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.
Best Answer
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