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Is anyone able to help me answer this question? Or point me in the right direction?

Use the squeezing theorem to find the limit of the sequence $\{a_n\}_{n=1}^{\infty}$ with $n$-th term $a_n=\dfrac{\cos(n)}{n!}$.

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$-1\leq\cos{n}\leq 1$ so $$\frac{-1}{n!}\leq\frac{\cos{n}}{n!}\leq\frac{1}{n!}$$ holds for every n, and then take the limit.

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  • $\begingroup$ $\cos n$ can be negative $\endgroup$ – ruler501 Mar 24 '14 at 4:30
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    $\begingroup$ Fixed that! thanks! $\endgroup$ – Stella Biderman Mar 24 '14 at 4:40
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-1 < cosn < 1 ==> -1/n! < cosn/n! < 1/n!. Next 0 < 1/n! < 1/n and clearly lim (1/n) = 0. So squeeze theorem gives: lim (1/n!) = 0. This means lim ( -1/n!) = 0. So squeeze theorem again says: lim cosn/n! = 0.

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