Prove a limit equals to $0$

I need to prove that $$\lim_{n\to\infty }{\frac 1n}\sin\left(\frac {n\pi}3\right) = 0.$$ This is what I did: $$\left|{\frac 1n}\sin{\frac {n\pi}3}-0\right|<\epsilon,$$ $$|n|>{\frac {\sin{\frac {n\pi}3}}{\epsilon}},$$ $$|n|>0,$$ but I think that my answer is not correct. Would you tell me whether my answer is right or not?

• product of a null sequence with a bounded sequence is a null sequence. Commented Jan 2, 2022 at 2:59

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.