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Math Help - Compound Fraction Limit

  1. #1
    Junior Member
    Joined
    Dec 2010
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    Dallas
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    Compound Fraction Limit

    Ok, so...

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

    I multiplied by the common denominator x(x+4) and got
    (x^2-4x)/(-x^2+2x+8)
    simplified to
    ((x-2)(x+2)) / (-(x-4)(x+2))
    simplified to
    (x-2) / -(x-4)
    simplified to
    2/0

    What did I do wrong?

    Thanks
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Hello, beanus!

    You "cancelled" incorrectly . . .


    \displaystyle \lim_{x\to4}\,\frac{\dfrac{x-4}{x+4}} {\dfrac{2}{x} -\dfrac{x}{x+4}}

    Multiply numerator and denominator by x(x+4)

    . . \displaystyle \frac{x(x+4)\cdot\dfrac{x-4}{x+4}} {x(x+4)\cdot\left(\dfrac{2}{x} - \dfrac{x}{x+4}\right)} \;=\;\frac{x(x-4)}{2(x+4) - x^2} \;=\;\frac{x(x-4)}{2x + 8 - x^2}

    . . \displaystyle =\;\frac{x(x-4)}{-(x^2 - 2x - 8)} \;=\;\frac{x(x-4)}{-(x+2)(x-4)} \;=\; -\frac{x}{x+2}


    \displaystyle \text{Therefore: }\:\lim_{x\to4}\left(-\frac{x}{x+2}\right) \;=\;-\frac{4}{6} \;=\;-\frac{2}{3}

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  3. #3
    Junior Member
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    Dallas
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    Ah I see I wasn't supposed to distribute the x(x-4) thanks!!
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