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Math Help - How would you about calculating this simple limit?

  1. #1
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    How would you go about calculating this simple limit?

    a(n) = (2^(n-1)) / ((2^n)-1)
    with n --> infinite.

    thank you
    Last edited by nappysnake; November 27th 2011 at 12:24 PM.
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  2. #2
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    Re: How would you about calculating this simple limit?

    Quote Originally Posted by nappysnake View Post
    a(n) = (2^(n-1)) / ((2^n)-1)
    with n --> infinite.
    \lim_{n \to \infty} \frac{2^{n-1}}{2^n - 1} =

    divide every term by 2^{n-1} ...

    \lim_{n \to \infty} \frac{1}{2 - \frac{1}{2^{n-1}}} = \frac{1}{2}
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  3. #3
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    Re: How would you about calculating this simple limit?

    thank you skeeter for letting me come to a crucial realization, it was very clear
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