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I'm stuck with the limit $\lim_{n\to\infty} (2^n)\sin(n) $. I've been trying the squeeze theorem but it doesn't seem to work. I can't think of a second way to tackle the problem. Any push in the right direction would be much appreciated.

Also, please don't just post the answer up because I want to try and get it.

Here's what I've got so far:

$$ \lim_{n \to \infty} (2^n)\sin(n) $$

So I know, $\ sin(n) $ is bound by -1 and 1, but multiplying the inequality by $\ 2^n $ will give me a negative and positive $\ 2^n $. So I am stuck here. This would mean that the function is bounded by limits that tend to negative and positive infinity, pretty useless.

So can I take the absolute value of each side of the inequality? Like this:

$$ \lvert -2^n\rvert \le \lvert 2^n \sin(n) \rvert \le \lvert 2^n\rvert $

If this works, I can say it tends to infinity, but it seems a bit dodgy to me.

Thanks for taking a look.

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Every interval $[n\pi/2-1/2, n\pi/2+1/2]$, $n$ odd contains an integer. Use this to show some subsequence tends to $\infty$ and some subsequence tends to $-\infty$. –  David Mitra Aug 12 at 9:30
1  
Limit does not exist –  Hamou Aug 12 at 9:30
    
Thanks guys. That should do it. +1's all around –  Travis Aug 12 at 9:32
    
Questions should be be marked as solved by editing the title, but rather by accepting an answer. (Admittedly, David Mitra's comment may be more informative than Graham Kemp's answer) –  Hagen von Eitzen Aug 12 at 10:01

2 Answers 2

up vote 1 down vote accepted

Every interval $[n\pi/2-1/2, n\pi/2+1/2]$, $n$ odd contains an integer. Use this to show some subsequence tends to $\infty$ and some subsequence tends to $-\infty$. -- David Mitra

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No.   You cannot take the absolute value when you are looking for the limit.   It's not a valid concept.

$2^n\sin n$ oscillates periodically between $-2^n$ and $+2^n$ as $n$ tends to infinitude.   It does not converge on a limit.   It does not even tend towards infinitude.   No limit exists.

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