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Could someone explain how to correctly prove that $$\lim_{n\to\infty}\sin\frac{1}{n}$$ where $n=1,2,\cdots,n$ doesn't exist. I have no problem with it if $\sin\frac{1}{x}$ where $x$ is real, because just taking values $x=\frac{2}{(2n-1)\pi}, x=\frac{1}{n\pi}, x=\frac{2}{(2n+1)\pi}$ it is clear, for example, by Cauchy criterion. But what about when $n$ takes natural values?

This is posterior edit. It corresponds to comments and the answers I got till now: I am pretty sure that this limit doesn't exist. The graph of this function goes from $-1$ to $0$ to $\,1$ and so on infinitely many times as $\frac{1}{n}$ approaches $0$. So there is no way you could find $n$ big enough so that $|\sin \frac{1}{n}-0|<\epsilon$ if $\epsilon<1$

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$\sin(1/n)$ tends to $\sin(0) = 0$ as $n\to\infty$. Do you mean to ask about $\sin(n)$? – Sean Eberhard Oct 7 '12 at 12:30
You statement is not right since $\sin(\frac{1}{n})\sim \frac{1}{n} \to 0$ as $n\to \infty$ – Patrick Li Oct 7 '12 at 12:33
Just a guess, you might mean to take the limit for $n \rightarrow 0$ since that limit does not exist (and has $\infty$ for the sin). – adam W Oct 7 '12 at 13:15
@Mykolas In the limit you indicate $n\to\infty$, so you're looking at large $n$, such as in this graph:… – Sean Eberhard Oct 7 '12 at 13:26
graph @Sean Eberhard – Mykolas Oct 7 '12 at 13:29
up vote 3 down vote accepted

You are asking about the limit: $$ \lim_{n\to \infty} \sin\left(\frac{1}{n}\right). $$ For this limit (as others have already noted, $\frac{1}{n} \to 0$ as $n\to\infty$. I.e., as $n$ gets very large $\frac{1}{n}$ becomes very small. So this limit is the same as the limit $$ \lim_{t \to 0} \sin(t) $$ where $t = \frac{1}{n}$. And this limit indeed exists and is equal to $0$.

If you want to do the whole $\epsilon - N$ thing, you would, given an $\epsilon > 0$ want a $N (>0)$ such that if $n > N$, then $\left\lvert\sin\left(\frac{1}{N}\right)\right\lvert < \epsilon$. Without saying too much, you simply pick $N$ (large enough) such that $\frac{1}{N} < \sin^{-1}(\epsilon)$.

I am guessing that you might be asking about another limit, or that you at least confuse the above limit with the following: $$ \lim_{n\to 0}\sin\left(\frac{1}{n}\right). $$ Here you have that $\frac{1}{n} \to \infty$ as $n\to 0$. You are then considering what happens to $\sin(t)$ as $t$ becomes arbitrarily large. But since, loosely speaking, $\sin$ just keeps oscillating between $-1$ and $1$, this limit indeed does not exist.

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Using sequences, we can characterize continuity at a point for real functions.

Function $f: A \rightarrow \mathbb{R}$ is continuous at a point $x_0 \in A$ if and only if for any sequence $(x_n)$ in $A$ converging to $x_0$, we have $$\lim_{n\to\infty}f(x_n) = f\left(\lim_{n\to\infty}x_n\right) = f(x_0)$$

Since $\sin$ is continuous at $0$, it follows that $$\lim_{n\to\infty}\sin\left(\frac{1}{n}\right) = \sin\left(\lim_{n\to\infty}\frac{1}{n}\right) = \sin(0) = 0$$

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From your discussion it seems likely that you actually meant to ask about the limit of $\sin n$. Here is a possible proof that $\sin n$ does not converge. Since

$$ \sin(n+1) = \sin(n)\cos(1) + \cos(n)\sin(1),$$

it follows that if $\sin n$ converges so does $\cos n$, and therefore $e^{in} = \cos n + i\sin n$ also converges. But $|e^{i(n+1)} - e^{in}| = |e^{i} - 1|$, which is a nonzero constant, so $e^{in}$ doesn't converge.

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I mean $sin\frac{1}{n}$. I edited the question, please take a look. @Sean Eberhard – Mykolas Oct 7 '12 at 13:25

In my opinion this limit does exist. It is 0 because $\sin(1/n)$ is continuous and so we have $$ \lim_{n \rightarrow \infty} \sin\left(\frac 1n\right ) = \sin \left(\frac 1 {\lim_{n \rightarrow \infty} n }\right) = \sin(0) = 0 $$

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I dislike the way you wrote it. I'd prefer $sin\left( \lim_{n \rightarrow \infty} \frac 1 { n }\right)$ to $sin \left( \frac 1 { \lim_{n \rightarrow \infty}n }\right)$. – xavierm02 Oct 7 '12 at 12:52

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