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Thread: Using L'Hopital's rule...have answer but...

  1. September 12th 2010, 07:59 PM #1
    janedoe
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    Using L'Hopital's rule...have answer but...

    need help getting to it! See pic: http://i55.tinypic.com/e1df2x.jpg

    thanks!
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  2. September 12th 2010, 08:56 PM #2
    Soroban
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    Hello, janedoe!

    \displaystyle \lim_{x\to\infty} \dfrac{\ln\left(1-\frac{3}{x}\right)}{\frac{1}{x}}

    Watch carefully . . .


    The numerator is: . \ln\left(1 - \frac{3}{x}\right) \;=\;\ln\left(1 - 3x^{-1}\right)

    Its derivative is: . \dfrac{1}{1-3x^{-1}}\cdot(3x^{-2}) \;=\;\dfrac{1}{1-\frac{3}{x}} \cdot\dfrac{3}{x^2}


    The denominator is: . \dfrac{1}{x} \:=\:x^{-1}

    Its derivative is: . -x^{-2} \;=\;-\dfrac{1}{x^2}


    Hence, the fraction is: . \dfrac{\dfrac{1}{1-\frac{3}{x}}\cdot\dfrac{3}{x^2}}{-\dfrac{1}{x^2}} \;=\; -\dfrac{3}{1-\frac{3}{x}}
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