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What is the value of $\sin(x)$ if $x$ tends to infinity?

As in wikipedia entry for "Sine", the domain of $\sin$ can be from $-\infty$ to $+\infty$. What is the value of $\sin(\infty)$?

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It says that the domain is $(-\infty,\infty)$ which means the endpoints $\infty$ and $-\infty$ are not included in the domain. That is, the function $\sin x$ is not defined at those points, $\infty$ and $-\infty$. However we can talk about the limit of this function as it tends to those points, only to be disppointed that the limits don't exist as well! –  user21436 Mar 4 '12 at 18:07
    
Can you define what is $\infty-1$? –  user21436 Mar 4 '12 at 18:11
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$(-\infty,\infty)$ is just another way to express the set of all real numbers. –  Patrick Mar 4 '12 at 18:12
    
@Kannappan: There's no way to make sense of it as a real number. –  Patrick Mar 4 '12 at 18:13
    
@Patrick Precisely. I was waiting for OP to realise that! –  user21436 Mar 4 '12 at 18:15

3 Answers 3

If it said that the domain was $[-\infty,\infty]$ as opposed to $(-\infty, \infty)$, then that means that $\infty$ would be an allowable argument of sine. But the standard sine function is defined on the real numbers, and $\infty$ is not a real number.

But let's do a thought experiment. Suppose we were trying to consider sine as a map (perhaps not a function now, as we'll see in a moment) from the extended real numbers to the extended real numbers. That is, $\sin: \overline{\mathbb{R}} \to \overline{\mathbb{R}}$, and where the extended reals are the standard reals, plus two points that we call $\pm \infty$ (see also wiki: it's not a trivial thing to consider). What would be a reasonable way to define sine at the infinities?

If $\lim_{x \to \infty} \sin x$ existed, that would be the clear candidate. But it's not. Perhaps the set of limit points should be the definition, i.e. saying $\sin{\infty} = [-1,1]$. That seems interesting to me, but it's no longer a function. This even feels like what the definition should be to me, but it loses many convenient properties, like continuity, and it's a bit nonsensical.

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Suppose $\lim_{x \to \infty} \sin(x) = L$. $\frac{1}{2} > 0$, so we may take $\epsilon = \frac{1}{2}$.

let N be any positive natural number. then $2\pi (N + \frac{1}{4}) > N$ as is $2\pi (N+\frac{3}{4})$.

but $\sin(2\pi (N + \frac{1}{2})) = \sin(\frac{\pi}{2}) = 1$.

so if $L < 0$, we have a $y > N$ (namely $2\pi (N + \frac{1}{4})$) with:

$|\sin(y) - L| = |1 - L| = |1 + (-L)| = 1 + |L| > 1 > \epsilon = \frac{1}{2}$.

similarly, if $L \geq 0$, we have for $ y = 2\pi (N+\frac{3}{4}) > N$:

$|\sin(y) - L| = |-1 - L| = |(-1)(1 + L)| = |-1||1 + L| = |1 + L| = 1 + L \geq 1 > \epsilon = \frac{1}{2}$.

thus there is NO positive natural number N such that:

$|\sin(y) - L| < \frac{1}{2}$ when $y > N$, no matter how we choose L.

since every real number L fails this test for this particular choice of $\epsilon$, $\lim_{x \to \infty} \sin(x)$ does not exist.

(edit: recall that $\lim_{x \to \infty} f(x) = L$ means that for every $\epsilon > 0$, there is a positive real number M such that $|f(y) - L| < \epsilon$ whenever $y > M$. note that there is no loss of generality by taking M to be a natural number N, since we can simply choose N to be the next integer greater than M.)

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Before writing "so we may take $\epsilon = \frac12$", you really should give the $\epsilon/\delta$ (or is that $\epsilon/L$ here?) definition of $\lim$ you're using, if only so that we can tell which symbols you're using for what purpose. –  Ilmari Karonen Mar 4 '12 at 19:34
    
for limits at infinity, there is no delta. that is what N is for. my use of epsilon is fairly standard. –  David Wheeler Mar 4 '12 at 19:43
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Yes, I know, but it would help if you actually gave the definition so that we wouldn't have to guess. (And so that readers who don't remember the definition by heart can still make sense of your answer.) In particular, I'm personally more likely to use $\epsilon/M$ for the definition of $\lim$ at $\infty$, since that's the convention I learned back in Real Analysis I. Unless, of course, I've already used $\epsilon$ or $M$ for something else, in which case I'll just pick some other letter to use. –  Ilmari Karonen Mar 4 '12 at 19:54
    
definition added for clarity's sake. hopefully this sheds some light on why I used "N" instead of M. –  David Wheeler Mar 4 '12 at 23:05

Mathematics has many different ways of talking about infinity. In particular, there is more than one way of adjoining infinite quantities to the real number line. One, which is similar in spirit to what the others have been talking about, is the extended real number line. Another is the hyperreals. In the hyperreals, we have many different sizes of infinity, functions such as the sine can be extended to the whole hyperreal line in a natural and uniquely defined way, and it makes sense to say that $\sin x$ has some value, where $x$ is a certain infinite number. There is also a notion of an infinite integer, and we can, for example, say that $\sin(\pi n)=0$ and $\sin(\pi (n+1/2))=\pm 1$ if $n$ is an infinite integer.

What this still does not allow is any conclusion about the value of $\sin x$ as $x$ tends to infinity: $\sin x$ takes on all values between $-1$ and $+1$ for various infinite values of $x$.

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whereas a function such as $f(x) = 1/x$ has a constant "standard part" on the infinite hyperreals, namely 0. –  David Wheeler Mar 4 '12 at 23:09

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