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Math Help - limit of infinite sequence

  1. #1
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    limit of infinite sequence

    not sure to put this in pre calculus or calculus, sorry if mistaken

    Find the following limit.
    Given that  \frac{\pi}{2}<\theta<\pi where  \theta\neq\frac{3}{4}\pi ,
    find  \lim_{n \to \infty} (\frac{sin^n2\theta}{cos^n\theta^-3})

    the answer given is:
    since  \frac{\pi}{2}<\theta<\pi, 0<|cos\theta|<1
    since  \pi<2\theta<2\pi \and \2\theta\neq\frac{3}{2}\pi, 0<|sin2\theta|<1
    therefore, when  n \to \infty, sin^n2\theta \to 0 \and \cos^n\theta \to 0
    therefore  \lim_{n \to \infty} (\frac{sin^n2\theta}{cos^n\theta^-3})=0

    i don't understand how 0<|cos\theta|<1 and 0<|sin2\theta|<1 is found.
    what i find instead is -1<sin2\theta<0 and -1<cos\theta<\frac{-\sqrt{2}}{2}, \frac{-\sqrt{2}}{2}<cos\theta<0
    using that, i also find  \lim_{n \to \infty} (\frac{sin^n2\theta}{cos^n\theta^-3})=0

    but please help explain the steps, as the steps in the answers are pretty different from those that i use, im afraid that i've mistaken
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  2. #2
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    Re: limit of infinite sequence

    Quote Originally Posted by muddywaters View Post
    the answer given is:
    since  \frac{\pi}{2}<\theta<\pi, 0<|cos\theta|<1

    as \theta goes from \frac{\pi}{2} to \pi, \cos{\theta} goes from 0 to -1, so the absolute value goes from 0 to 1.

    since  \pi<2\theta<2\pi \and \2\theta\neq\frac{3}{2}\pi, 0<|sin2\theta|<1

    as 2\theta goes from \pi to 2\pi, \sin{\theta} goes from 0 to -1 and back to zero, but the point where it is -1, 2\theta=\frac{3\pi}{2}, is excluded so the absolute value is between 0 to 1.
    Since \sin{2\theta} and \cos{\theta} have absolute value strictly less than 1, their n-th powers go to zero as n goes to infinity. So the numerator goes to zero and the denominator goes to 0-3 = -3, and therefore the limit is zero.

    - Hollywood
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    Re: limit of infinite sequence

    "as  \theta goes from  \frac{\pi}{2} to \pi, cos\theta goes from 0 to -1, so the absolute value goes from 0 to 1"
    that's how the answer says, but how can we say that 'so the absolute value goes from 0 to 1'? that would be  0<|cos\theta|<1 , which is the same as -1<cos\theta<0, 0<cos\theta<1, but what we have found so far is actually only -1<cos\theta<0, not  0<cos\theta<1 as well.
    and the question gives \theta\neq\frac{3}{4}\pi, so taking that into account, cos\theta\neq-\frac{\sqrt{2}}{2}, and that's how i get  -1<cos\theta<-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}<cos\theta<0 for the cos\thetapart of the question.
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    Re: limit of infinite sequence

    All of the things you say are true. But when we say that 0<|\cos\theta|<1, we're not saying that \cos\theta takes on all possible values that fit the inequality. We're only saying that whatever values of \cos\theta takes on, the inequality holds.

    So for every value of \theta, \frac{\pi}{2}<\theta<\pi, it is true that 0<|\cos{\theta}|<1, even though, as you said, there are other values of \cos{\theta} which would also make the statement true.

    - Hollywood
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