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.
 A: Before answering the question, it's worth pointing out why any expression involving $\infty$ might be sensible to write in the first place. In particular, we should look at the extended reals, which is $\bar{\mathbb R} = \mathbb R \cup \{-\infty,\infty\}$. This is a topological space - which, for our purposes, we'll just say is some space where we can meaningfully talk about limits - and it has the property that
$$\lim_{n\rightarrow\infty}x_n=\infty$$
whenever, for all $M\in\mathbb R$ there exists an $N_0$ such that if $n> N_0$ then $x_n > M$ - and $-\infty$ behaves similarly. In all other respects, the space behaves like $\mathbb R$.
Generally, we are interested in continuous functions - which preserve all limits. Whenever we have a continuous function $f:\mathbb R^n\rightarrow\mathbb R$ (or from a subset thereof), we can consider trying to extend it to a function $\bar f:\bar{\mathbb R}^n\rightarrow \bar{\mathbb R}$ in a way that preserves continuity - and we may call the tuples $(x_1,\ldots,x_n)$ in $\bar{\mathbb R}^n$ for which this is possible "determinate forms" - and these are precisely those locations at which the limit of $f(a_1,\ldots,a_n)$ tends to a fixed value as each $a_i$ tends to $x_i$ along any sequence.
So, if we took our function to be multiplication, we can say things like $\infty\cdot \infty = \infty$ and $\infty\cdot -2 = -\infty$, but  $\infty\cdot 0$ doesn't make sense because, while $n$ and $2n$ both approach $\infty$ and $1/n$ approaches $0$, the products $n\cdot 1/n$ and $2n\cdot 1/n$ approach different values - so we can't determine $\infty\cdot 0$. Thus, $\sin(\infty)$ is indeterminate because approaching $\infty$ by the sequence $0,\pi/2,\pi,3\pi/2,2\pi,\ldots$ gives a sequence of values of $\sin(x)$ that oscillates $0,1,0,-1,0,\ldots$ and hence fails to converge.

HOWEVER! This is more of a problem with $\infty$ than a problem with $\sin$. While you will necessary lose properties like an ordering when you do this, it is possible to extend $\mathbb R$ to include other kinds of $\infty$. Let's take an example where we can take the $\sin$ of an infinite value. The example I will give is somewhat trivial, but do not be fooled into thinking that every example is like this. Let's define $R=\mathbb R \cup \{\infty_y:y\in [-1,1]\}$ - that is, we have a whole spectrum of infinities added, which we're calling $\infty_y$ - which is nothing more than a symbol. We should probably add some negative infinities too, but let's not for the sake of brevity.
$$\lim_{n\rightarrow\infty}x_n = \infty_y$$
if and only if $\lim_{n\rightarrow\infty}x_n = \infty$ in $\bar{\mathbb R}$ and $\lim_{n\rightarrow\infty}\sin(x_n)=y$. This defines another topological space! An in this space, we see that expressions like $\sin(\infty_0)$ do make sense (and evaluate to $0$), because no matter how we approach $\infty_0$, the sequence of sines will approach $0$.  Note that approaching $\infty_0$ has a clear meaning too: our sequence not only gets arbitrarily large, but it gets arbitrary close to multiples of $\pi$ as well - so the properties of $\infty_0$ are somehow encoding more about how it is approached than the properties of $\infty$ did.
This is not a standard definition to make - but if your goal is to be able to consider sets of sequences with more granularity, defining new spaces can accomplish that and then let you say things about functions that you couldn't before - and it seems to fall along the lines you were thinking of in the chat transcript, since it is true that, as long as we approach "$\infty$" the right way, we can get $\sin$ to have a limit.
(Still, if you just write $\sin \infty$, this makes no sense because $\infty$ already means the standard infinity - you need to communicate new definitions before you can work with $\sin$ at $\infty$ in any sensible way. Also note that there are drawbacks here: for instance, now $\infty_0\cdot \infty_0$ is indeterminate, which is somewhat unfortunate... - and $\infty_0+\infty_0=\infty_0$ and $\infty_{1}+\infty_1 = \infty_0$, but $\infty_1+\infty_0$ and $\infty_0 + 1$ are indeterminate... which is really unfortunate - you have to work harder to get anything nice to happen with this kind of reasoning. You can get a somewhat more interesting object by making $\infty_{\theta}$ something which is approached whenever a sequence goes to $\infty$ and the angles the sequence represents approach $\theta$ (mod $2\pi$) - then addition works out okay, though multiplication is still bad)

Post-script: The objects we're talking about here can be described best as "compactifications" of $\mathbb R$ - meaning they sort of tie up the loose ends of $\mathbb R$. The extended reals can be imagined by taking each end and putting a single element there. There's another way I didn't mention called the projective line (or the one-point compactification), where we take the two ends of the reals, and wrap them together into a circle - giving just one $\infty$ - that one's sort of nice because then you can topologically justify writing $1/0=\infty$ - although it will screw up all your algebra if you do. There are some really crazy compactifications - such as one called the Stone-Cech compactification which, by some really weird trickery, does not lead to a single indeterminate form in a single variable... but it is not exactly an object you hope to encounter frequently.
A: 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 $[-\infty,\infty]$.
A: By $ \sin(\infty) $, I assume you mean $ \lim_{x\to\infty} \sin(x) $. This limit simply doesn't exist. Going by the wikipedia article you linked regarding "indeterminate", I'd say it is not indeterminate since we can confidently say the limit doesn't exist. 
A: I'd give a generalization along the lines of:

An indeterminate form is an octuple $(X,\tau,G,Y,\rho,S,c,f)$ such that:
  
  
*
  
*$(X,\tau)$ and $(Y,\rho)$ are topological spaces
  
*$S$ is a dense subset of $Y$
  
*$c\in X$ and $c\in G'$
  
*$f:G\to S$ is a function such that there is no function $\overline f:G\cup \{c\}\to Y$ such that $\overline f$ is continuous.
  

Then, it should be rather clear that $\left([0,\infty],\text{standard},[0,\infty),[-\infty,\infty],\text{standard},\Bbb R,\infty,\sin\right)$ is undetermined just like it happens (with other entries, but analogously) for $\frac\infty\infty$ and $1^\infty$.
Added: So that I am clear, here's a version of l'Hopital.

Let $I\subseteq [-\infty,\infty]$ be an interval, let $c$ be an accumulation point of $I$ and let $f,g:I\setminus \{c\}$ be differentiable functions such that there is a neighbourhood $U$ of $c$ such that $g'^{-1}(0)\cap U\subseteq\{c\}$ and such that $\lim_{x\to c}\frac{f'(x)}{g'(x)}$ exists in $[-\infty,\infty]$. If $$\left([-\infty,\infty]\times[-\infty,\infty],\text{std},(-\infty,\infty)\times((-\infty,\infty)\setminus\{0\}),[-\infty,\infty],\text{std},\left(\lim_{x\to c}f(x),\lim_{x\to c}g(x)\right),\text{division}\right)$$ is an indeterminate form, then $\lim_{x\to c}\frac{f(x)}{g(x)}=\lim_{x\to c}\frac{f'(x)}{g'(x)}$.

A: 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.
