Is $x^{-0}$ defined?

Question

In the form of mathematics that most of humanity is taught, the following operation is undefined:

$\Large{\frac{x}{0}}$

But, how about the following operation?

$\Large{x^{-0}}$

Is the following statement true?

$\Large{x^{-0}=\frac{1}{x^0}=\frac{1}{1}=1}$

Answer 1

Yes. $-0=0$, hence $x^{-0}=x^0=1$.

Addendum: see DonAntonio's comment below.

Answer 2

$x^0=1$ for all $x$ except $x=0$, which some conventions have undefined while others set equal to $1$ (see Zero to the zero power - is $0^0=1$?). This means that $x^{-0}=1$ for all $x$ except $x=0$, for which it is (depending on convention) either undefined or $1$. This has nothing to do with the fact that $x\over 0$ is always undefined.

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I should add that I am strongly in favor of "$0^0=1$" - to my mind there's no question that this is the "right" definition - but there are definitely people out there who disagree, and I have seen some professional mathematicians treat $0^0$ as undefined, so I'm erring on the side of caution above.

Answer 3

$x^y=e^{ylog(x)}$ so $x^0=x^{-0}$ are defined for $x>0$.

Answer 4

A bit off discussion, but it may be dangerous to assign a value to $0^0$ as two non-commuting limits are involved. Any positive real number is possible. Given $a>0$ and 'define' (which I wouldn't recommend doing, though): $$ 0^0 = \lim_{x\rightarrow 0+} x^{y(x)}=\lim_{x\rightarrow 0+} \exp(\log x \; y(x)),$$ where $y(x)$ goes to zero as $x$ goes to zero. Choosing $y(x)=\log a/\log x$ (which indeed goes to zero as $x$ goes to zero): $$ 0^0 = \lim_{x\rightarrow 0+} x^{y(x)} = \exp\left(\log x \frac{\log a}{ \log x}\right) = a.$$ Later edit: This suggests better leave this case undefined when context indicates that powers are in ${\Bbb R}$ or ${\Bbb C}$. There is a broad general concensus (of algebraic and practical nature, binomial formula, Taylor series,...) of defining $x^0 =1$ for any real or complex $x$ whenever the context indicates that powers are in ${\Bbb N}$.

Answer 5

There are two principles: "anything to the power $0$ is $1$" and "$0$ raised to any positive power is $0$". It makes sense to allow the first principle to define $0^0:=1$, because there is no reason for the second principle to extend to $0^0$, as it clearly does not extend to any negative power, and there are abundant examples, such as $\lim_{x\to0+}x^x=1$, which would violate it. That said, one can easily find examples where such a limit may not be $1$: Consider $\lim_{x\to0+}y^x$, which is $0$ in the case that $y=\exp(-1/x^2)$ although both $x$ and $y$ go to $0$. It is probably because of cases like this that some mathematicians are uncomfortable with the convention that $0^0=1$, and prefer to consider it undefined.