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 $\sin(\infty)$?

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.) 
