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I know this post may be covering a subject that is considered 'low quality' but I wanted to try and cover it in a more advanced manner (before writing I also searched if there were duplicate posts).
I have to teach some students about analysis, but I already wanted to give them tricks to solve limits quickly, just not to be the usual teacher explaining the usual things that everyone can find online.
Having to start the course by explaining the indeterminate forms, I wanted a confirmation regarding a notion that I know, but that I have not found anywhere.
When we have indeterminate forms $\left[\dfrac{0}{0}\right],\;[0^0],\; [1^{\infty}],[0\cdot\infty], [\infty^0]$, we can build the following tables:
($\textbf{ind.}$=indeterminate form, n.d.=not defined)
$$\begin{array}{|c|c|c|c|}\hline \left[\frac{0}{0}\right]&\frac{0^+}{\cdot}&\frac{0}{\cdot}&\frac{0^-}{\cdot}\\\hline \frac{\cdot}{0^+}&\textbf{ind.}&0&\textbf{ind.}\\\hline \frac{\cdot}{0}&\text{n.d.}&\text{n.d.}&\text{n.d.}\\\hline \frac{\cdot}{0^-}&\textbf{ind.}&0&\textbf{ind.}\\\hline \end{array}\qquad \begin{array}{|c|c|c|c|}\hline [0^{0}]&(0^+)^{\cdot}&(0)^{\cdot}&(0^-)^{\cdot}\\\hline (\cdot)^{0^+}&\textbf{ind.}&0&\textbf{ind.}\\\hline (\cdot)^{0}&1&1&1\\\hline (\cdot)^{0^-}&\textbf{ind.}&\text{n.d.}&\textbf{ind.}\\\hline \end{array}$$ $$ [1^{\infty}]\;\Rightarrow\;\begin{cases} (1^+)^{\infty}&\textbf{ind.}\\ (1)^{\infty}&1\\ (1^-)^{\infty}&\textbf{ind.}\\ \end{cases}\qquad [0\cdot\infty]\;\Rightarrow\begin{cases}0^{+}\cdot\infty&\textbf{ind.}\\ 0\cdot\infty&0\\ 0^{-}\cdot\infty&\textbf{ind.} \end{cases}\qquad[\infty^0]\;\Rightarrow\begin{cases}\infty^{0^+}&\textbf{ind.}\\ \infty^{0}&1\\ \infty^{0-}&\textbf{ind.} \end{cases}$$
The sense of these schemes is to show that the indeterminate forms are not all the forms where $0$ and $1$ appear brutally, but only the cases in which the numbers tend to $0$ and $1$.
My doubt is the fact that I have never found these tables in books but honestly they seem very valid, especially in cases where it's asked "If I multiply $1$ infinite times by itself, shouldn't it give $1$?", even to give a sense of the value of $0^0$.
P.s.
For all those who could write me "$0^0$ is indeterminate because $0^0=\exp(0\ln(0))$" I point out that this example is not valid since with the same method I could say that $0=0^2=\exp(2\ln(0))$

