As $x$ tends to infinity, sin x oscillates rapidly between $1$ and $-1$. So we are not able to pinpoint exactly what the limit is. But whatever it is, can we say that it would be finite? Or do we always have to say that its limit doesn't exist or it's undefined?
[Math] Is limit of $\sin x$ at infinity finite
trigonometry
Best Answer
The limit does not exist in the usual sense. If $\lim\limits_{x\to\infty} \sin x = L$ and $L$ is some particular number, then it would be possible to assure that $\sin x$ is between $L\pm0.001$ by making $x$ big enough, and we would probably be able to figure out how big is big enough in this case. But $\pm1$ both occur as values of $\sin x$ no matter how big $x$ gets and $\pm1$ cannot both differ from the same number $L$ by less than $0.001$.
But I wrote "in the usual sense". I cannot rule out the possibility that in some contexts an unusual sense might be appropriate. For example $$ \lim_{A\to\infty} \Big( \text{average value of $\sin x$ in the interval } 0\le x\le A \Big) = 0, $$ etc.