Answer
See proof below.
Work Step by Step
If we are given an $\epsilon\gt 0$,
we want to find an N such that, for $n\gt N,$
$|\displaystyle \frac{\sin n}{n}-0|\lt \epsilon.$
\begin{aligned}
|\displaystyle \frac{\sin n}{n}-0|&=\displaystyle \frac{|\sin n|}{n} \\
& \displaystyle \leq\frac{1}{n}\leq\epsilon\\
& \text{... so, we take}\\
N&\displaystyle \gt\frac{1}{\epsilon} \end{aligned}
Since we can find N= any integer greater than $\displaystyle \frac{1}{\epsilon}$,
by the definition, $\displaystyle \lim_{n\rightarrow\infty}\frac{\sin n}{n}=0$
