Notice that \left| {\frac{1}{{x + 4}} - \frac{1}{2}} \right| = \frac{1}{2}\frac{{\left| {x + 2} \right|}}{{\left| {x + 4} \right|}}.

    If |x+2|<1 what can you say about \frac{1}{|x+4|}~?
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  3. #3
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    Re: Epsilon - Delta Proof Question

    Sigh... I'm confused -.-

    I guess that
    -1<|x+2|<1
    -3<x<-1
    1<x+4<3

    sooooo uh yeah... haha
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  4. #4
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    Re: Epsilon - Delta Proof Question

    Quote Originally Posted by poipoipoi10 View Post
    Sigh... I'm confused -.-
    Here a hint.
    |x+4|\ge 1
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  5. #5
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    Re: Epsilon - Delta Proof Question

    So would the answer be D???
    Because if you plug 1 as the |x+4|, from the equation you gave me: 1|x+2|/2|x+4|<epsilon
    I would get |x+2|<2*epsilon
    I don't really know if i've done that correctly, I've been doing work all day and my head is swimming with different subjects
    Last edited by poipoipoi10; October 10th 2011 at 02:07 PM.
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  6. #6
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    Re: Epsilon - Delta Proof Question

    Quote Originally Posted by poipoipoi10 View Post
    lim ((1/(x+4)) = 1/2
    x-> -2

    or

    a=-2
    L=1/2
    f(x)= (1/(x+4))

    The question says we will need a constraint on |x+2|, so it says to assume |x+2|<1.
    This is the part that is confusing me because I can solve any other limit question.

    The options for answers are:
    a)[0,min{1,4epsilon})
    b)(0,min{1,6epsilon}]
    c)[0,min{1,2epsilon})
    d)(0,min{1,2epsilon}]
    e)(0,min{1,4epsilon})
    To show \displaystyle \lim_{x \to c}f(x) = L you need to prove \displaystyle 0 < |x - c| < \delta \implies |f(x) - L| < \epsilon. So you need to show \displaystyle 0 < |x + 2| < \delta \implies \left|\frac{1}{x + 4} - \frac{1}{2}\right| < \epsilon.

    Scratch work:

    \displaystyle \begin{align*} |f(x) - L| &< \epsilon \\ \left|\frac{1}{x + 4} - \frac{1}{2}\right| &< \epsilon \\ \left|-\left[\frac{x+2}{2(x + 4)}\right]\right| &< \epsilon \\ \frac{|x + 2|}{2|x + 4|} &< \epsilon \\ |x + 2| &< 2|x + 4|\epsilon  \end{align*}

    But we can't let \displaystyle \delta = 2|x + 4|\epsilon because \displaystyle \delta is only supposed to be a function of \displaystyle \epsilon, not a function of \displaystyle x. But if we remember that the act of taking a limit means that we are showing that moving \displaystyle x closer to the value of \displaystyle c in both directions moves \displaystyle f(x) closer to the value of \displaystyle L in both directions, that means we can choose a value of \displaystyle x as close to \displaystyle c as we like, then make it move closer. So say we want to start off with \displaystyle x no more than \displaystyle 1 unit away from \displaystyle c, then \displaystyle |x + 2| < 1. Therefore

    \displaystyle \begin{align*} |x + 2| &< 1 \\ -1 < x + 2 &< 1 \\ 2 < x + 4 &< 3 \\ |x + 4| &< 3 \\ 2|x + 4| &< 6 \\ 2|x + 4|\epsilon &< 6\epsilon \end{align*}

    So we can let \displaystyle \delta = \min\left\{1, 6\epsilon\right\} and reverse each step to get your proof.
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