Show that $\lim_{n\to \infty}\sin{n\pi x} =0$ if $x\in \mathbb{Z},$ but the limit fails to exist if $x\notin \mathbb{Z}.$ Show that $\lim_{n\to \infty}\sin{n\pi x} =0$ if $x\in \mathbb{Z},$ but the limit fails to exist if  $x\notin \mathbb{Z}.$
1st part
If $x\in \mathbb{Z}$ then $\sin{n\pi x}=0$ for all $n,$ giving the first part.
Edit:

2nd part
If $x\notin \mathbb{Z},$
I want to show that the limit doesn't exist.
How to do that?


Fix $0<x<1.$ Define $A$ to be the closed arc on the unit circle centerd at $(0,1),$ having arc length $\pi x.$ Good to draw a picture here, because it makes things transparent. A little geometry shows that if $e^{it}\in A,$ then $\sin t \ge \sin [(\pi/2)(1-x)] >0.$
Claim: $e^{in\pi x}\in A$ for infinitely many $n.$
Proof of claim: The points $e^{in\pi x}, n=1,2,\dots$ travel around the circle infinitely many times, in steps of arc length $\pi x.$ The closed arc $A$ has arc length $\pi x.$ Thus there is no way $e^{in \pi x}$ can "jump over" $A$ in this process, so it will land in $A$ at least once in every orbit of the circle. That proves the claim.
The claim implies $\sin (n\pi x) \ge \sin [(\pi/2)(1-x)]$ for infinitely many $n.$
But exactly the same kind of reasoning applies to the arc $B,$ the closed arc on the unit circle centerd at $(0,-1),$ having arc length $\pi x.$ The conclusion of the claim holds for $B,$ hence $\sin (n\pi x) \le - \sin [(\pi/2)(1-x)]$ for infinitely many $n.$
Now in general, if $a>0$ and $y_n$ is a sequence such that $y_n\ge a$ for infinitely many $n,$  and $y_n \le -a$ for infinitely many $n,$ then $y_n$ cannot converge. Since that's the situation with $\sin (n\pi x)$ here, we see $\sin (n\pi x)$ diverges.
Now $0<x<1$ above, but $\pi$-periodicity, or negative $\pi$-periodicity, shows the same result holds if $x\in (m,m+1)$ for any $m\in \mathbb Z.$ Thus $\sin (n\pi x)$ diverges for all $x\notin \mathbb Z.$
