A question about the limit of a composite function: $\lim_{n\to\infty}\frac{\sin n}{n}$ 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?
 A: \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}
is CORRECT, BUT it needs to be explained right in my opinion. To make the argument more clear, here is basically the complete argument:
Let $t_n=\frac{1}{n}$. Then it is true that $t_n \to 0$ and 
\begin{align}\tag{*}
\lim_{n\to\infty}\frac{\sin n}{n}=\lim_{n\to \infty}t_n\sin\frac{1}{t_n}
\end{align}
Now, here is the subtle thing:
$$\lim_{t\to 0}t\sin\frac{1}{t}$$
is more general than $\lim_{n\to \infty}t_n\sin\frac{1}{t_n}$, since in both the $t'$s go to zero, but in the first one $t$ can be any real number, while in the second there are few particular $t'$s.
So, knowing that
$$\lim_{t\to 0}t\sin\frac{1}{t}=0$$
one can conclude that
$$\lim_{n\to \infty}t_n\sin\frac{1}{t_n}=0$$
but the other implication is not true.

In General
If $f(x)$ is any function, if you know that $\lim_{t \to 0} f(t)=L$ then you can conclude that $\lim_{n \to \infty} f(\frac{1}{n})=L$.
Anyhow, if you know that $\lim_{n \to \infty} f(\frac{1}{n})=L$, you cannot conclude that $\lim_{t \to 0} f(t)=L$.
Any conclusion of the type 
\begin{align}\tag{*}
\lim_{n\to\infty}f(\frac{1}{n})\overset{\frac{1}{n}=t}{=}\lim_{t \to 0} f(t)
\end{align}
is true if you know that the second limit exists, but not otherwise.
