# Subsequential limit of sequence

I'm trying to determine all subsequential limit points of the following sequence

X_n = cos(n)

Not sure how to decompose this into subsequences.

Anyone know how to tackle this problem?

Thanks!

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It is rather well known that the image of the integers under sine and cosine is dense in $[-1, 1]$. For a reference, see the related question here.
With the knowledge that $\cos(n)$ is dense in $[-1, 1]$, we shall show that any $x\in[-1,1]$ is a sub-sequential limit point. Indeed, given any $\epsilon > 0$ and some $n_0 > 0$, we can find $n > n_0$ such that $$\cos(n) \in (x - \epsilon, x + \epsilon)$$ so we can iteratively pick a sub-sequence that lies within this $\epsilon$-neighborhood. This shows that any point is a sub-sequential limit point of the sequence.
It means that every point in $[0,1]$ is a sub-sequential limit point. I will make an edit to make this more explicit. – EuYu Aug 28 '12 at 0:50
My mistake, it should be $[-1, 1]$. For some reason I was interpreting the sequence as positive. The argument still holds though. Every point in $[-1, 1]$ is a limit point of the sequence. – EuYu Aug 28 '12 at 1:00