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Math Help - Evaluation of a limit as n approaches infinity using L'Hopital's Rule

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    Evaluation of a limit as n approaches infinity using L'Hopital's Rule

    PROBLEM:

    -series of numbers where:

    x_0 = 1

    and,




    Find:

    ATTEMPT:

    So far,







    L'Hopital's rule:







    My application of L'Hopital's rule is incorrect. It's an indeterminate form, but not resolved in the way I indicated. Any ideas on how I could resolve the problem?
    Last edited by Lambin; December 21st 2012 at 06:37 PM.
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    Re: Evaluation of a limit as n approaches infinity using L'Hopital's Rule

    To use LaTeX on this forum, you need to put the code in tex tags, which look like [ tex ] [ / tex ] without the spaces.
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    Re: Evaluation of a limit as n approaches infinity using L'Hopital's Rule

    I stripped out the mimetex stuff and was able to make the question readable:

    Quote Originally Posted by Lambin View Post
    PROBLEM:

    (x_{n})-series of numbers where:

    x_0 = 1

    and,

    x_{n+1} = sqrt(2+x_{n})

    Find: \lim_{n\rightarrow \infty } 4^{n}(x_{n} - 2) = ?

    ATTEMPT:

    So far,
    y\,=\,\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\,.\,.\,.\sq  rt{3}}}}}

    y^2\,=\,2\,+\,y

    (y\,-\,2)(y\,+\,1)\,=\,0\,\Rightarrow\,y\,=\,2

    L'Hopital's rule:

    \lim_{n\to\infty}\,\frac{x_n\,-\,2}{\frac{1}{4^n}}

    \lim_{n\to\infty}\,\frac{0}{\frac{-4^n\ln(4)}{4^{2n}}}\,=\,0

    \lim_{n\to\infty } 4^{n}(x_{n}\,-\,2) = 0

    My application of L'Hopital's rule is incorrect. It's an indeterminate form, but not resolved in the way I indicated. Any ideas on how I could resolve the problem?
    In the last step you substituted 0 for x_n-2. But it is not true that x_n-2=0 for all n, so you can't make that substitution.

    I put the problem into Excel, and it looks like the limit should be about -1.097, but I can't see how to get to that result algebraically. Can someone else help?

    - Hollywood
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    Re: Evaluation of a limit as n approaches infinity using L'Hopital's Rule

    Sorry I tested on my maple the answer is 0:


    \begin{align*}\therefore \lim_{n\to \infty} 4^n (x_n - 2)=&  \lim_{n\to \infty} 4^n \left(\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+ \cdots +1}}}}} - 2\right) \\ =& 0\end{align*}


    Is there a way to show that this limit 0 can be found mathematically?

    How can one find limit of \lim_{n\to \infty } 4^{n}(x_{n} - 2) = \lim_{n\to \infty} 4^n \left(\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+ \cdots +1}}}}} - 2\right)

    using math(without the help of computer) where x_{n} = \sqrt{2 + x_{n-1}} \text{ and } x_0 = \sqrt{(2+1)}?
    Last edited by x3bnm; December 22nd 2012 at 08:40 PM.
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    Re: Evaluation of a limit as n approaches infinity using L'Hopital's Rule

    Thank you all for trying to help.

    And sorry for the trouble of reading this post; I just now started learning LaTeX.

    Here's the solution:

    Put x_n=2y_n to get  y$_{n+1}$=\sqrt{\frac{1+y_n}{2}} and y_0=\frac{1}{2}=cos(\frac{\pi}{3})

    Therefore, y_n=cos(\frac{\pi}{3*2^n})\implies x_n=2cos\frac{\pi}{3*2^n}

    Thus, z$_{n}=4$^{n}(x_n-2)=\frac{-2(1-cos\frac{\pi}{3*2^n})}{(\frac{\pi}{3*2^n})^2}\frac  {\pi}{3*2^n}^2}(4^n)

    Hence, $\lim_{n\to\infty}z_n=-2*\frac{1}{2}*\frac{\pi^2}{9}=\frac{-\pi^2}{9} \approx -1.097$
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