$\lim_{x\rightarrow\infty}\sin(x)$? In physics I came across these kind of equations when I am trying to find the asymptotic behaviour of some function.
Can anyone explain if there is any sense in talking about $\sin(x)$ or $\cos(x)$ as $x$ tends to infinity?
$$\lim_{x\rightarrow\infty}\;\sin(x)?$$
 A: I think the easiest way to express what a limit really means, is to say that you get arbitrarily close to the limit as you get closer and closer to your desired input.   
As $x$ goes to infinity, $\sin(x)$ and $\cos(x)$ take the values $-1$ and $1$ infinitely often, and therefore do not get as close as we might like to anything. We therefore say that the limit does not exist.
A: If we take $x_n=2\pi n$ and $x'_n=2\pi n+\frac{\pi}{2}$ then we have 
$$\lim_{n\to\infty}x_n=\lim_{n\to\infty}x'_n=+\infty$$
but 
$$\lim_{n\to\infty}\sin(x_n)=0\neq1=\lim_{n\to\infty}\sin(x'_n)$$
hence $\displaystyle\lim_{x\to\infty}\sin x$ does not exist.
A: The limit
$$
\lim_{x\to \infty} \sin(x)
$$
does not exist. What we mean by saying that it doesn't exist is that there is not an $L$ such that $\sin(x)$ can be made arbitrarily close to $L$ for $x$ "large enough". You could try to prove that such an $L$ doesn't exist by assuming that such did exist and then getting a contradiction.
Another way so see that the limit doesn't exist is to note that for $x_n = 2\pi n$ you have
$$
\lim_{n\to \infty} \sin(x_n) = 0
$$
while for $y_n = \frac{\pi}{2} + 2\pi n$
$$
\lim_{n\to \infty} \sin(y_n) = 1.
$$
Since you have found two sequences that both go to infinity at which $sin$ approaches two different values, the limit $\lim_{x\to \infty} \sin(x)$ can't exist. 
