If $ x_{n}$ is unbounded sequence then the $\lim_{n \to \infty}\sin(\frac{1}{x_{n}})=0$

Question

I tried to say that if $x_{n}$ is unbounded then it is divergent...

Consider the three possible ways for $x_n$ to diverge:

1)$x_{n} \to \infty$.

2)$x_{n} \to -\infty$.

3)$x_{n}$ has multiple limit points.

I thought in either cases (1) or (2), the statement would be true.

In case (3) I took the following sequence:

$x_{n}=\begin {cases} n & n \text{ is even} \\ 1 & n \text{ is odd} \end{cases}$

then $x_{n}$ is unbounded and the statement would be false in this case.