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Math Help - Nasty limit...

  1. #1
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    Nasty limit...

    I have this really nasty limit that I've been banging my head on all night. The question asks me to calculate the following:

    \lim_{n\to\infty}\frac{n \times \left(\frac{2n-2}{2n-1}\right)^{2n-2}}{2n-1}


    The limit evaluates to \frac{\infty}{\infty}, so my first instinct was to use L'Hospital's Rule and take the limit of \frac{f'(x)}{g'(x)}. But that didn't really simplify it; doing that just gave another nasty limit. Is there a simpler approach?


    According to Wolfram Alpha, the limit is equal to \frac{1}{2e}. But it just says "series expansion at n=\infty" without showing any of the steps.


    I know that e is the sum of an infinite series. But how do you recognize that from the limit? And how do you get to \frac{1}{2e}?


    Any pointers would be greatly appreciated!

    - A really puzzled math student.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Express

    L=\lim_{n\to +\infty}\dfrac{n}{2n-1}\cdot \left(1-\dfrac{1}{2n-1}\right)^{2n-2}

    and apply the definition of the number e .
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  3. #3
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    Write the expression under the limit (I wonder what it's called?) as:

    \left( \frac{n}{2n-1}\right)\left(\frac{2n-1}{2n-2}\right)\left(1-\frac{1}{2n-1}\right)^{2n-1}

    EDIT: far too late! O_o
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  4. #4
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    You guys are amazing. Thanks so much!
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  5. #5
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    \displaystyle \begin{align*}\lim_{n \to \infty}\frac{n\left(\frac{2n - 2}{2n - 1}\right)^{2n-2}}{2n - 1} &= \lim_{n \to \infty}\left(\frac{n}{2n - 1}\right) \lim_{n \to \infty}\left(\frac{2n - 2}{2n - 1}\right)^{2n - 2}\\ &= \frac{1}{2}\lim_{n \to \infty}\left(\frac{2n - 2}{2n - 1}\right)^{2n - 2} \\ &= \frac{1}{2}\lim_{n \to \infty}e^{\ln{\left[\left(\frac{2n-2}{2n - 1}\right)^{2n-2}\right]}} \\&= \frac{1}{2}e^{\lim_{n \to \infty}\ln{\left[\left(\frac{2n-2}{2n-1}\right)^{2n-2}\right]}}\\ &= \frac{1}{2}e^{\lim_{n \to \infty}\left[(2n - 2)\ln{\left(\frac{2n - 2}{2n - 1}\right)}\right]} \\ &= \frac{1}{2}e^{\lim_{n \to \infty}\left[\frac{\ln{(2n - 2)} - \ln{(2n - 1)}}{(2n - 2)^{-1}}\right]}\\ &= \frac{1}{2}e^{\lim_{n \to \infty}\left[\frac{\frac{1}{n - 1} - \frac{2}{2n - 1}}{-2(2n - 2)^{-2}}\right]}\,\,\textrm{by L'Hospital's Rule}\end{align*}

    Go from here.
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