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?

 A: If $x = \frac{p}{q} \in \mathbb{Q} \setminus \mathbb{Z}$
Let $k$ be an integer.


*

*If $n = 2qk$, then $\sin(n \pi x) = \sin(2\pi pk) = 0$.

*If $n = 2qk+1$, then $\sin(n \pi x) = \sin(2\pi pk+\pi x) = \sin(\pi x) \neq 0$


So the sequence $n \to \sin(n \pi x)$ contains two different numbers infinitely often, so it has no limit.
If $x \in \mathbb{R} \setminus \mathbb{Q}$
For convenience we define the irrational number $y = x/2$.
The sequence $a_n = ny \operatorname{mod} 1$ is dense on $[0, 1)$. 
(The equidistribution theorem gives the stronger result that it's uniformly distributed on that interval, but we don't need that for our proof) 
This implies that for every $n_0 \in \mathbb{N}$ and $0 \le a < b \le 1$, you can find an $n>n_0$ such that $a_n \in (a, b)$. (because otherwise, there would only be a finite number of elements in the interval $(a, b)$, and then it wouldn't be dense)
The idea of this proof is that for every $n_0 \in \mathbb{N}$, we can always find an $n > n_0$ such that $\sin(n \pi x) > c_1$, and we can also find an $n > n_0$ such that $\sin(n \pi x) < c_2$, where $c_1 > c_2$.
We take $c_1 = \sin(2 \pi \cdot 0.1)$ and $c_2 = \sin(2 \pi \cdot 0.6)$.
Given an $n_0$, we can find an $n > n_0$ such that $a_n \in (0.1, 0.2)$. We have:
$$n \pi x \operatorname{mod} 2\pi = 2\pi n y \operatorname{mod} 2\pi =
2\pi(n y \operatorname{mod} 1) =
2\pi a_n \\
\sin(n \pi x) = \sin(2\pi a_n) > c_1
$$
Similarly, for every $n_0$ we can find an $n > n_0$ such that $a_n \in (0.6, 0.7)$ and
$$\sin(n \pi x) < c_2$$
So, the sequence $n \to \sin(n \pi x)$ has no limit.
A: Take $x=1/2$, then
$$
\sin(n\pi x)=\left\{
\begin{array}{rlc}
0 & \text{if $n$ is even},\\
1 & \text{if $n=4k+1$ for some $k\in\mathbb N$,} \\
-1 & \text{if $n=4k+3$ for some $k\in\mathbb N$.}
\end{array}
\right.
$$
Thus the sequence $\sin(n\pi x)$ DOES NOT converge, even if $x$ is rational!
A: For $x \in \mathbb{R}$, for $n \in \mathbb{N}$, we have $$\sin ((n+1) \pi x) - \sin (n \pi x) = \sin (n \pi x) \big( \cos(\pi x) - 1) + \cos(n \pi x)\sin(\pi x).$$
Denote $A = \cos(\pi x)-1$ and $B = \sin(\pi x)$. We have $A \neq 0$ and $B \neq 0$, and $$\sin \big( (n+1) \pi x \big) - \sin (n \pi x) = A \sin(n \pi x) + B \cos(n \pi x).$$
Now denote $C = \sqrt{A^2+B^2}$ ; classically, there exists $\phi \in \mathbb{R}$ such that $$\forall n \in \mathbb{N},\ \sin \big( (n+1) \pi x \big) - \sin (n \pi x) = C \sin (n \pi x + \phi).$$
$ $
Now we assume that $\lim \limits_{n \to +\infty} \sin(n \pi x)$ exists. Thus $C \sin (n \pi x + \phi) \underset{n \to +\infty}{\longrightarrow} 0$. Then, you can find here a short proof that $$\forall y \in \mathbb{R},\ |\sin (y)| \ge \frac{2}{\pi}d(y,\pi \mathbb{Z})$$ where $d(t,A) = \inf \{ |t-a|,\ a\in A\}$ stands for the distance to the set $A$.
As $C > 0$, we have that $d(n\pi x + \phi, \pi \mathbb{Z}) \underset{n \to +\infty}{\longrightarrow} 0$. Using the continuity of the distance yields $d\big( ((n+1)\pi x + \phi)-(n\pi x + \phi), \pi \mathbb{Z} \big) \underset{n \to +\infty}{\longrightarrow} 0$, i.e. $d(\pi x, \mathbb{Z}) \underset{n \to +\infty}{\longrightarrow} 0$, so $d(\pi x, \pi \mathbb{Z}) = 0$. As $\pi \mathbb{Z}$ is closed in $\mathbb{R}$, we conclude $\pi x \in \pi \mathbb{Z}$, and thus $x \in \mathbb{Z}$.
$ $
Hence, if $\lim \limits_{n \to +\infty} \sin(n \pi x)$ exists, then $x \in \mathbb{Z}$ (and thus, if $x$ is not an integer, then $\big( \sin (n \pi x) \big)_{n \ge 0}$ does not converge).
Note that if $x$ is irrational, you can even prove that $\big( \sin (n \pi x) \big)_{n \ge 0}$ is dense in $[0,1]$.
A: Hint: Suppose the limit exists, fix its value as $f(x)$, then use the formal definition of limit and find two values of $n$ that satisfy the definition of limit buthave a different enough value.
A: 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 $x\notin \mathbb Z.$
A: If x is a rational the sequence rotation $n\pi x$ radians of the unit circle will form an equidistributed (see equidistribution theorems) periodic orbit. The length of the orbit will be exactly two times the denominator. Hence will form a periodic orbit of order 2 if x is an integer, luckily these two positions both give sin=0. If x is irrational, then notice $xn\pi$ is dense and topologically transitive to use some terminology from dynamical systems, hence will never stabilize.
