Evaluating $\lim_{n\to \infty} \sin(\pi n x)$ I have the sequence $f_n = \sin (\pi n x)$ and I want to calculate the limit of this sequence, but I struggle with it.
$\lim_{n\to \infty} \sin(\pi) = 0$
$\lim_{n\to \infty}\sin(\pi n) = 0$ 
but if $x$ isn't an integer $\lim_{n\to \infty}\sin (\pi n x)$ doesn't have to be $0$. So $f_n$ just diverges or should I do a discussion? I don't know. 
 A: It seems that you have some amount of mathematical competency so would like to leave the following hint,
$$ \sin(\pi n x) = \sin\left(\pi n \left(\lfloor x\rfloor + x_{frac} \right) \right) $$
Where $x_{frac}$ is the fractional part of $x$ and $\lfloor x\rfloor$ is the greatest integer less than equal to $x$.
Let me know if you want the complete solution(I would have to work behind it as well).
edit: The main idea is that you want to break it into parts that is a multiple of $\pi$ and something that is less than $\pi$. After doing so you can focus solely on the later part.
A: Assume $x\in (0,1).$ Let $A$ be the arc $\{e^{it}: t\in [\pi/2-\pi (x/2), \pi/2+\pi (x/2)]\}.$ Note $A$ is centered at $i$ and has length equal to $\pi x.$ It's good to draw a picture and think of the situation geometrically.
Now the points $e^{in\pi x}, n = 1,2,\dots,$ march around the unit in steps of arc length $\pi x.$ Because the length of $A$ is exactly the same as the length of any step, at least once in every orbit of the circle we must have $e^{in\pi x}\in A$ for some $n.$ It follows that $e^{in\pi x}\in A$ for infinitely many $n.$
With exactly the same argument, you can see this is also true of the arc $A'$ of the same length centered at $-i.$ It follows that $\sin(n\pi x)\ge \sin(\pi/2-\pi (x/2))$ for infinitely many $n,$ and $\sin(n\pi x)\le -\sin(\pi/2-\pi (x/2))$ for infinitely many $n.$
This shows that the full sequence $\sin(n\pi x)$ diverges. Again, this was for $0<x<1.$ But using a little bit of symmetry and $2\pi$-periodicity shows $\sin(n\pi x)$ diverges for every $x\in \mathbb R \setminus \mathbb Z.$  
