Talk:Indeterminate form

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comment by Taku

Isn't 0 over 0 is a determinate form because if you keep multiplying very small number very few times you just get 0. -- Taku 02:25, Nov 15, 2003 (UTC)

0/0 is indeterminate since if it had some definite value, say x, so 0x=0. But any number satisfies that equation for x, so 0/0 has no definite or determinable value. Dysprosia 02:28, 15 Nov 2003 (UTC)

Umm, elegant! Thanks a lot, Dysprosia. -- Taku
But outside the realm of limits, there is very little practical use in calling 0/0 indeterminate, versus just grouping it with a/0 undefined forms (a != 0)

In the sense of this article, 0/0 is indeterminate because if f(x) and g(x) both approach 0 as x approaches something, then f(x)/g(x) can approach a number that depends on which functions f and g are. (That does not detract from the elegance of Dysprosia's observation, however.) Michael Hardy 20:18, 15 Nov 2003 (UTC)

I've just moved some material from zero divided by zero to this page performing significant changes to it to make it apropriate for this page. I am planning to move more material in a few days time but not necessarily to this page and maybe to a subpage of this page. Barnaby dawson 19:56, 20 Sep 2004 (UTC)

After an edit clash, I've added to the second paragraph. Charles Matthews 11:05, 18 Mar 2005 (UTC)

I've now done further work on that discussion. Charles Matthews 11:16, 18 Mar 2005 (UTC)

Missing Infinity ^ Infinity

The form oo^oo is missing from the list.

How can ∞ be considered an indeterminate form? If f(x) and g(x) both approach ∞ as x approaches a, then f(x)g(x) can only approach one thing, ∞, regardless of which functions f and g are. So it's not an indeterminate form. Michael Hardy 23:34, 18 December 2005 (UTC)

It can be indeterminate in a weak sense, like 1/0 (but perhaps a bit stronger). For example, as x approaches 0 from the right, (1/x)1/x approaches infinity (from the left), while (1/x)-1/x approaches zero (from the right). But nothing of this form can converge to any non-zero finite value (because the logarithm must converge to infinity). I'm not sure if this should count as indeterminate or not; the precise definition that I've just written says that it is, but one could easily fix that so that it's not. --Toby Bartels 01:57, 23 May 2006 (UTC)

Remember that when one says ∞, one means "positive infinity" (i.e. increasing without bound). So your example is not the same because it raises "positive infinity" to "negative infinity". --Spoon! 11:33, 31 August 2006 (UTC)
What one means depends on context; compare the extended real line to the real projective line. See also this formal definition, which uses the Riemann sphere. —Toby Bartels 22:03, 20 September 2006 (UTC)

On merging with defined and undefined

User:Michael Hardy, at 15:37, 18 December 2005, wrote in this diff:

This merger proposal has no merit. Too many readers keep failing to see that "indeterminate form" does NOT mean something that is undefined. Maybe we should hit that point much harder.

(copied here for future reference by Oleg Alexandrov (talk) 00:16, 19 December 2005 (UTC))

To be honest I'm not exactly sure what MH's getting at there. An indeterminate form per se is not undefined, but its value is, and they're pretty much coextensive with the expressions treated at defined and undefined. The latter is just a bad article with a bad title, and I think the real point of the merger was to figure out some way to get rid of it, a project I'd support. --Trovatore 16:58, 13 February 2006 (UTC)
No -- they're NOT coextensive, since 1/0 is NOT an indeterminate form. 0/0 is an indeterminate form; 5/0 is not. That means that if f and g both approach 0 as x approaches something, then f/g could approach anything. That is not true if f approaches 5 and g approaches 0. Michael Hardy 22:08, 13 February 2006 (UTC)
OK, true, 1/0 is not an indeterminate form, but it's not enough by itself to justify a separate article from this one, particularly one with as bad a name as "defined and undefined". And all the indeterminate forms are denotationally undefined (unless you're going to take the 00=1 approach; I'd be against presenting that as the default approach, though it ought to be mentioned). --Trovatore 22:16, 13 February 2006 (UTC)

Determinate and inderminate discussion

The "Discussion" section had a problem paragraph (reading "By contrast "1/0" is not an indeterminate form because there is no range of different values that f/g could if f approaches 1 and g approaches 0."). There was at least a missing word (between could and if). I attempted to correct that and add more detail to clarify it, but then realized I still don't understand the distinction which is the whole point of the article.

