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Let $ x_n=(1-\frac{1}{n})\sin (\frac{n\pi}{3})$ for n $\ge $1. denote l=limit inferior and s=limite superior Then

  1. -$\sqrt{3}/2\le l\lt s \le \sqrt{3}/2$

  2. -$1/2\le l\lt s \le 1/2$

  3. $l=-1 $,$ s=1$

  4. $l=0=s=0$

My attempt: I manually plotted the sequence terms for n=1,2,... for few n. And then saw that this sequence has 3 limit points namely -$\sqrt{3}/2$, 0 , $\sqrt{3}/2$. So least one is limit inf , largest of limit point is limit sup. But I need suggestions for more efficient way. Thanks

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Did you only see that the sequence has three limit points or did you actually prove that it is so?

Hinit: First, simplify the expression $\sin\left(\frac{n\pi}{3}\right)$. Then, it will be much easier to

  1. Find the three subsequences converging to the three convergence points.
  2. Show that these are the only three convergence points.
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  • $\begingroup$ I only saw it, by plotting values of this sequence on paper $\endgroup$ – BigBang Feb 5 '15 at 8:05
  • $\begingroup$ Plz give some himt on simplyfying sin term $\endgroup$ – BigBang Feb 5 '15 at 8:07
  • $\begingroup$ @BigBang Calculate the term for $n=1,2,3,4,5$. Can you see a pattern? $\endgroup$ – 5xum Feb 5 '15 at 9:53

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