Prove a limit equals to $0$

Hint: $\left|\sin\left(\dfrac{n\pi}{3}\right)\right|\le 1$.

The Archimedean property of the real numbers states that for all $\epsilon > 0$ one can always find $N \in \mathbb{N}$ so that $\dfrac{1}{N} < \epsilon$. Now, since $| \sin \alpha | \leq 1 $ for all $\alpha$, then it must be the case that

$$ \left| \dfrac{1}{n} \sin \left( \dfrac{ n \pi }{3} \right) \right| \leq \dfrac{1}{n} $$

Now, Let $\epsilon > 0$ be arbitrary. Choose $N \in \mathbb{N}$ so that $\dfrac{1}{N} < \epsilon$. Therefore, for all $n > N$, one has

$$ \left| \dfrac{1}{n} \sin \left( \dfrac{ n \pi }{3} \right) \right| \leq \dfrac{1}{n} < \dfrac{1}{N} < \epsilon $$

In other words, $\lim \dfrac{1}{n} \sin \left( \dfrac{ n \pi }{3} \right) = 0$

$\textbf{Comments:}$

It is actually easier to use the squeeze rule to check this limit. This is a special case of the squeeze theorem. One can always rely on the definition and the archimedean property, but it is an o-v-e-r-k-i-l-l.