Several times 1/0 is contrasted with 0/0, but the only characterization of this contrast is that there is "no range of different values" without elaboration. But f/g could have a limit of either +∞ or -∞ depending on whether the function g approaches 0 from the + or - side. So there seems to be more than one "value" possible among limits approaching 1/0. Just how is this determinate while 0/0 is not? Whatever the answer, I think the article needs improvement in explaining this. -R. S. Shaw 00:58, 13 March 2006 (UTC)

I think that the whole article is rather unclear in a few ways, first of all that this is fudnamentally about limits (although of course, that is not your problem, R. S. Shaw). With this in mind, I'm going to rewrite it a good deal right now, and I should be able to address your (Shaw's) problem at the same time. -- Toby Bartels 20:29, 22 May 2006 (UTC)

Lead sentence

The lead sentence says

In calculus and other branches of mathematical analysis, an indeterminate form is an algebraic expression whose limit cannot be evaluated by substituting the limits of the subexpressions.

This is clearly wrong - an indeterminate form may be obtained by substituting limits of subexpressions, but the form itself is just a few symbols and doesn't have a "limit", it has a "value" (or more precisely, it does not have a value). That is, the question "What is the limit of 0/0?" is meaningless. CMummert 02:22, 29 December 2006 (UTC)

Yep. How about:
…an indeterminate form is an algebraic expression representing a limit that cannot be evaluated by substituting the limits of the subexpressions.
EdC 17:02, 29 December 2006 (UTC)

Untitled comment

I think the situation is being looked into the wrong way. Say you had (x^2-9)/(x+3). Someone that didn't know that x+3 is a conjugate of x^2-9 would tell you that -3 gives an undefined answer. Yet someone who factored would get x-3, and say that all values are defined. I take this to mean that in a situation like x^2/x, 0/0 is 0. However, in the cases where it's 0(a)/0, where a is any number except 0, the result is a as 0/0 would negate eachother and equal one. An example is above. And don't gimme a "but my calculator says no." k, I want to know what you guys think. Also, another example is x/x, every number outside of zero gives a 1 answer, so it would be expected that 0/0 gives a one answer. This even works with x^2/x. The reason (0,0) is a point, and not (0,1), is because when you expand ((0)^2)/0 you get 0*0/0. 0/0 is one, and that leaves 0*1, which gives 0. —Preceding unsigned comment added by (talk) 07:11, 3 January 2007

