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Thread: Limit question

  1. #1
    Senior Member Vinod's Avatar
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    Limit question

    Let $x_1 = 100, $and for n ≥ 1, let $ x_{n+1}=\frac12(x_n +\frac{100}{x_n}).$ Assume that $L = lim_{n→∞}x_n$ exists, and calculate L.

    Answer provided is 10. But I don't understand how it was calculated?
    Last edited by Vinod; Apr 23rd 2018 at 06:11 AM.
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  2. #2
    MHF Contributor
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    Re: Limit question

    $x_{n+1} = \dfrac{1}{2}\left( x_n + \dfrac{100}{x_n} \right)$

    Take limit of both sides as $n \to \infty$:

    $\displaystyle \begin{align*}\lim_{n \to \infty} x_{n+1} & = \lim_{n \to \infty} \dfrac{1}{2} \left( x_n + \dfrac{100}{x_n} \right) \\ L & = \dfrac{1}{2}\left( \lim_{n \to \infty} x_n + \dfrac{100}{\displaystyle \lim_{n \to \infty} x_n } \right) \\ L & = \dfrac{1}{2}\left( L + \dfrac{100}{L} \right)\end{align*}$

    Solve for L.
    Last edited by SlipEternal; Apr 23rd 2018 at 06:22 AM.
    Thanks from Vinod and topsquark
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