# 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:

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