Well, some of your terminology is nonstandard: you say "conjugate" when you mean "factor" and "negate" when you mean "cancel". But the "0"s in the expression 0/0 do not cancel each other. When you have, e.g., 3/3, the two "3"s cancel each other because you can divide both the numerator and the denominator by 3. But in 0/0 you cannot divide both the numerator and the denominator by 0. But I'm not sure what your question is otherwise. It is certainly true that x/x approaches 1 as x approaches 0, and x2/x approaches 0 as x approaches 0, and (x2 − 9)/(x + 3) approaches −6 as x approaches 3. But what is in the article that is being "looked into the wrong way"? Michael Hardy 00:45, 4 January 2007 (UTC)
Conjugate is being used the right way, just outside of complex numbers. I'm using the other factor to remain with a 2 variable-number combination (or however you want to say it). The article says that 0/0 is 0 when it's X^2/x (or just call it x if you want), but x/x is not 1 when x=0. I say it's being looked at the wrong way because if you have 0^2/0, it expands to (0/1 * 0/1) / 0/1. and since a top and the bottom 0/1 cancel out, you get (0/1 * 1) / 1 which is zero. However, when you have x/x, any real number / any real number is 1. It would also make sense that if you had (0/1) / (0/1) it would equal one. And if that doesn't make sense, a perfect example is limits. If you do the differential equation, you have to cancel out the h on the bottom (if my equation's the same as yours) with the hs' on top. This gives you the formula for any point. But in any situation, if you used the original formula with h=0, You get the numerator over 0, which be definition is indeterminate or at the minimum undefined, which refutes the ability to use limits on points. Therefore this means that division comes before multiplication when using the order of operation because otherwise, with the above x/x and differential equation, the equations wouldn't be true in any situation, so the only thing can be true is 0/0 is 1 —The preceding unsigned comment was added by (talk) 23:12, 8 January 2007 (UTC).
No, the article says that x^2/x has the limit 0 as x tends to 0, and x/x has the limit 1. Neither formula has a value at x=0.
You can't cancel out zeroes from a fraction; you can only cancel out non-zero common factors.
I don't understand which differential equation you're referring to.
I don't think refute means what you think it means.
I don't understand your last point. In mathematics the order of operations is explicit. –EdC 00:32, 9 January 2007 (UTC)
You can cancel out zeroes, which is why the limits are what they are. The differential equation I'm referring to is the basic slope formula or whatever other areas call it. When you use f(x) instead of y, the equation is dy/dx = (f(x+h) - f(x)) / h. This is the equation used to find the slope of a line between two points. When this is done with any equation with a leading x power > 1, you result has xs and hs. To determine the "instantaneous veloctiy", "slope", whatever you want to call it, 0 is substituted for every h value because you want the slope at that point. Now this part is crucial. Slopes work, and they are proven to work, otherwise we wouldn't use them because they wouldn't work. To say that you can't divide by 0 makes the whole beginning equation moot. And yes, the definition of refute is "to prove to be false or erroneous, as an opinion or charge." You would be refuting the whole ability to find the slope on a given point, which is a significant part of calculas. I know you will rant about "Slopes work, and they are proven to work, otherwise we wouldn't use them because they wouldn't work." but there is no other way to word it. It's like telling me to prove adding works. It works because it does, it doesn't need a reason. —The preceding unsigned comment was added by (talk) 02:17, 17 January 2007 (UTC).
No. As the article on differentiation describes, the slope of a line is found by taking the limit of the equation (f(x+h) - f(x)) / h as h tends to 0: . At no point is a division by 0 performed. Imagining that taking a derivative involves cancelling zeroes leads to the same mistakes that motivated the development of mathematical analysis on a rigorous footing. –EdC 03:11, 17 January 2007 (UTC)
Note that you can, if you believe in non-standard analysis, set h to be an infinitesimal. However, that's not the same as zero. –EdC 03:19, 17 January 2007 (UTC)

List given in introduction

I've changed "The indeterminate forms include [...]" to "The indeterminate forms are [...]" on the assumption that the list is exhaustive. Previous wording admitted interpretation of list as a subset.

If the given list is not exhaustive, I believe the best compromise is "The indeterminate forms (as described this article) are [...]". That is to say, whether or not the list is exhaustive, the current wording is needlessly ambiguous. 23:29, 7 March 2007 (UTC)

List given in introduction (second concern)

A separate critique is that it seems better to say "are expressed by/using/as" or "are referred to by/as" in place of "are". It may be confusing to many readers to say such-and-such is 0/0 as it might seem to imply (as we almost always intend when writing such expressions) that 0/0 represents a number rather than a form. I have not made this change; I don't know what wording best resolves this potential misreading. Please discuss here. 23:35, 7 March 2007 (UTC)

The real problem is that although numerous books use the term "indeterminate form", I have never seen published a definition of one. Can you think of a textbook that defines them? CMummert · talk 00:25, 8 March 2007 (UTC)

Indeterminate form, but continuous

Example: Define f as f(x)=(x+3)(x-3)/(x-3) for x other than 3, and f(x)=6 for x=3. This function is continuous at x=3, even though its limit as x goes to 3 is of the indeterminate form 0/0. FilipeS 18:06, 11 June 2007 (UTC)

