# Why 0^0 is undefined

Here’s an explanation for why $0^0$ is undefined that should be within the grasp of pre-algebra students:

Part 1.

• What is $0^3$? Of course, it’s $0$.
• What is $0^2$? Again, $0$.
• What is $0^1$? Again, $0$.
• What is $0^{1/2}$, or $\sqrt{0}$? Again, $0$.
• What is $0^{1/3}$, or $\sqrt[3]{0}$? In other words, what number, when cubed, is $0$? Again, $0$.
• What is $0^{1/10}$, or $\sqrt[10]{0}$? In other words, what number, when raised to the 10th power, is $0$. Again, $0$.

So as the exponent gets closer to $0$, the answer remains $0$. So, from this perspective, it looks like $0^0$ ought to be equal to $0$.

Part 2.

• What is $3^0$. Of course, it’s $1$.
• What is $2^0$. Again, $1$.
• What is $1^0$. Again, $1$.
• What is $\left( \displaystyle \frac{1}{2} \right)^0$? Again, $1$
• What is $\left( \displaystyle \frac{1}{3} \right)^0$. Again, $1$
• What is $\left( \displaystyle \frac{1}{10} \right)^0$? Again, $1$

So as the base gets closer to $0$, the answer remains $1$. So, from this perspective, it looks like $0^0$ ought to be equal to $1$.

In conclusion: looking at it one way, $0^0$ should be defined to be $0$. From another perspective, $0^0$ should be defined to be $1$.

Of course, we can’t define a number to be two different things! So we’ll just say that $0^0$ is undefined — just like dividing by $0$ is undefined — rather than pretend that $0^0$ switches between two different values.

Here’s a more technical explanation about why $0^0$ is an indeterminate form, using calculus.

Part 1. As before,

$\displaystyle \lim_{x \to 0^+} 0^x = \lim_{x \to 0^+} 0 = 0$.

The first equality is true because, inside of the limit, $x$ is permitted to get close to $0$ but cannot actually equal $0$, and there’s no ambiguity about $0^x = 0$ if $x >0$. (Naturally, $0^x$ is undefined if $x < 0$.)

The second equality is true because the limit of a constant is the constant.

Part 2. As before,

$\displaystyle \lim_{x \to 0} x^0 = \lim_{x \to 0} 1 = 1$.

Once again, the first equality is true because, inside of the limit, $x$ is permitted to get close to $0$ but cannot actually equal $0$, and there’s no ambiguity about $x^0 = 1$ if $x \ne 0$.

As before, the answers from Parts 1 and 2 are different. But wait, there’s more…

Part 3. Here’s another way that $0^0$ can be considered, just to give us a headache. Let’s evaluate

$\displaystyle \lim_{x \to 0^+} x^{1/\ln x}$

Clearly, the base tends to $0$ as $x \to 0$. Also, $\ln x \to \infty$ as $x \to 0^+$, so that $\displaystyle \frac{1}{\ln x} \to 0$ as $x \to 0^+$. In other words, this limit has the indeterminate form $0^0$.

To evaluate this limit, let’s take a logarithm under the limit:

$\displaystyle \lim_{x \to 0^+} \ln x^{1/\ln x} = \displaystyle \lim_{x \to 0^+} \frac{1}{\ln x} \cdot \ln x$

$\displaystyle \lim_{x \to 0^+} \ln x^{1/\ln x} = \displaystyle \lim_{x \to 0^+} 1$

$\displaystyle \lim_{x \to 0^+} \ln x^{1/\ln x} = 1$

Therefore, without the extra logarithm,

$\displaystyle \lim_{x \to 0^+} x^{1/\ln x} = e^1 = e$

Part 4. It gets even better. Let $k$ be any positive real number. By the same logic as above,

$\displaystyle \lim_{x \to 0^+} x^{\ln k/\ln x} = e^{\ln k} = k$

So, for any $k \ge 0$, we can find a function $f(x)$ of the indeterminate form $0^0$ so that $\displaystyle f(x) = k$.

In other words, we could justify defining $0^0$ to be any nonnegative number. Clearly, it’s better instead to simply say that $0^0$ is undefined.

P.S. I don’t know if it’s possible to have an indeterminate form of $0^0$ where the answer is either negative or infinite. I tend to doubt it, but I’m not sure.