Question

Q. what is the result of 0 power of o ?

Answer 1

Answer on Question #57064 – Math – Combinatorics | Number Theory

Question

What is the result of 0 power of 0?

Solution

In fact, $0^0$ is not defined.

Nevertheless, there is an agreement in mathematics that $0^0 = 1$ like $0! = 1$. It is also accustomed in many different calculators and software applications.

These arguments arise in algebra.

On the one hand, it is true that $0^0 = 1$, because by the definition of power:

$$a^n = 1 \cdot \underbrace{a \cdots a}_{n \text{ times}}.$$

When $a = 0$ and $n = 0$ we have:

$$0^0 = 1 \cdot \underbrace{0 \cdots 0}_{0 \text{ times}} = 1.$$

On the other hand, it is true that $0^0 = 0$:

$$0^x = 0^{1 + x - 1} = 0^1 \cdot 0^{x - 1} = 0 \cdot 0^{x - 1} = 0,$$

which is true since anything times 0 is 0. That means that

$$0^0 = 0$$

The next arguments arise in calculus.

On the one hand, the limit of $x^x$ as $x$ tends to zero from the right is 1. In other words, if we want the $x^x$ function to be right continuous at 0, we should define it to be 1.

On the other hand, the function $f(x,y) = y^x$ has a discontinuity at the point $(x,y) = (0,0)$.

In particular, when we approach $(0,0)$ along the line with $x = 0$ we get

$$\lim_{y \to 0} f(0, y) = 1$$

But when we approach $(0,0)$ along the line segment with $y = 0$ and $x > 0$ we get

$$\lim_{x \to 0^+} f(x, 0) = 0.$$

Therefore, the value of $\lim_{(x,y)\to (0,0)}f(x,y)$ depends on the direction that we take the limit. This means that there is no way to define $0^0$ that will make the function $y^{x}$ continuous at the point $(x,y) = (0,0)$.

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