limit of infinite sequence

**muddywaters** · December 12th 2012, 04:44 AM

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

**hollywood** · December 12th 2012, 09:06 AM

Originally Posted by muddywaters

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

**muddywaters** · December 12th 2012, 10:30 PM

"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 $cos\theta$ part of the question.

**hollywood** · December 13th 2012, 11:00 AM

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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