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?


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 $$\forall n \in \mathbb{N},\ \sin \big( (n+1) \pi x \big) - \sin (n \pi x) = C \sin (n \pi x + \phi).$$
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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 $x \in \mathbb{Z}$.
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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 $[0,1]$.
