0^0 wishes it were equal to 1 (but it’s not)
Hey folks. It’s been a long time. I’ve had lots of things on the brain, along with lots of new responsibilities, and I just kind of fell out of the posting spirit around here (though I’ve still been reading others’ blogs regularly).
Anyway, maybe a short and easy math post will get me back into the swing of things (no promises on that). Here’s something irksome. It’s common knowledge that for a real number x, we have x^0 = 1. Well, not just any real number - x can’t itself be zero. People since the time of Cauchy have called 0^0 an “indeterminate form”. Ugly, ugly.
0^0 wants to equal 1, and here’s one of several compelling reasons why:
Nevermind the orange graph - for almost all negative x, we need complex numbers to make sense of x^x, but for simplicity we can just restrict our attention to positive x.
Well, I won’t go through the computation of this limit - you can just go to Wolfram Alpha and click “show steps”. But doesn’t the fact that this limit equals 1 suggest to us, at the top of its lungs, so to speak, that we should go ahead and define 0^0 := 1?
Unfortunately doing so would be hasty and lead to problems. The key point is we have only demonstrated one instance in which 1 is a sensible definition for 0^0. Although x^x admittedly provides us a very nice model for thinking about 0^0, it’s not the only available one, and by no means is it special, natural, canonical, universal or other such adjectives. It’s merely pulled out of thin air.
There’s still a gleam of hope in favor of defining 0^0 := 1, though! Perhaps if we take any two functions f(x) and g(x) such that both have limits of 0 as x approaches 0, then we’ll find the limit as x goes to 0 of f(x)^g(x) equals 1. This would be tremendous evidence in favor of 0^0 := 1.
Sadly, it doesn’t work. I invite you to come up with two such functions f and g which break our goal. I’ll reveal a solution below. But first, a couple examples of menacing looking pairs f and g which do work as nicely as we’d like. First we take f(x) := tan(x) and g(x) := x^3 - x^2 - x, both of which go to 0 as x does.
Splendid. Now let’s try taking f(x) := ln(1 + x^2) and g(x) := sin(x^(1/3)). Again, both functions go to 0 as x goes to 0.
Wow! If our hopeful identity holds for such gnarly functions, it must take a really hideous pair to break it, right?
Yes and no. We will use one somewhat hideous function, taking our f(x) to be e^(-1/x). As x approaches 0 through positive numbers, the limit is 0 (since -1/x is going to negative infinity). The left-sided limit comes out different (namely to infinity), which means f has no hope of being continuous at 0, but for us that doesn’t matter. Now take g(x) := x, as nice and simple as can be. We look at (e^(-1/x))^x - but by simplifying, this is just equal to e^(-1) for nonzero x. And so the limit as x approaches 0 for this pair is 1/e.
…ugly, ugly. Nobody told poor 0^0 about such freakish occurrences back when it first dreamed of equaling 1.
It seems we have no choice but to leave 0^0 undefined, since on some occasions it wants to be 1 and on others it wants to be 1/e (and you can be sure there will be other times it wants to be something else). And after all, there are times we want to use l’Hôpital’s rule to evaluate a limit of the form 0^0 and get some nontrivial function of x as the answer; it would sure be stupid and break a lot of calculus if such limits were automatically just equal to 1.
