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 $n > N$ $|\sin(1/n)| < \varepsilon$
 A: $$\lim\limits_{n\to\infty}\sin\left(\frac{1}{n}\right)=\lim\limits_{n\to\infty}\frac{\sin\left(\frac{1}{n}\right)}{\left(\frac{1}{n}\right)}\left(\frac{1}{n}\right)=\left(\lim\limits_{n\to\infty}\frac{\sin\left(\frac{1}{n}\right)}{\left(\frac{1}{n}\right)}\right)\left(\lim\limits_{n\to\infty}\frac{1}{n}\right)=1\times0=0$$
A: Because $\sin$ is continuous, we can take:
$$\lim\limits_{n\to\infty}\sin\left(\frac{1}{n}\right)=\sin\left(\lim\limits_{n\to\infty}\frac{1}{n}\right)=\sin{(0)}=0$$
To answer the new question:
For each $\epsilon>0$ we need to find how big $n$ (for $N>0$, $n>N$) must be to ensure that $|\sin\left(\frac{1}{n}\right)-0|<\epsilon$. Thus we have:
$$|\sin\left(\frac{1}{n}\right)|<\epsilon$$
Let's only consider positive instances of $\sin1/n$ as negative ones will always be less than $\epsilon$. Then:
$$\frac{1}{\arcsin\epsilon}<n$$
So we put $N=1/\arcsin\epsilon$. 
Then, formally:
Let $\epsilon>0$ and let $N=1/\arcsin\epsilon$. Then $n>N$ implies $n>1/\arcsin\epsilon$, hence $\arcsin\epsilon>\frac{1}{n}$, hence $\epsilon>\sin1/n$, hence $|\sin1/n|<\epsilon$. Finally, we have $|\sin1/n-0|<\epsilon$ which proves that the limit is 0.
A: Hint
$$\quad0 \leq\sin\left(\frac{1}{n}\right)\leq \frac{1}{n}$$
A: 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 $N$.
A: Consider the Taylor series expansion of $\sin(1/n)$, then
$$ \frac{1}{n}- \frac{1}{n^3} \leq\sin \left(\frac{1}{n}\right) \leq \frac{1}{n}. $$ 
