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?


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.
