Undetermined or indeterminate forms: $\frac{0}{0}, \frac{\infty}{\infty}, 0\cdot\infty, 1^\infty, 0^0, +\infty-\infty$ I wanted to know who has decided that for the calculation of the limits of the following forms,
$$\color{orange}{\frac{0}{0},\quad \frac{\infty}{\infty},\quad  0\cdot\infty,\quad  1^\infty,\quad  0^0,\quad  +\infty-\infty}$$
are called indeterminate forms. For example, it would be spontaneous to me to say that
$$1^{\infty}=1\cdots 1 \cdots 1 =1$$
or
$$0^0=1$$
 A: It depends where you look. Many people define $0^0 = 1$. The problem is that there is not one solution that makes sense. Take for example $1^\infty$. You would like to have some kind of continuity, but for every $x > 1$, you have $x^\infty = \infty$ and for $0 < x < 1$, you have $x^\infty = 0$. Why should $1^\infty$ have one specific value in between?
For $0^0$, you can plot $x^y$ on wolframalpha and you will see that there are many possibilities to define $0^0$ as limit $x \to 0$ and $y\to 0$, depending on the direction you take.
As for fractions of the form $0/0$ or $\infty/\infty$, how do you want to define them? As limits? How would you like to distinguish
$$
\lim_{x\to 0} \dfrac{x}{x} = 1
$$
or
$$
\lim_{x\downarrow 0} \dfrac{x}{x^2} = +\infty
$$
or
$$\lim_{x\uparrow 0} \dfrac{x}{x^2} = -\infty$$ or
$$\lim_{x\to 0} \dfrac{x^2}{x} = 0?$$ There are so many limits which could be interpreted as $0/0$ that it makes no sense to choose between them. And this is true for every indeterminate form.
A: The following expressions are all called indeterminate forms:
$${\frac{0}{0},\quad \frac{\infty}{\infty},\quad  0\cdot\infty,\quad\infty^0,\quad  1^\infty,\quad  0^0,\quad  \infty-\infty}$$
Remember that $\infty$ is not a real, honest number, but a shorthand for a limiting process. Here are more precise statements of what it means that each of these forms are indeterminate:

*

*$\frac{0}{0}$ is indeterminate, because knowing that ${\underset{x\to c}{\lim}}~{f(x)}=0$ and ${\underset{x\to c}{\lim}}~{g(x)}=0$ is not enough information to determine $\underset{x\to c}{\lim}\frac{f(x)}{g(x)}$.

*$\frac{\infty}{\infty}$ is indeterminate, because knowing that ${\underset{x\to c}{\lim}}~{f(x)}=\infty$ and ${\underset{x\to c}{\lim}}~{g(x)}=\infty$ is not enough information to determine $\underset{x\to c}{\lim}\frac{f(x)}{g(x)}$.

*$0\cdot{\infty}$ is indeterminate, because knowing that ${\underset{x\to c}{\lim}}~{f(x)}=0$ and ${\underset{x\to c}{\lim}}~{g(x)}=\infty$ is not enough information to determine $\underset{x\to c}{\lim}{f(x)}\cdot{g(x)}$.

*$\infty^0$ is indeterminate, because knowing that ${\underset{x\to c}{\lim}}~{f(x)}=\infty$ and ${\underset{x\to c}{\lim}}~{g(x)}=0$ is not enough information to determine $\underset{x\to c}{\lim}{f(x)}^{g(x)}$.

*$0^0$ is indeterminate, because knowing that ${\underset{x\to c}{\lim}}~{f(x)}=0$ and ${\underset{x\to c}{\lim}}~{g(x)}=0$ is not enough information to determine $\underset{x\to c}{\lim}{f(x)}^{g(x)}$.

*$1^\infty$ is indeterminate, because knowing that ${\underset{x\to c}{\lim}}~{f(x)}=1$ and ${\underset{x\to c}{\lim}}~{g(x)}=\infty$ is not enough information to determine $\underset{x\to c}{\lim}{f(x)}^{g(x)}$.

*$\infty-\infty$ is indeterminate, because knowing that ${\underset{x\to c}{\lim}}~{f(x)}=\infty$ and ${\underset{x\to c}{\lim}}~{g(x)}=\infty$ is not enough information to determine $\underset{x\to c}{\lim}(f(x)-g(x))$.

