What is the limit of $n\sin \left(\frac{1}{n}\right)$ as $n \to \infty$

Question

The textbook I'm reading gives no proof of the limit being 1 apart from a flimsy 0/0, and has covered only the "epsilon" and sandwich test for convergence to a particular limit.

Do not rearrange to turn $n$ into $\frac{1}{n}$, that is the source of the flimsy $0/0$ argument.

Answer 1

Hint: $$\lim_{x\to 0}\frac{\sin x}x=1.$$

Answer 2

Hint: $\sin \left(\dfrac{1}{n}\right) \approx_{0} \dfrac{1}{n} - \dfrac{1}{2!}\cdot \dfrac{1}{n^2} + \dfrac{1}{3!}\cdot \dfrac{1}{n^3}- ......\Rightarrow n\sin \left(\frac{1}{n}\right) \approx_0 1 - \dfrac{1}{2n} + \dfrac{1}{6n^2} - ....--$