As we all know, $\lim_{n\to \infty}\frac{\sin n}{n}=0$. But recently I found that one of my student calculate this limit like this, and I have difficulty in judging if it is right: \begin{align}\tag{*} \lim_{n\to\infty}\frac{\sin n}{n}\overset{\frac{1}{n}=t}{=\mkern-3mu=}\lim_{t\to0}t\sin\frac{1}{t}=0, \end{align} assuming that my students have studied and all known that $\lim_{t\to0} t\sin \frac{1}{t}=0.$
I know that from $\lim_{t\to0} t\sin \frac{1}{t}=0$ and E.Heine's result, $\lim_{n\to \infty}\frac{\sin n}{n}=0$. But the solution $(*)$ does not manifest this. Even by the limit of composite function (Cf: Zorich, Mathematical Analysis, Vol I, Page 133, Theorem 5), I do not know if my student's solution is totally right. I know that student want to calculate by using substitution $t=\frac{1}{n}$, but because $ n \in \mathbb{N}$, the notation $\lim_{t\to0}$ here is not as it should indicate, thus $t \in \{ t\in\mathbb{R} | \exists n\in\mathbb{N}, t=\frac{1}{n}\}$. Hence I have doubt the righteouness of $(*)$. But I can not give the proper reason to explain this. Can anyone help me?