Is limit of $\sin x$ at infinity finite? 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? 
 A: “Limit” has a particular, technical meaning.  According to this meaning, the sine function has no limit as its argument goes to infinity.  There are several different ways a function can fail to have a limit, and the limit concept is not discerning enough to be able to tell the difference between the reasons.
The notion I think you are looking for is that the sine function is bounded.  

(Details…)
I think actually what you really want me be not simply “$\sin x$ is bounded” but “$\sin x$ is bounded as $x$ goes to infinity”.  This means that there is some constant $C$ so that, if $x$ is large enough, it is always true that $$-C \le \sin x \le C.\tag{$\star$}$$
In this case $C=1$ will do, as will $C=17$.
To pin down the meaning of “if $x$ is large enough" we can say that there is a second constant, say $x_0$, so that $(\star)$ is true for all $x > x_0$.  The sine function is bounded everywhere, so for the sine function any $x_0$ will work.
For more a interesting example, consider the function $\frac1x$.  This function is not bounded everywhere, or even on the positive reals, because $\frac 1x$ becomes infinitely large when $x$ is close to $0$.  But it is bounded as $x$ goes to infinity, because if we take $x_0 = 2$ (say) then $$-\frac12 \le \frac1x \le \frac12$$ for all $x>x_0$.
A: 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.
A: We would say that the limit does not exist.
As it does not exist you cannot say that it is finite or not.
Good luck with you study.
