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Limit of infinite sines

  1. Feb 1, 2010 #1
    Hello,
    1. The problem statement, all variables and given/known data
    Calculate the limit of [tex] lim_{n-> /infty} sinsinsin..._{n times } sin(x)[/tex] where x is any real number


    My attempt:

    Using the Taylor expansion we know that [tex]sin(x)=x- \frac{x^3}{3!} +\circ(x^3)[/tex] and that [tex]1 \geq sin(x) \geq -1 [/tex]
    If the expression has a limit then every subsequence aproaches the same limit. specifically we can define a sequence
    [tex] a_{n+1}=|a_{n}- \frac {a_{n}^3}{3!}| [/tex] and so the limit of this sequence must be the limit of the above function.
    We can restrict our inquiry to [tex] a_1\in[-1,1] [/tex] since tex]1 \geq sin(x) \geq -1 [/tex]

    We will show by induction that the sequence is montonic decreasing:
    [tex] a_{2}=|a_{1}- \frac {a_{1}^3}{3!}| \geq |a_{1}| - |\frac {a_{1}^3}{3!}| /leq a_{1} [/tex] which holds for any value of [tex] a_1 [/tex]
    We will assume for n and prove for n+1
    [tex] a_{n+1}=|a_{n}- \frac {a_{n}^3}{3!}| \leq^{?} a_{n} \iff | \frac {3!a_{n} - a_{n}^3}{3!} < a_{n}| \iff |3!a_{n} - a_{n}^3| \leq 3!a_{n} -> |3!a_{n} - a_{n}^3| \geq |3!a_{n}| -| a_{n}^3| \leq 3!a_{n} \iff -| a_{n}^3| \leq 0 [/tex]
    Or in words, we used the traingle inequality and the fact that the minimum of an absolut value is zero to show that the sequence decreases.

    There is no need to show that it is bounded since an absolute value is lowerly bounded by 0 and we have shown that the sequence is monotonically decreasing.
    So we knwo the sequence converges to a finite limit and so the original function does as well. but how do i show that the limit is zero?

    Thanks for the help,
    Tal
     
  2. jcsd
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