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Math Help - Limits

  1. #1
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    please help -urgent

    Find the trigonometric limit: lim (as x approaches infinity) (x-tan 2x)/sin 2x
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  2. #2
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    We have \frac{x - \tan 2x}{\sin 2x} = \frac{x}{\sin 2x} - \frac{1}{\cos 2x}. Now consider the subsequences x = 2\pi n + \frac\pi2 and x = 2\pi n + \pi + \frac\pi2 to see that there is no limit.
    Last edited by rgep; December 30th 2005 at 11:15 PM. Reason: Oops - misread original post
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by rgep
    We have \frac{x - \tan 2x}{\sin 2x} = \frac{x}{\sin 2x} - \frac{1}{\cos 2x}. Now consider the subsequences x = 2\pi n + \frac\pi2 and x = 2\pi n + \pi + \frac\pi2 to see that there is no limit.
    Surly; it is sufficient to consider x=n.\pi for n=1,2,.. to show that there can be no finite limit,
    and x=\pi/4+n.\pi for n=1,2,.. to show that it does not go to +\infty?

    RonL
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  4. #4
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    Quote Originally Posted by bobby77
    Find the trigonometric limit: lim (as x approaches infinity) (x-tan 2x)/sin 2x
    Are you sure that it is not as x approaches zero?

    If it is as x approaches infinity, then what is tan(infinity)? Or, sin(infinity)?

    If x approaches zero, limit is (-1/2).
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