Advanced methods to explain indeterminate forms

Question

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 $0=0^2=\exp(2\ln(0))$

Answer 1

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.

Answer 2

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:

• $\infty-\infty$
• $\infty/\infty$