Limit point of sin function in real analysis If a set $S= \sin(n\pi/2)$ and $n$ is Natural number, then what would be its limit point. I assume that it's limit point should be Null set because this set is finite set having only $3$ real values {$-1,0,1$}. Can any one bags to differ, because answer seems to be that these $3$ points are limit points.

Your answer is right. Consider $a_n=\sin\left(\frac{n \pi}{2}\right) $ for $n=k$ ,for $n=k+1$ , for $n=k+2$  and for $n=k+3$. Then you split your sequence into 3 subsequences which one of them converges to 1,0,-1 respectively. In this kind of problems you need to take the subsequences $mod 2\pi $  in order to find alla limit points. For example if you have $b_n=\cos(n \pi / 6 ) $ you have to consider 12 subsequences in order to cover all the cases . On the other hand if you have $c_n=\sin(n) , n\in \mathbb{N}$  then this is a real problem. Someone needs to use special techniques to show that every point of $[-1,1]$ is a limit point of $c_n$ 
