finding the limits for a sin function: u_n = sin (n*pi)/4
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Any help finding the limit of a sin function?
I have an exercise like this http://prntscr.com/d1b66x and I have no idea how to seqence it, u8k i mean how it oscillates, there are 5 limits to it. Also I have the same issue for cos function. http://prntscr.com/d1b7wm
I have an exercise like this http://prntscr.com/d1b66x and I have no idea how to seqence it, u8k i mean how it oscillates, there are 5 limits to it. Also I have the same issue for cos function. http://prntscr.com/d1b7wm
Okay, so if I'm understanding you correctly, the two problems you're attempting to solve are something like the following:
\(\displaystyle \displaystyle \lim _{n\to \infty }\left(sin\left(\frac{n\pi }{4}\right)\right)\)
\(\displaystyle \displaystyle \lim _{n\to \infty }\left(-cos\left(\frac{n\pi }{2}\right)\right)\)
If the above is not correct, please reply with the necessary corrections, ideally with the full and exact text of the exercise, quoted word-for-word. Also, regardless of if the above is the correct problem statement or not, please include with your reply any and all work you've done, even if you know it's wrong. Thank you.
\(\displaystyle \displaystyle \lim _{n\to \infty }\left(sin\left(\frac{n\pi }{4}\right)\right)\)
\(\displaystyle \displaystyle \lim _{n\to \infty }\left(-cos\left(\frac{n\pi }{2}\right)\right)\)
If the above is not correct, please reply with the necessary corrections, ideally with the full and exact text of the exercise, quoted word-for-word. Also, regardless of if the above is the correct problem statement or not, please include with your reply any and all work you've done, even if you know it's wrong. Thank you.
Steven G
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- Dec 30, 2014
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If in fact there are multiple limits then the limit does not exact.
Assuming you mean find the limit (accumulation) points of something like the sequence
an = tan(\(\displaystyle n\, \frac{\pi}{b}\))
where b is an integer, we first start out with the period of the tangent function which is \(\displaystyle \pi\).
That is, the number of accumulation points might be k=1, k=2, k=3, ..., k=b because when k=b, we have completed a complete cycle of the tangent function. That is, let n = mb+k and let m go to infinity so that we have
tan[\(\displaystyle (mb + k) \frac{\pi}{b}\)] = tan[mb*\(\displaystyle \frac{\pi}{b}\)] + tan[\(\displaystyle k\frac{\pi}{b}\)] = tan[\(\displaystyle m\,\pi\)] + tan[\(\displaystyle k\frac{\pi}{b}\)] = tan[\(\displaystyle k\frac{\pi}{b}\)]
EDIT: Of course, for different trig functions, you need to use the proper cycle and, for all such problems, some of those accumulation points may be the same (such as in your problem).
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