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:

  1. State the definition of indeterminate form, either from an online or textbook source, or a definition you came up with on your own.

  2. 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.

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    $\begingroup$ For my humble opinion to write $\sin \infty$ is wrong. If I put the limit operator, IMHO $\sin x$ have not limit because is a limitate function. $\endgroup$ – Sebastiano Sep 22 at 22:49
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    $\begingroup$ @Sebastiano Would you then say that to write $0/0$ is wrong? Because that is not well-defined either. Yet we call that an indeterminate form. The point is, what does "indeterminate form" mean? And does it apply to $\sin \infty$? $\endgroup$ – 6005 Sep 22 at 22:55
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    $\begingroup$ "Would you then say that to write 0/0 is wrong? " Well, I would..... Who'd say it is "right"? And what do you mean by "right". $\endgroup$ – fleablood Sep 22 at 23:00
  • $\begingroup$ Always for my humble opinion: $0/0$ not have sense in the real field. I have not sense in the classic contest. It is have a sense in the limit calculus. $\endgroup$ – Sebastiano Sep 22 at 23:01
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    $\begingroup$ I personally don't find the expression "indeterminate form" to be useful. But ...I'd have to say if we take $\sin \infty$ to mean $\lim_{f(x) \to \infty} \sin(f(x))$ then it must be indeterminate as $f_1(x) = x$ or $f_{2,3}:\mathbb N\to \mathbb R$ via $f_2(x)=2\pi*x$ or $f_3(x)=(2x+\frac 12)\pi$ yield different results ... $\endgroup$ – fleablood Sep 22 at 23:15

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 $[-\infty,\infty]$.

  • $\begingroup$ Please, can you explain with further explanation the proof of the first and the second sequence? Thank you very much. I vote positively your answert. The last edit reminds me of de Hopital's theorem. But in the context of succession it cannot be applied. $\endgroup$ – Sebastiano Sep 22 at 23:07
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    $\begingroup$ @Sebastiano proof of the first and second sequences? They are the constant sequences $0,0,0,\dots$ and $1,1,1,\dots$ respectively. This is made obvious by the well known fact that $\sin$ is periodic with period $2\pi$. $\endgroup$ – JMoravitz Sep 22 at 23:09

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.

  • $\begingroup$ Thanks! I enjoyed the extended example about a spectrum of infinities, and the comment about Stone-Cech at the end. $\endgroup$ – 6005 Sep 23 at 0:43

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.

  • $\begingroup$ Actually, we can't confidently say that: $\lim_{k \to \infty} \sin (k \pi x) = 0$, and does exist. $\endgroup$ – 6005 Sep 22 at 22:51
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    $\begingroup$ That assumes that $k$ and $x$ are integers. $\endgroup$ – Bladewood Sep 22 at 22:53
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    $\begingroup$ The issue is that for $ \lim_{x\to\infty} f(x) = L $, we must have for every sequence $ \{x_n\}_{n\geq 1} $ with $ x_n\to x $, that $ f(x_n)\to L $. While the specific sequence $ \{k\pi x\}_{k\geq 1} $ gives convergence to $ 0 $, not every sequence does. See Theorem 4.2 in baby Rudin for the sequence definition for the limit of a function. $\endgroup$ – poopstraw Sep 22 at 22:55
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    $\begingroup$ @Bladewood Right. But when I write $\sin \infty$, I don't mean $\lim_{x \to \infty} \sin x$, just like when you write the indeterminate form $0/0$, you don't mean $\lim_{x \to 0} x/x$. Do you see? The indeterminate form is supposed to suggest a limit of some function, not necessarily just the function $x$. $\endgroup$ – 6005 Sep 22 at 22:58
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    $\begingroup$ @poopstraw Can you state your preferred definition of an indeterminate form and see what it says about the expression $\sin \infty$ (is it an indeterminate form or not?) I strongly suspect you will not arrive at the same answer. As with other indeterminate forms like $\infty/\infty$, $\sin \infty$ (if it makes any sense) would not mean the limit of $\sin x$, but rather the limit of some thing which approaches $\infty$. $\endgroup$ – 6005 Sep 22 at 23:01

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:

  1. $(X,\tau)$ and $(Y,\rho)$ are topological spaces
  2. $S$ is a dense subset of $Y$
  3. $c\in X$ and $c\in G'$
  4. $f:G\to S$ 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 $\lim_{x\to c}\frac{f(x)}{g(x)}=\lim_{x\to c}\frac{f'(x)}{g'(x)}$.

  • $\begingroup$ The definition is not pleasing to me, but it's a definition :) $\endgroup$ – 6005 Sep 23 at 0:38
  • $\begingroup$ I mean, it's more or less the same thing as liking my car. $\endgroup$ – Gae. S. Sep 23 at 1:04

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.

  • $\begingroup$ Please try to format in MathJax. $\endgroup$ – N. Bar Sep 22 at 23:23

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