Is $\sin \infty$ an indeterminate form? I had a long discussion in chat over what seems like a simple question:


  Is $\sin \infty$ an indeterminate form?


What do you think? :)



I'm labelling this as (soft-question) to be safe, but it should have a clear answer depending on the definition of indeterminate form used.
The term "indeterminate form" is seldom used in post-calculus mathematics, but I believe it has one or more accepted definitions, which are either informal or formal. So any answer which takes a standard definition and argues the case would be interesting to me.

How to make a good answer:


State the definition of indeterminate form, either from an online or textbook source, or a definition you came up with on your own.
Determine, using your stated definition, whether $\sin \infty$ is indeterminate or not.


EDIT: What do I mean by $\sin (\infty)$?

It's not a well-defined expression, but neither are any of the other indeterminate forms: $\frac{0}{0}$ doesn't exist, $1^\infty$, doesn't exist, and so on. So the question is whether this expression -- which is not well-defined, just like any other indeterminate form -- is an indeterminate form.
Yes.  An indeterminate form is an expression such that if you replace the constants appearing in the expression with sequences which approach those values, then the limit of the expression is not uniquely determined.  In the case of $\sin(\infty)$, if we replace $\infty$ by the sequence $a_n=\pi n$ we get a limit $$\lim_{n\to\infty}\sin(\pi n)=0.$$  If we instead take the sequence $b_n=2\pi n+\pi/2$ we get $$\lim_{n\to\infty}\sin(2\pi n+\pi/2)=1.$$  If we instead take the sequence $c_n=n$ then we get a limit $$\lim_{n\to\infty}\sin(n)$$ which does not exist.

The point here is that if you are trying to evaluate a limit of the form $$\lim \sin(\text{something})$$ where you know that the "something" is approaching $\infty$, you cannot tell what the answer is from just this information--the limit could be any number between $-1$ and $1$, or it could not exist.  This is just like the more familiar indeterminate forms like $\frac{0}{0}$ which are taught in calculus: if you have a limit $$\lim\frac{\text{something}}{\text{something else}}$$ where both "something" and "something else" approach $0$, that is not enough information to determine the limit.



It is perhaps worth remarking that with this definition, an indeterminate form is equivalently an expression such that if you replace the constants with variables ranging over the real numbers, then the limit as these variables jointly approach their values does not exist (where "exist" includes the possibility of being $\pm\infty$).  In other words, to say that $\sin(\infty)$ is an indeterminate form really just means that $\lim\limits_{x\to\infty}\sin(x)$ does not exist (where here $x$ approaches $\infty$ in the real numbers, in contrast to the sequential limits we had before).  Indeed, if you can have multiple different sequential limits, then the limit over the real numbers cannot exist.  Conversely, if the limit over the real numbers does not exist, then by compactness of $[-\infty,\infty]$ the limit must still accumulate somewhere in $[-\infty,\infty]$ and so must accumulate at two different values, and then you can choose sequences for which the limit approaches two different values in $[-\infty,\infty]$.
since oo is not a real number sin(oo) is meaningless. If you meane the limit for x->oo , it is indeterminate, since the value can not be determined.