What is the limit inferior and limit superior of sequence,$x_n=\left(1-\dfrac{1}{n}\right)\sin \left(\dfrac{n\pi}{3}\right),n\geq1$?

Question

Let,$x_n=(1-\frac{1}{n}) \sin(\frac{n\pi}{3}),n\geq1$.Then what is the limit inferior and limit superior if this sequence?

I know that the limit inferior of $x_n$ and limit superior of $x_n$ are respectively the smallest and the largest accumulation point.

Writing the sequence in the explicit form,

{$x_1$,$x_2$,,$x_3$,$x_4$,$x_5$,$x_6$,$x_7$,...,...}

={$(1-\frac{1}{1}) \sin(\frac{1\pi}{3})$,$(1-\frac{1}{2}) \sin(\frac{2\pi}{3})$,$(1-\frac{1}{3}) \sin(\frac{3\pi}{3})$,$(1-\frac{1}{4}) \sin(\frac{4\pi}{3})$,$(1-\frac{1}{5}) \sin(\frac{5\pi}{3})$,$(1-\frac{1}{6}) \sin(\frac{6\pi}{3})$,$(1-\frac{1}{7}) \sin(\frac{7\pi}{3})$}={$0,\frac{1}{4}$,$(1-\frac{1}{n}) \sin(\frac{n\pi}{3}),...,...$}

={$0,\frac{1}{4}$,$\frac{-2}{3}$,$\frac{-3\sqrt3}{8}$,$\frac{-2\sqrt3}{5}$,$0$,$\frac{3\sqrt3}{7}$,$\frac{-7\sqrt3}{16}$,...,...}

Now i collected the subsequences;{$x_1$,$x_6$,$x_{11}$,$x_{16}$,$x_{17},,,$}={$0,0,0,0,...,...,...$}.

I got struck here.I'm unable to select other suitable subsequences of ($x_n$) by which we can choose the smallest and the largest accumulation point of the

Answer 1

Check that

$$\sin\frac{n\pi}3=\begin{cases}&0,&n=0,\,3\pmod 6\\{}\\&\cfrac{\sqrt3}2,&n=1,\,2\pmod3\\{}\\&-\cfrac{\sqrt3}2,&n=4,\,5\pmod 6\end{cases}$$

IT may also be worthwhile to take into account that $\;\left\{\left(1-\frac1n\right)\right\}\;$ is a descending sequence...