Math Attack
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  • $\exp(2\ln(0^+))$ tends to $\exp(2\cdot-\infty)=\exp(-\infty)\to 0$, but $\exp(0\ln(0^+))$ tends to $\exp(0\cdot -\infty)$ where another indeterminate form of $0\cdot \infty$ appears. – Тyma Gaidash Apr 30 '23 at 11:24
  • Correct, but I was talking about $0$ puntcual. With this example you can't even use $0^{-}$, and in you example in "$0\cdot -\infty$", that "$0$" is a puntctual $0$, so it tends to $1$ (you also confused the punctual number with the neighborhood, I have noticed that this is a very common mistake) – Math Attack Apr 30 '23 at 11:35
  • In general I noted also that people treat $\infty$ like a number. $\infty$ is not a number but a concept of "a number that grows indefinitely". If I have $0$ puntctual, the product $0\cdot\infty$ means "zero repeted for a number that grows indefinitely", but this product is $0$, instead if I have $0$ as neighborhood means "a number that tends to 0 repeted for a number that grows indefinitely" – Math Attack Apr 30 '23 at 11:40
  • $\infty$ can be defined as a number in the extended real numbers. Do you mean defining $0\cdot\infty=0$ like in this question so that $\exp(0\cdot\infty)=\exp(1)=1$? – Тyma Gaidash Apr 30 '23 at 11:56
  • In teaching I often used the term determinate form when there was "no indeterminacy". Possibly of interest is this 28 Dec 2007 sci.math post where I mention 4 arithmetic indeterminate forms, 3 exponential indeterminate forms, and 5 logarithmic indeterminate forms. See also these follow-ups: 29 Dec 2007 & 7 Jan 2008. – Dave L. Renfro Apr 30 '23 at 12:10
  • @TymaGaidash sorry but you are saying something really wrong. In $\overline{\mathbb{R}}$ the infinities aren't inserted as number (because infinity is not a number, and I invite you to find a single book that says the opposite), but the neighborhoods of the $\infty$ are inserted. The fact they write $\overline{\mathbb{R}}=\mathbb{R}\cup{-\infty,\infty}$ is only a notation. If $a=\infty$ is a number you should be able to compare $a$ and $2a$. In that question there is an answer that say $"\infty\cdot 0=0"$ in Measure Theory, why? Because in Measure Theory that 0 is punctual 0 and not a limit – Math Attack Apr 30 '23 at 12:11
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    The notion $\overline{\mathbb{R}}=\mathbb{R}\cup{-\infty,\infty}$ is correct when interpreted as equality of "pure" sets (and not as groups, fields, complete ordered fields, metric spaces, topological spaces, or whatever), which is nearly always (if not always) the intention. – Dave L. Renfro Apr 30 '23 at 12:16
  • @DaveL.Renfro you are completely right, I wanted to write it but I was't able to insert it in the comment because it was too long ahahah – Math Attack Apr 30 '23 at 12:20

2 Answers2

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Personally I think that we should stop this "indeterminate form nonsense". Basically when we say $0/0$ is an "indeterminate form" we are just saying that knowing that:

  • The limit of the numerator is $0$,
  • The limit of the denominator is $0$,

is simply not enough information to determine whether the limit exists (and of course, if it exists, its value). So some further analysis has to be done. What you are trying to say is that if the numerator is constantly zero and the denominator converges to zero but it's (almost) never zero, then the limit is $0$ (and this fact is true, but it's not in contradiction with the fact that $0/0$ is an "indeterminate form").

Maybe it's an unpopular opinion, but I think that indeterminate forms are a pedagogical mess.

Kandinskij
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    Exactly, because in my opinion it is explained badly and there is confusion between a number that tends to x and x. Basically when indeterminate forms are explained, numbers are confused with their neighborhoods. This is why in my opinion those tables are valid, because they separate the three cases. – Math Attack Apr 30 '23 at 10:08
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    If a function is constantly equal to $a$ then it tends to $a$. These facts are not mutually exclusive: it's more correct to say that one is a special subcase of the other. So I don't know why you want to emphatize this special subcase. It's no different from saying that a function that decreases exponentially to zero "wins" over a function that decreases linearly. Basically the (almost) constant zero is "the most powerful" zero. – Kandinskij Apr 30 '23 at 10:13
  • The fact is that when dealing with limits, being able to say which of the two zeros "is more powerful" affects you in the result. Basic example: $\lim_{x\to 0}\frac{x}{x^3}=\infty$, $\lim_{x\to 0}\frac{x}{x}=1$ and $\lim_{x\to 0}\frac{x^3}{x}=0$ but if hypothetically in the numerator or denominator you have the punctual 0 in the first case the value is 0 regardless of x, in the second case the division by 0 is not defined, so the limit is not defined – Math Attack Apr 30 '23 at 10:40
  • That's just what I said in the previous comment. – Kandinskij Apr 30 '23 at 10:54
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Indeterminate forms arise in analysis in the context of limits, but it does not mean they cannot be assigned values algebraically. Here is a list of values that create the least troubles algebraically and break the least rules:

  • $0^0=1$ - this is more or less universally accepted, follows from algebraic property of empty product.
  • $\infty^0=1$ - from the same property as the previous one.
  • $1^\infty=1$ - this follows from the algebraic properties of multiplicative unity (it is adsorbing element of exponentiation).
  • $0\cdot\infty=0$ - this follows from the absorbing property of zero.
  • $0/0=0$ - follows from the previous one and associativity.

The remaining indeternimate forms cannot be assigned a value algebraically:

  • $\infty-\infty$
  • $\infty/\infty$
Anixx
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