The indeterminate form is because the division function is discontinuous at (0,0). Note that the indeterminate form is "0/0", not 'f(0)', which is not an indeterminate form. Also, the sentence you are editing is describing a binary operation, not a unary function as your example above provides. — Carl (CBM · talk) 20:49, 11 June 2007 (UTC)
Yeah, but the passage about continuity and indeterminate forms in general is problematic anyway. I'd go so far as to say it's original research. As far as I know, the literature dealing with indeterminate forms never gives a general definition of "indeterminate form"; it simply enumerates them. --Trovatore 21:44, 11 June 2007 (UTC)
It's more a question of how closely you want the sources to match the phrasing of the article. Searching on google books, I can find several sources that are very close to the statement here: [1] [2] [3] [4]. I don't think it's an unacceptable deduction from the statements in these references and what is in the article. I'll clarify that the term is ordinarily only used for algebraic expressions. — Carl (CBM · talk) 23:42, 11 June 2007 (UTC)
Carl, that is needlessly obscure. "Indetermination" is a purely formal, and not entirely rigorous concept, yet a useful one. 0/0 is an indetermination, right? Trying to compute directly the limit of f as x goes to 3 gives 0/0, right? Then it's an indetermination! And yet the function is not discontinuous, as one can easily see by solving the indetermination. The article was wrong. FilipeS 11:56, 12 June 2007 (UTC)
0/0 is an indeterminate form because the binary division operation is discontinuous. Your function f is not a binary operation. Your revert also removed the clarification about what is meant by algebraic operations and the important fact that the extended real line is being considered. I added the word binary in a few more places to further clarify that this paragraph is referring to binary operations only. — Carl (CBM · talk) 13:42, 12 June 2007 (UTC)

What is your source for claiming that being an indeterminate form has anything to do with the arity of the operations involved? I've never heard of such a notion! Exponentiation is a unary operation, yet 1 to the power of infinity is still an indetermination. FilipeS 13:53, 12 June 2007 (UTC)

Exponentiation (a^b) is a binary operation. Every indeterminate form involves a binary algebraic operation: 0^0, 0/0, 0 * infty, etc. — Carl (CBM · talk) 14:02, 12 June 2007 (UTC)

It depends on how you define it: the exponential function is unary. And you can regard division as binary, f(x, y) = x/y, or unary, g(x) = x/b, or h(x) = a/x. Which is why bringing arity into the conversation makes no sense to me. Indeterminate forms have to do with calculus and limits, not algebra and arity. FilipeS

But there are no indeterminate forms of the form exp(_). And for an indeterminate form of the type to exist, both divisor and dividend must be free (to range as a function of x); if one is fixed as in your h(x), g(x), the indeterminate form does not exist. –EdC 22:41, 12 June 2007 (UTC)
f(x)=1^x and g(x)=0^x can also be called exponential functions.
And what do you suppose you get if you apply exp( ) to 0/0?...
As for your other objection, the numerator and the denominator in f(x)/g(x) are both free to vary. You don't need a two-variable function to get that. FilipeS 23:28, 12 June 2007 (UTC)
If you apply exp() to 0/0, you get to deal with the indeterminate form 0/0 and then apply exp(). JumpDiscont (talk) 04:15, 5 January 2010 (UTC)

Generality, abstraction, sophistication

(This goes back to Carl's remark that "It's more a question of how closely you want the sources to match the phrasing of the article.") I don't think that's the whole issue. While no doubt someone has done it, I don't believe it has ever been standard to give a general, abstract definition of "indeterminate form". If you want to discuss discontinuous funcions, you talk about discontinuous functions; the "indeterminate form" terminology is a fifth wheel when you get to that level of abstraction.

Historically the phrase has been used to give calculus students a way of detecting situations where they have to apply extra care when computing a limit, without bringing up anything as sophisticated as the definition of a continuous function from R2 to R. It's an enumerated collection of situations; no rule is given for extending it. I think that should be the focus of this article. Something like, say, ei could easily be added to the list, but historically, it has not been.

If generalizations can be sourced, they can be added, but they should be clearly marked as generalizations of the notion, not part of the standard interpretation of "indeterminate form". --Trovatore 20:13, 12 June 2007 (UTC)

Another example of an expression which could easily be argued to be an indetermination, but corresponds to a clearly unary operation, is arg(0), where 'arg' stands for the complex argument. Although some authors apparently define it as zero by convention, arg(z) can actually approach any real number as z goes to zero, depending on the path followed by z in the complex plane. FilipeS 20:38, 12 June 2007 (UTC)
But you're still arguing about the details of the presentation of the general case. My point is that, historically, there is no general case. There are exactly seven indeterminate forms. --Trovatore 20:51, 12 June 2007 (UTC)
I'm agreeing with you, if you haven't noticed. As for the number of indeterminate forms, it depends on whether you regard infinity and minus infinity as separate entities, or identify them with each other. FilipeS 20:53, 12 June 2007 (UTC)

