# Limit of $(2^n)\sin(n)$ as $n$ goes to infinity

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.

Thank you for taking a look at this problem.

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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 '14 at 9:30
Limit does not exist – Hamou Aug 12 '14 at 9:30
Thanks guys. That should do it. +1's all around – Travis Aug 12 '14 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 '14 at 10:01

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
$2^n\sin n$ 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.