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Math Help - Proving a lim

  1. #1
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    Proving a lim

    I am trying to find the limit of this sequence:  s_n = \dfrac{sin (n)}{n}. Any help would be appreciated.
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  2. #2
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    consider |sin(n)| =< 1 for all n.
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  3. #3
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    Thanks, that helps some, but I am still not sure about what to take as N.
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  4. #4
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    y not greater than 1 over epsilon?
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  5. #5
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    lim \frac{sin(n)}{n}
    Have: \frac{sin(n)}{n} \leq \frac{1}{n}
    But lim \frac{1}{n} = 0
    So: lim \frac{sin(n)}{n} =0
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  6. #6
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    Quote Originally Posted by thedoctor818 View Post
    I am trying to find the limit of this sequence:  s_n = \dfrac{sin (n)}{n}. Any help would be appreciated.
    For that limit you have to use the following theorem:

    If  lim_{n\to\infty} x_{n} = 0 and the sequence  y_{n} is bounded ,then lim_{n\to\infty} x_{n}y_{n}=0.

    And in our case :

     y_{n} = sin(n) where |sin(n)|\leq 1 for all n ,and

     x_{n} =\frac{1}{n},where lim_{n\to\infty}\frac{1}{n}=0

    Hence lim_{n\to\infty}\frac{sin(n)}{n} = 0
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