I would like to clarify the article, but apparently I'm not succeeding. I agree there are only seven indeterminate forms. If the purpose of that paragraph was only to generalize them, it should be removed. But I think it can be salvaged, and has merit, as an explanation of why these particular seven expressions are indeterminate but the rest aren't. The point is that they are the only expressions involving one of the operations of addition, subtraction, multiplication, division, and exponentiation and two values from the extended real numbers such that the value of the operation on those two values is not determined by continuity, and therefore it isn't possible to commute the limit across the operation. That point ought to be conveyed by the article, but if the current paragraph isn't doing it, I'll be glad to rewrite it from scratch. — Carl (CBM · talk) 23:44, 12 June 2007 (UTC)

My advice: don't try to explain. It's fairly obvious why, and I see nothing to be gained from making the notion of 'indeterminate form' strict. If some mathematician somewhere decided to state that arg(0) is indeterminate, what would be wrong with that?... FilipeS 23:52, 12 June 2007 (UTC)
Please note I am not in any way trying to make that paragraph a definition, only an explanation. Also there is complete agreement at present about the term "indeterminate form" in every reference I have looked at, and I have looked at quite a few in the process of discussing this article in the last few months. — Carl (CBM · talk) 00:25, 13 June 2007 (UTC)
I don't feel that the status of the indeterminate forms as the discontinuities of the standard arithmetical operations on the extended reals is in any way obvious, especially to the lay reader. I agree that the seven indeterminate forms are pretty much fixed, and would support User:CBM's proposed rewrite. –EdC 21:58, 17 June 2007 (UTC)
I'm against getting overly clever at explanations. Note also that if we refer to discontinuities, we have to be careful, as there is no such thing as a function being discontinuous (or continuous) at a point not in its domain. The proper phrasing is that the function does not have (respectively, has) a continuous extension containing that point. And I think that's getting too involved for this article. --Trovatore 22:04, 17 June 2007 (UTC)
I agree. Plus, f(x, y)=x/y does not have a continuous extension for y=0, x nonzero either, yet the article claims that x/0 is not an indeterminate form. FilipeS 17:38, 19 June 2007 (UTC)

I'm confident that the four people in this discussion all understand perfectly what's going on with indeterminate forms. The question is: if someone in 11th grade asked you for a one-sentence explanation of why these seven forms are called indeterminate but the others are not, what would you say? — Carl (CBM · talk) 18:58, 19 June 2007 (UTC)

I would say that indeterminate forms are mathematical dead ends. When one shows up in a limit you're trying to compute, that's a sign that you must turn back and take a different route to the calculation. FilipeS 01:03, 22 June 2007 (UTC)
Surely her next question would be why they are dead ends? — Carl (CBM · talk) 01:13, 22 June 2007 (UTC)
To convince her, I might give a few examples of different functions whose limit at a given point leads to some indeterminate form (say, 0/0), but that when computed in a different way turn out to be different numbers. FilipeS 01:18, 22 June 2007 (UTC)
You're beating around the bush. It's fully reasonable, and expected, for an encyclopedia article on indeterminate forms to do more than give a list, an example of two limits, and no insight whatsoever. Doing so is really a disservice. I'm planning to remove the paragraph in question - in fact, I'll do it right now - but it should be repalced by something that conveys some intuition about why these forms are indeterminate but others are not. — Carl (CBM · talk) 01:31, 22 June 2007 (UTC)
It's not very civil of you to ask a question and then say you don't like the answer. This is your pet article, not mine. You sort out the OR mess you've made of it. Don't expect any help from me. FilipeS 01:33, 22 June 2007 (UTC)
I don't agree there was OR in the previous version - I've read a lot of published sources and it was a relatively clear statement of the general principle. If you don't wish to help, I encourage you not to respond to my requests for comment. — Carl (CBM · talk) 01:39, 22 June 2007 (UTC)
Funny, then, how you can't seem to be able to come up with any references, or even a coherent definition. If you'd taken any longer to delete that paragraph, I would have done it myself. FilipeS 01:51, 22 June 2007 (UTC)

false statment that should be deleted/fixed (by BrainScratcher)

in the 0^0 form it sayed that if f & g both analytic and g in not fixed zero in the nighberhood of the limit than f(x)^g(x) is always 1 in that limit which is just false it is often the case but consider the example: f(x)=3^x which is positive for all x & g(x)=1/x both zeroing at negative infinity but f(x)^g(x) is clearly 3 whats up with that ?

