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Math Help - problem of limit

  1. #1
    kin
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    problem of limit

    i try L'Hopital at the first glance,

    but i make it more complicated....

    \lim_{n\to\infty}(2x -sin 2x)/(x (sinx)^2)
    Last edited by kin; September 23rd 2009 at 08:59 PM.
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  2. #2
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    Quote Originally Posted by kin View Post
    i try L'Hopital at the first glance,

    but i make it more complicated....

    \lim_{n\to\infty}(2x -sin 2x)/(x (sinx)^2)
    It probably requires three applications of l'Hopital's Rule. Personally, I'd substitute the power series for sin(x) and sin(2x), simplify and then take the limit.
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  3. #3
    kin
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    Quote Originally Posted by mr fantastic View Post
    It probably requires three applications of l'Hopital's Rule. Personally, I'd substitute the power series for sin(x) and sin(2x), simplify and then take the limit.
    it doesnt work...........
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  4. #4
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    Quote Originally Posted by kin View Post
    it doesnt work...........
    Sorry, I misread the limit as x --> 0.

    Are you sure it's x \rightarrow + \infty? (By the way, you have {\color{red}n} \rightarrow + \infty .... ) Because I doubt the limit exists if that's the case.
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  5. #5
    kin
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    Quote Originally Posted by mr fantastic View Post
    Sorry, I misread the limit as x --> 0.

    Are you sure it's x \rightarrow + \infty? (By the way, you have {\color{red}n} \rightarrow + \infty .... ) Because I doubt the limit exists if that's the case.

    sorry, it should be x -->infinity

    there are two cases:

    1) x tends to 0: my answer = 4/3 ( am i correct?)

    2) x tends to infinity: i hv no idea..
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  6. #6
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    Quote Originally Posted by kin View Post
    there are two cases:

    1)limit tends to 0: my answer = 4/3 ( am i correct?) Mr F says: Yes.

    2)limit tends to infinity: i hv no idea.. Mr F says: I doubt the limit exists.
    ..
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