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.

Why 0^0 is undefined

Published by John Quintanilla

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