Your function f is not analytic in a neighborhood of infinity (in fact, it's not even continuous at infinity), and in any case the statement talks about neighborhoods of a point on the complex plane. Infinity is not a point on the complex plane (although it is a point on the Riemann sphere). --Trovatore (talk) 00:50, 25 September 2009 (UTC)

negative infinity over negative infinity

I didn't see any mention of negative infinity in the article. I only mention this because my Calculus book (Calculus - Early Transcendentals 6th Edition by James Stewart) says that -∞/-∞ is also indeterminate. Also, what about the forms of -∞/∞ and ∞/-∞? Are they all indeterminate forms? Thanks. BuddhaBubba (talk) 00:03, 20 July 2010 (UTC)

Yes they're all indeterminate,they're all just variants of ∞/∞, the signs make no difference in a multiply or divide except to determine the final sign. Dmcq (talk) 09:26, 20 July 2010 (UTC)

Tan(90 degrees)

Would Tan(90 degrees) be an indeterminate form. The tan wave goes to infinity at 90. Also if you imagine a right angled triangle and imagine the adjacent side getting smaller and smaller theta gets closer and closer to 90 making tan(theta) closer and closer to tan(90) and therefore undefined. To have a right angled triangle with Tan(90) it's adjacent side would have a length of zero and the hypotenuse and opposite sides would be the same length. 90 would be theta, the angle between the adjacent and opposite sides would be 90 and the third angle would be 0. In effect it would be a straight line. However basic trigonometry still works.

For example, if the hypotenuse and opposite had both got a length of 1 then;

Sin(90)=1/1 which is correct. Cos(90)=0/1 which is correct. Sin(0)=0/1 which is correct. Cos(0)=1/1 which is correct. Tan(0)=0/1 which is correct.

However, Tan(90)=1/0. Which is an indeterminate form. This means that Tan(90) can be converted to 1/0 (and other determinate forms using l'hopitals rule).

If we were trying to work out the length of the hypotenuse and opposite it wouldn't work because we'd get Tan(90)=Hypotenuse/0. This happens because the hypotenuse and opposite could be any length as long as they are the same, the angles would be unchanged.

As well as this Tan(theta)=sin(theta)/cos(theta) so Tan(90)=Sin(90)/Cos(90). Sin(90) is 1 and Cos(90) is 0. Therefore Tan(90) again equals 1/0.

So is Tan(90) an indeterminate form? It tends towards infinity on the Tan graph. The limit of Tan theta as theta approaches 90 is ±∞. Also are all of the above valid arguments? — Preceding unsigned comment added by (talk) 14:40, 20 July 2012 (UTC)

1/0 is not regarded as an indeterminate form, as the limits of quotients that lead to it, if they exist, can only be positive or negative infinity, which isn't an infinite number of possible values. Double sharp (talk) 10:06, 26 October 2012 (UTC)

1^∞ and 0 × ∞ Indeterminate?

These don't seem to be indeterminate based ont he way the article explains the term.


A power is the number times itself that many times. 1 times itself a unlimited number of times would be 1.

0 × ∞

Standard rule is any number times zero is zero ie 0 x X = 0 so it doesn't matter what X is.

If these are indeterminate the article should explain why they are. (talk) 05:31, 18 January 2013 (UTC)

It doesn't look to me like you've read the article all the way through. FilipeS (talk) 12:37, 24 February 2013 (UTC)

What is the definition of "indeterminate form" (is there a widely accepted definition?)

