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Math Help - Limit of a sequence

  1. #1
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    Limit of a sequence

    Hi everyone,

    Prove the following:
     \lim_{n \to \infty} (n+2)^2 sin(\frac{1}{n}) = \infty

    I must show that for each M>0 we can find N such that (n+2)^2 sin(\frac{1}{n})> M for every n>N.

    Thing is that I can't figure out how to "get rid of" sin(\frac{1}{n}). Is sin(\frac{1}{n}) bigger than anything for every n?
    Thanks!
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  2. #2
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    Re: Limit of a sequence

    For small x, sinx is approximately equal to x. So as n increases sin(1/n) is approximately equal to 1/n
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  3. #3
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    Re: Limit of a sequence

    True, |sin(\frac{1}{n})|<|\frac{1}{n}| but this doesn't really help me. I need to get rid of sin(\frac{1}{n}) by showing that it's bigger than something else, not smaller, no?
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  4. #4
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    Re: Limit of a sequence

    I was thinking of replacing sin(1/n) by 1/n and then the limit is clearly infinite, but I suspect that may not do as a solution.
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  5. #5
    Junior Member ignite's Avatar
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    Re: Limit of a sequence

    Yes you can approximate sin(\frac{1}{n}) by \frac{1}{n} as n tends to infinity.There is nothing wrong in that.
    As an example to enlighten you,even (n+2)^2 can be approximated by n^2 as n tends to infinity.
    Last edited by ignite; April 29th 2012 at 05:01 AM.
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  6. #6
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    Re: Limit of a sequence

    Yes but how does that help me? I need to prove the limit by definition.
    For example, I can do this:

    (n+2)^2sin(\frac{1}{n}) > n^2sin(\frac{1}{n}) > nsin(\frac{1}{n}) > ...

    but then what?
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  7. #7
    Junior Member ignite's Avatar
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    Re: Limit of a sequence

    Quote Originally Posted by loui1410 View Post
    Yes but how does that help me? I need to prove the limit by definition.
    For example, I can do this:

    (n+2)^2sin(\frac{1}{n}) > n^2sin(\frac{1}{n}) > nsin(\frac{1}{n}) > ...

    but then what?
    (n+2)^2sin(\frac{1}{n}) > n^2sin(\frac{1}{n}) > n^2(1/n-1/6n^3) = n -1/6n

    Now given any M can you find your N?Think about it.
    Last edited by ignite; April 29th 2012 at 05:15 AM.
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  8. #8
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    Re: Limit of a sequence

    but how did you get to

    ... > n^2sin(\frac{1}{n}) > n^2(\frac{1}{n} - \frac{1}{6n^3}) > ... ?

    This is the move I'm looking for, but why is this true?
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  9. #9
    Junior Member ignite's Avatar
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    Re: Limit of a sequence

    There are many ways to prove it.But simplest is to look at the Taylor series exapansion of sin(x).
    sin(x)=x-x^3/3!+x^5/5!-....
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  10. #10
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    Re: Limit of a sequence

    We haven't learned Taylor series yet so I'm not supposed to use it, there must be another solution that they had in mind. Anyway, I'll ask the teaching assistant tomorrow for guidance and see what happens..
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