Are there any set of numbers into which any of the indeterminate forms we see in a calculus course, like 00, n/0, 1infinity, etc has an answer?

I'm asking that because, thanks to the Net, I took notice of other kinds of numbers besides those commonly seen in the high school and most of the university courses: Real and Complex numbers.

There are Hyperreals, Surreals, Quaternions, ... so I thought that some indeterminate form could be have an answer among them, just like we have an answer for the x-th root, x being par, of negative numbers in the set of complex numbers.

$0^0$ and $1^\infty$ are indeterminant forms because when you have limits where the pieces approach those parts, any value is possible. They aren't usually defined even in other number systems because it doesn't mesh well with the limits (although some advanced real analysis books will define $0*\infty$ to be $0$, they just do it so they don't have to write separate cases for some formulas).

However, $n/0$ for nonzero $n$ is not an indeterminate form; the only things limits that look like that can do is approach $\infty$ or $-\infty$ (or both due to oscillation, but the point is that the absolute value approaches $\infty$ in any case). As such, there is a number system where that makes total sense: the real projective line, which adds a single "unsigned infinity" so that things which could only be $\pm\infty$ can be defined.

$0^0$ is an interesting example.

Since

$$\lim_{x->0} x^x = 1$$

it makes sense to define

$0^0 := 1$

This also has advantages concerning the binomial theorem.

But some mathematicians set $0^0 := 0$ and some say it is indetermined. There seems to be no consense what $0^0$ should be.

The other examples you mentioned are even more problematic.

• Not too good. What is $$\lim_{x\to 0^+} x^{1/\ln x}$$ or $$\lim_{x\to 0^+} x^{-1/\ln x}?$$ – Ted Shifrin Feb 1 '14 at 21:07

Indeterminate forms are indeterminate because they could be defined in a different manner and reach different values.

The simplest example is $0\cdot\infty$. This is indeterminate because it depends on what $0$ means and what $\infty$ means.

Notice that in all indeterminate form we take that values may be defined through different limits, $0$ does not have to be $0$ but some expression that evaluates to $0$. So in essence they are indeterminate because some common operations like switching limits does not yield the same result - necessarily, or similar other limitations.

If you however take $0$ literary as $0$ then anything times $0$ is $0$. So from this perspective there is nothing that is undetermined. However, in this case we have defined that $0$ means $0$. This is actually what is meant when we say an indeterminate form. It says: the expression depends on the definition of $0$ and $\infty$.

So no, it is not the situation like $\sqrt{-1}$ where this expression is not a part of the set of real numbers.

It is like the situation when you specify an additional condition for the expressions like $\frac{\sin(x)}{x}$ for $x=0$ in order to achieve a goal of having a nice behaving function.

Take for example that $0 \cdot \infty=5$. This requires $0=\frac{5}{\infty}$ Take a function that reaches $0$ for whatever like $f(1)=0$ and have $\frac{5+f(x)}{f(x)+\infty}$ I ask you: what is $\lim\limits_{x \to 1} \frac{5+f(x)}{f(x)+\infty}$?

There is the catch.

This last expression evaluation totally depends on the game between $f(x)$ and $\infty$. It can be $0$ but it does not have to be.