In contast, forms like $\frac{0}{\infty},\infty+\infty,$ and $0^\infty$ are
determinate, because if they arise in a limit, the answer is
unambiguous (in these cases, these forms resolve to $0,\infty, 0,$ respectively).
A: Well, $\infty$ is not a number, so we call the things you listed "forms".  They are short hand notations for various limits.   The trouble with saying
$$1^\infty = 1$$
is that you can find a function $f(x)$  whose limit at infinity is $1$ and a function $g(x)$ whose limit at infinity is $\infty$, but
$$\lim_{x\to \infty} f(x)^{g(x)} = 5.$$
And you can find functions so that $5$ can be replaced by any number you please.
So if you have a limit of one of these "forms", you can't know the answer right off the bat.  Something that goes to $1$ raised to something that goes to $\infty$ could turn out to be anything.
A: To see these are indeterminate forms consider the respective $x\to0^+$ behaviours of$$\frac{\pm cx}{x},\,\frac{c/x}{1/x},\,\pm x\cdot\frac{c}{x},\,(1+x)^{c/x},\,\left(e^{-1/x^2}\right)^{\pm cx^2},\,\frac1x-\left(\frac{1}{x}\pm c\right)$$for $c\ge0$ (with $c\ne0$ whenever we have a $c/x$). These can achieve any value in $\Bbb R\bigcup\{-\infty,\,\infty\}$, except:

*

*$\infty/\infty$ can only be $\ge0$;

*$1^\infty$ can only be $\ge0$, and the above example doesn't show how to achieve $0$ (for that, consider $(1+x)^{-1/x^2}$);

*$0^0$ can only be $\ge0$, and the above example doesn't show how to achieve $0$ (for that, consider $(e^{-1/x^4})^{1/x^2}$).

The $c/x$ uses are also worth explaining further:

*

*To get $\infty/\infty=0$, consider $x^2/x$;

*To get $0\cdot\infty=0$, consider $x^2\cdot\frac1x$;

*To get $1^\infty=1$, consider $(1+x^2)^{1/x}$.

Lastly, let me expand on an ambiguity @MushuNrek's answer discusses. So far, I've looked at $x\to0^+$ limits. But in combinatorics, it makes sense to say $0^0=1$. If in an abuse of notation we take $\infty$ to be a transfinite cardinality, we can similarly give combinatorial meaning to $0\cdot\infty=0,\,1^\infty=1$.
A: Let me say, that in some books, for example in real variable(I.P. Natanson, Theory of Functions or real variable, 80p, Russain ed.) or measure theory (Halmos P.R. - Measure Theory, 1p), is allowed to have sense for $0\cdot (\pm \infty) =\pm \infty \cdot 0 = 0$, as author(s) count it more convenient.
A: A simple answer without too much technical detail:
With a lot of expressions, Mathematicians have to try to figure out exactly what value they should assign to them that still preserves "nice properties" that they want. An example is square rooting a number; if we start with the simple definition that if ${n}$ is a positive integer then you can define
$${x^n=\underbrace{x\cdot x\cdot ...\cdot x}_{\text{$n$ times}}}$$
Then one can quickly show that
$${x^ax^b=x^{a+b}}$$
So if we want to extend and try to find something like
$${x^{\frac{1}{2}}=?}$$
for example, what is the answer? Clearly from the original definition this doesn't make any sense. But from the property above we can define ${x^{\frac{1}{2}}}$ to be any number satisfying
$${x^{\frac{1}{2}}\times x^{\frac{1}{2}}=x^{\frac{1}{2}+\frac{1}{2}}=x^1=x}$$
(and you know this as the square root). In the same way, expressions like ${\frac{0}{0},\frac{\infty}{\infty}}$... are expressions that don't have any meaning by current definition, so we have to try and extend them in a "nice way" - but it turns out there is no nice way to extend them. A really easy example without limits is ${\frac{0}{0}}$. If we claim ${\frac{0}{0}=1}$, Then using just two rules of Algebra we can prove an inconsistency:
$${2\times \frac{0}{0}=\frac{2\times 0}{0}=\frac{0}{0}=1}$$
But on the other hand
$${2\times \frac{0}{0}=2\times 1=2}$$
So we have gotten ${2=1}$, which is obviously nonsense. There are contexts where maybe it makes sense to define ${\frac{0}{0}=1}$ - but in this example, if you want to preserve standard Algebra so that everything remains consistent with our current rules - no extension is possible (substitute ${\frac{0}{0}=a}$ for any number $a$ and you can find an inconsistency in one way or another) As others have also pointed out - you can also show problems with a lot of these expressions by taking limits of functions that approach these expressions, and get different answers depending on the functions you pick.
