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Finding limit - given x_n

  1. Nov 1, 2013 #1
    1. The problem statement, all variables and given/known data
    Let ##x_0=2\cos(\pi/6)## and ##x_n=\sqrt{2+x_{n-1}}##, n=1,2,3,....

    Find $$\lim_{n \rightarrow \infty} 2^{n+1}\cdot \sqrt{2-x_n}$$


    2. Relevant equations



    3. The attempt at a solution
    I found ##\displaystyle x_n=2\cos\left(\frac{\pi}{6(n+1)}\right)##

    $$\Rightarrow \sqrt{2-x_n}=2\sin\left(\frac{\pi}{12(n+1)}\right)$$
    The limit to be evaluated is
    $$\lim_{n \rightarrow \infty} 2\cdot 2^{n+1}\cdot \sin\left(\frac{\pi}{12(n+1)}\right)$$
    Evaluating it results in infinity but the given answer is ##\pi/3##. :confused:

    Any help is appreciated. Thanks!
     
  2. jcsd
  3. Nov 1, 2013 #2

    ehild

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    Are you sure????? :devil: How did you arrived at it?

    ehild
     
  4. Nov 1, 2013 #3
    I just realised that it's wrong. :redface:

    The correct one is
    $$x_n=2\cos\left(\frac{\pi}{6 \times 2^n} \right)$$

    This gives the answer. Thank you ehild! :smile:
     
  5. Nov 1, 2013 #4

    ehild

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    You see how fast you find a mistake? You do not really need us .:tongue2:

    ehild
     
  6. Nov 1, 2013 #5
    No, I do. Without you people, I would never have been able to solve such problems. :)
     
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