# limit of infinite sequence

• Dec 12th 2012, 04:44 AM
muddywaters
limit of infinite sequence
not sure to put this in pre calculus or calculus, sorry if mistaken

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

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

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

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
• Dec 12th 2012, 09:06 AM
hollywood
Re: limit of infinite sequence
Quote:

Originally Posted by muddywaters
since $\displaystyle \frac{\pi}{2}<\theta<\pi, 0<|cos\theta|<1$

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

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

as $\displaystyle 2\theta$ goes from $\displaystyle \pi$ to $\displaystyle 2\pi$, $\displaystyle \sin{\theta}$ goes from 0 to -1 and back to zero, but the point where it is -1, $\displaystyle 2\theta=\frac{3\pi}{2}$, is excluded so the absolute value is between 0 to 1.

Since $\displaystyle \sin{2\theta}$ and $\displaystyle \cos{\theta}$ 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
• Dec 12th 2012, 10:30 PM
muddywaters
Re: limit of infinite sequence
"as $\displaystyle \theta$goes from $\displaystyle \frac{\pi}{2}$ to $\displaystyle \pi$, $\displaystyle cos\theta$ 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 $\displaystyle 0<|cos\theta|<1$, which is the same as $\displaystyle -1<cos\theta<0, 0<cos\theta<1$, but what we have found so far is actually only $\displaystyle -1<cos\theta<0$, not $\displaystyle 0<cos\theta<1$ as well.
and the question gives $\displaystyle \theta\neq\frac{3}{4}\pi$, so taking that into account, $\displaystyle cos\theta\neq-\frac{\sqrt{2}}{2}$, and that's how i get $\displaystyle -1<cos\theta<-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}<cos\theta<0$ for the $\displaystyle cos\theta$part of the question.
• Dec 13th 2012, 11:00 AM
hollywood
Re: limit of infinite sequence
All of the things you say are true. But when we say that $\displaystyle 0<|\cos\theta|<1$, we're not saying that $\displaystyle \cos\theta$ takes on all possible values that fit the inequality. We're only saying that whatever values of $\displaystyle \cos\theta$ takes on, the inequality holds.

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

- Hollywood