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