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$\lim_{n\to\infty} \dfrac{\ 1+|\sin n|}{2n}$.
I tried this: \begin{align*} &-1\le \sin n \le 1,\\ &0\le |\sin n |\le 1,\\ &1\le 1+|\sin n|\le 2,\\ &\frac1{2n}\le \frac{1+|\sin n|}{2n} \le \frac2{2n}. \end{align*} Is the way I'm using squeeze theorem correct?

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  • $\begingroup$ Impossible to tell from what you've written. So, what is the limit, and why? $\endgroup$ – Andrés E. Caicedo Oct 6 '13 at 7:06
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    $\begingroup$ So far so good. It would be cleaner to use $0\le \frac{1+|\sin n|}{2n}\le \frac{2}{2n}$. Then you need to come to a conclusion. $\endgroup$ – André Nicolas Oct 6 '13 at 7:08
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Since $\lim \frac{C}{n} = 0 $ for constant $C$, then the limit of your sequence must go to zero by the squeeze theorem.

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