Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\epsilon<1$

share|improve this question
    
$\sin(1/n)$ 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: wolframalpha.com/input/… –  Sean Eberhard Oct 7 '12 at 13:26
    
graph @Sean Eberhard –  Mykolas Oct 7 '12 at 13:29

4 Answers 4

up vote 2 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.

share|improve this answer

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 $$\lim_{n\to\infty}\sin\left(\frac{1}{n}\right) = \sin\left(\lim_{n\to\infty}\frac{1}{n}\right) = \sin(0) = 0$$

share|improve this answer

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.

share|improve this answer
    
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 $$ \lim_{n \rightarrow \infty} \sin\left(\frac 1n\right ) = \sin \left(\frac 1 {\lim_{n \rightarrow \infty} n }\right) = \sin(0) = 0 $$

share|improve this answer
11  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.