# A question about the limit of a composite function

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?

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I suggest that you do something about your 0% accept rate first. People might be discouraged from answering your question if you have a habit of not accepting answers. –  Haskell Curry Oct 26 '12 at 1:58

\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.

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Thank you very much! N.S.'s answer benefits me much! –  azhi Oct 26 '12 at 2:20
Then it'd be a good idea if you accept it as answer and thus enhance your accept rate, @azhi –  DonAntonio Oct 26 '12 at 3:20
@azhi: If you think that the answer benefits you, please accept it. This is basic etiquette. –  Haskell Curry Oct 26 '12 at 19:54
I am sorry, but I do not know how to "accept" it. I've noticed that "0% accept rate", how can I increate it? Please tell me how! I am just a newby here. –  azhi Oct 27 '12 at 2:55