Prove $\sin \left(1/n\right)$ tends to $0$ as $n$ tends to infinity. I'm sure there is an easy solution to this but my mind has gone blank! Any help on proving that $\sin( \frac{1}{n})\longrightarrow0$ as $n\longrightarrow\infty$ would be much appreciated.
This question was set on a course before continuity was introduced, just using basic sequence facts. I should have phrased it as:
Find an $N$ such that for all $n > N$ $|\sin(1/n)| < \varepsilon$

Hint: $\sin$ is a continuous function.
Edit: Since $\dfrac{1}{n}\longrightarrow 0$ the statement below is true: $$(\forall \delta>0)(\exists M\in \Bbb N)(\forall n\in \Bbb N)\left(n\ge M\implies \left\vert \dfrac{1}{n}\right\vert<\delta\right)$$
Using $M$ from above and the fact that $\left \vert \sin\left(\dfrac{1}{n}\right)\right \vert<\left \vert \dfrac{1}{n}\right \vert$ for all $n\in \Bbb N$ try to find a suitable $N$.
