Solution: Note that for every natural number $n$, we have \[ \left\vert \sin \frac{1}{n} \right\vert \le 1. \] So, \begin{align*} -1 \le \sin \frac{1}{n} \leq 1 & \implies -\frac{1}{n} \leq \frac{1}{n}\sin \frac{1}{n} \leq \frac{1}{n}. \end{align*}

Here we have, \[ a_n = -\frac{1}{n},~b_n = \frac{1}{n}\sin \frac{1}{n}\text{ and } c_n = \frac{1}{n}. \] We also have \[ \lim_{n \to \infty} a_n = \lim_{n \to \infty} c_n = 0, \] hence using the Sandwich theorem, we conclude that \[ \lim_{n \to \infty} \frac{1}{n}\sin \left( \frac{1}{n} \right) =0. \] Thus, the limit point of the given set is $\{0\}$.