Find the subsequential limit points of $x_n$. 
Let $x_n=(1+\frac{1}{n})\sin \frac{n\pi}{3}$ for $n\ge 1$. Find the subsequential limit points of $x_n$.

My effort:
On finding the first few terms of the sequence I found that the values of the sequence shows some repeat of the terms.
The three sub-sequences that come up are $\{(1+\frac{1}{n})\frac{\sqrt 3}{2}\},\{0\},\{(1+\frac{1}{n})\frac{-\sqrt 3}{2}\}$
Hence the limits are $\{\frac{\sqrt 3}{2},-\frac{\sqrt 3}{2},0\}$
But I will have to show there are no other subsequential limits.
If $a$ be another one then there exists a subsequence $x_{n_k}\to a \implies (1+\frac{1}{n_k})\sin \frac{n_k\pi}{3}\to a$
But I am unable to complete the proof that $a\in $$\{\frac{\sqrt 3}{2},-\frac{\sqrt 3}{2},0\}$.
Please give some hints.
 A: First you can argue that the term $(1+\frac{1}{n})$ can be disconsidered , since it converges to $1$. Now look at the term $\sin\frac{n\pi}{3}$.
If we have a subsequence $(n_k)$, then there are infinitely many $n_k$'s, but only finitely many possible values for $\sin\frac{n_k\pi}{3}$, namely $\pm\frac{\sqrt{3}}{2}$, $\pm\frac{1}{2}$ and $0$, so at least one of these values will be attained infinitely many times.
Now show that all these values are indeed subsequential limits.
A: Hint: Your working for the first bit is correct and you've identified $3$ limits. To continue, notice that this problem is the same as finding the convergent subsequences to $y_n = \sin(\frac{n\pi}{3})$ (why?). To find out if there are anymore, think about what values $\sin(\frac{n\pi}{3})$ can take. 
Once you've thought about that, think about sequences that only take finitely many values. If such a sequence converges, what eventual properties do we know about the sequence? What happens if we assume the such a sequence isn't eventually constant?
