Show that $\lim_{n\to \infty}\sin{n\pi x} =0$ if $x\in \mathbb{Z},$ but the limit fails to exist if $x\notin \mathbb{Z}.$

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.

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!