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Math Help - Question concerning delta epsilon proof

  1. #1
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    Question concerning epsilon-delta proof

    Lim\frac{\1}{x^2}=\frac{\1}{4}

    x\rightarrow2

    So I start with :


    \frac{\1}{x^2}-\frac{\1}{4}

    which breaks down to :

    \frac{\(x-2)(x+2)}{4x^2}

    But I'm confused how to bound this to get a delta.
    (PS: I cant figure out how to put the absolute value signs in here, but I know they are supposed to be there)

    Do I break it down to :

    (x-2)*\frac{\(x+2)}{4x^2}=(x-2)*\frac{\1}{4x}+\frac{\1}{2x^2}

    Then what?
    Last edited by Jerry99; August 19th 2012 at 11:31 AM.
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  2. #2
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    Re: Question concerning epsilon-delta proof

    Quote Originally Posted by Jerry99 View Post
    Lim\frac{\1}{x^2}=\frac{\1}{4}
    x\rightarrow2
    So I start with :
    \frac{\1}{x^2}-\frac{\1}{4}
    which breaks down to :
    \frac{\(x-2)(x+2)}{4x^2}
    But I'm confused how to bound this to get a delta.
    (PS: I cant figure out how to put the absolute value signs in here, but I know they are supposed to be there)
    [TEX]\left|{\frac{\1}{x^2}-\frac{\1}{4}\right|[/TEX] gives \left|{\frac{\1}{x^2}-\frac{\1}{4}\right|

    \left|{\frac{\1}{x^2}-\frac{\1}{4}\right|=\frac{|x-2||x+2|}{|4x^2|}

    If 0<\delta<1~\&~0<|x-2|<\delta then |x+2|<5 and \left|{\frac{\1}{4x^2}\right|<\frac{1}{4}

    See what you can do with that.
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  3. #3
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    Re: Question concerning epsilon-delta proof

    I get how \left|x+2\right|<5

    But how do you get the \left|\frac{\1}{4x^2}\right|<\frac{\1}{4}

    and thanks for the absolute value sign tip
    Last edited by Jerry99; August 19th 2012 at 12:03 PM.
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  4. #4
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    Re: Question concerning delta epsilon proof

    I think I get what is going on now actually. that has to be less than 1/4 basically

    thanks for the help
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    Re: Question concerning epsilon-delta proof

    Quote Originally Posted by Jerry99 View Post
    I get how \left|x+2\right|<5
    But how do you get the \left|\frac{\1}{4x^2}\right|<\frac{\1}{4}
    If 1<x<3 then 4<4x^2<36 or \frac{1}{36}<\frac{1}{4x^2}<\frac{1}{4}.
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  6. #6
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    Re: Question concerning epsilon-delta proof

    Ok cool I get it. Thanks for the help
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  7. #7
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    Re: Question concerning delta epsilon proof

    Sigh , well, I thought I did. What am I missing here that allows you to make the statement of  4 < 4x^2<36 =\frac{\1}{36}<\frac{\1}{4x^2}<\frac{\1}{4}


    My min \deltafrom this : 1, \frac{\epsilon}{3},\frac{\epsilon}{5} ??
    Last edited by Jerry99; August 19th 2012 at 02:01 PM.
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  8. #8
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    Re: Question concerning delta epsilon proof

    Quote Originally Posted by Jerry99 View Post
    Sigh , well, I thought I did. What am I missing here that allows you to make the statement of  4 < 4x^2<36 =\frac{\1}{36}<\frac{\1}{4x^2}<\frac{\1}{4}
    My min \deltafrom this : 1, \frac{\epsilon}{4},\frac{\epsilon}{5} ??
    \begin{array}{l}1 < x < 3\\ 1 < {x^2} < 9\\ 4 < 4{x^2} < 36\\ \dfrac{1}{{36}} < \dfrac{1}{{4{x^2}}} < \dfrac{1}{4} \end{array}
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  9. #9
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    Re: Question concerning delta epsilon proof

    if 0 < a < b < c, then 0 < a2 < ab < b2 < bc < c2, or in short:

    0 < a2 < b2 < c2.

    similarly, if 0 < a < b < c then:

    0 < 1/c < 1/b < 1/a

    the basis for the last inequality is this:

    we define for two fractions a/b, c/d: a/b < c/d if and only if ad < bc. note this agrees with our usual notion if b = d = 1.

    this means that if a < c, that when we compare:

    1/a and 1/c, we have 1(c) > a(1), so 1/a > 1/c.

    for example: 4 is bigger than 2, so 1/4 is smaller than 1/2.
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