Some sources say that 0/0 is an indeterminate form, but not all. For instance, in "Mathematical Analysis I, Volume 1 By Claudio Canuto, Anita Tabacco", when f(x),g(x) converge to 0, the expression f(x)/g(x) is an indeterminate form when x -> 0, while 0/0 denotes an indeterminate form.

In the subsection "Evaluating indeterminate forms", the word "indeterminate form" refers to f(x)/g(x) (with x -> 0), just like in the definition in the above mentioned book.

In many books, the indeterminate forms are frequently simply listed, without giving any definition. If the main page gives a definition, it should probably cite something for its definition. Moreover, when writing the definition, keep in mind that in some definitions, floor(0) is an indeterminate form, whereas in other definitions, it is not (I have no preference between those options; this is not a form that a calculus student is likely to encounter. But it would be good if the definition is clarified one way or the other). MvH (talk) 16:06, 12 February 2014 (UTC)MvH

My view is that there is no abstract definition, just a list. No, floor(0) is not an indeterminate form; at least, I have never seen it included. --Trovatore (talk) 18:56, 12 February 2014 (UTC)
floor(0) satisfies the current definition on the main page: "if the expression obtained after this substitution does not give enough information to determine the original limit, it is known as an indeterminate form". Many books give no clear definition of "indeterminate form"; in that case, the phrase "intermediate form" probably means the same as "member of the list: 0/0, inf/inf, ...". The current wording on the main page (the list "includes"..) leaves both possibilities open. MvH (talk) 19:16, 12 February 2014 (UTC)MvH
Well, the "does not give enough information" bit refers to "limits involving algebraic operations", and floor is not an algebraic operation. Of course, strictly speaking, neither is exponentiation.
There is a strong temptation in articles like this to go beyond the literature and abstract a definition where in fact none has been given, and this temptation needs to be strenuously resisted. (Oh, I'm not saying a source can't be found somewhere that gives an abstract definition; it probably can, but it's not standard.)
That said, I don't object too strongly to the current wording in the first sentence. It doesn't come across to me as a once-and-for-all "definition", to which I would be sharply opposed, but more as an explanation. Maybe there's some way to reword it to make it even more clear that it's not intending to be a definition. I would support such a clarification as long as it doesn't make it harder to read. --Trovatore (talk) 19:27, 12 February 2014 (UTC)

Revisiting 00 over complex numbers

The article claims

If the functions f and g are analytic and f is not identically zero in a neighbourhood of c on the complex plane, then the limit of f(z) g(z) will always be 1.

This strikes me as patently false: it depends on the "operation" of exponentiation that is ill-defined on the complex numbers and cannot be defined as a continuous function in a punctured neighbourhood of (0,0). Any sensible interpretation will rely on the multivalued complex logarithm, which yields no limit even if one allows limits of multifunctions. Reliance on a single-valued definition requires a "sensible" branch of the exponentiation to be taken, as it seems to me that spiral branches exist that yield any limit that one chooses. A complex function f(z)g(z) in a bland statement like this without defining the exponentiation or mentioning the mathematically esoteric aspects seems out of place here. I notice that this has survived edits by not-so-clueless editors, so perhaps someone could explain why this claim should remain in the article, or where I've gone wrong? —Quondum 17:04, 22 February 2014 (UTC)

There are two quick solutions: (1) The 0^0 section in exponentiation states it better (i.e. no "complex plane"), so you can copy what it says. Or (2) delete this line alltogether. I'm OK with either option. MvH (talk) 16:02, 26 February 2014 (UTC)MvH
I corrected it, and also shortened it a bit. I also fixed one other issue, it said that the indeterminate form 00 has been discussed for a long time, but that's misleading (the debate was whether 0^0 has a value or not; that's not the same as debating if it is an indeterminate form or not, because there is no standard definition of "indeterminate form"). There were two links on this issue in that section, I moved the second one, in order to combine it with the first. MvH (talk) 19:56, 27 February 2014 (UTC)MvH
Much better, thanks. Since this is about indeterminate forms, the debate about the value of the expression is IMO irrelevant here. I might strip that bit out. —Quondum 23:37, 27 February 2014 (UTC)
That's fine (I merged two mentions of the debate into one, but I overlooked that there was a third spot where this was mentioned). MvH (talk) 14:35, 28 February 2014 (UTC)MvH