Why 0^0 is undefined

TI00

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.

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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.

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