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Math Help - Finding the limit of ((x-1)/x)^x as x->infinity

  1. #1
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    Cool Finding the limit of ((x-1)/x)^x as x->infinity

    Dear all,
    I am trying to find, analytically, the limit of ((x-1)/x)^x as x->infinity.

    I tried l'Hopital's rule but it seems to make it more complicated. I know that the limit is 1/e but need to show my working and understand why it is 1/e.

    Hope you can help!
    Many thanks
    Mike
    Attached Thumbnails Attached Thumbnails Finding the limit of ((x-1)/x)^x as x->infinity-fullscreen-capture-10082009-161959.jpg  
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  2. #2
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    We know \lim_{x \to 0} (1+x)^\frac {1}{x}=e

    Let's simplify this first.

    \lim_{x \to \infty} \bigg ( \frac {x-1}{x}\bigg )^x = \lim_{x \to \infty} \bigg ( 1-\frac{1}{x}\bigg )^x

    Let -\frac {1}{x}=h, as x \to \infty, h \to 0

    \therefore \lim_{x \to \infty} \bigg ( \frac {x-1}{x}\bigg )^x=\lim_{h \to 0} (1+h)^{-\frac {1}{h}}=\frac {1}{e}
    Last edited by chengbin; August 10th 2009 at 07:53 AM.
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  3. #3
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    Take note that \left( {\frac{{x - 1}}{x}} \right)^x  = \left( {1 + \frac{{ - 1}}{x}} \right)^x .
    That is a standard e\text{-form}.
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  4. #4
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     \lim_{x \to \infty} \ln \Big(\frac{x-1}{x}\Big)^{x}

     = \lim_{x \to \infty} x \ln \Big(\frac{x-1}{x}\Big)

     = \lim_{x \to \infty} \frac{\ln \Big(\frac{x-1}{x}\Big)}{\frac{1}{x}}

     = \lim_{x \to \infty} \frac{\frac{x}{x-1} \frac{1}{x^{2}}}{\frac{-1}{x^{2}}}

     = \lim_{x \to \infty} \frac{-x}{x-1}

     = \lim_{x \to \infty} \frac{-1}{1-\frac{-1}{x}} = -1

    so  \lim_{x \to \infty} \Big(\frac{x-1}{x}\Big)^{x} = e^{-1}= \frac{1}{e}
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  5. #5
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    Quote Originally Posted by sevenquid View Post
    Dear all,
    I am trying to find, analytically, the limit of ((x-1)/x)^x as x->infinity.

    I tried l'Hopital's rule but it seems to make it more complicated. I know that the limit is 1/e but need to show my working and understand why it is 1/e.

    Hope you can help!
    Many thanks
    Mike
    let y = \left(\frac{x-1}{x}\right)^x

    \ln{y} = x \cdot \ln\left(\frac{x-1}{x}\right) = \frac{\ln\left(\frac{x-1}{x}\right)}{\frac{1}{x}} = \frac{\ln\left(1 - \frac{1}{x}\right)}{\frac{1}{x}}

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

    use L'Hopital ...

    \lim_{x \to \infty} \frac{\frac{\frac{1}{x^2}}{1 - \frac{1}{x}}}{-\frac{1}{x^2}}

    \lim_{x \to \infty} \frac{-1}{1 - \frac{1}{x}} = -1

    as x \to \infty , \ln{y} \to -1 ... y \to e^{-1}<br />
    Last edited by skeeter; August 10th 2009 at 07:55 AM. Reason: sucking on the hind tit for sure on this one.
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