After 6 years I still don't understand this

Also, using limits says nothing about the actual value at 0. So the limit way doesn't help. We at least get to show that those functions tend to different values, but those functions aren't continuous at x=0... so the two limits you showed don't prove that it's undetermined... technically. Guess it's a nice way of trying to make someone realize that that is the case though.

Oops, can't prove it, but can at least show that there is some weird things going around!

If you approach this from 0-, you get 0/0 (indeterminate). Only if you approach it from 0+ do you get 0. Therefore, because there is a discontinuity here, it would make more sense to define 0^0 using the first limit you provided,

It's a very good idea to try and approach this from 0- to check out what the actual value is, but that's where we head into some problems: 0^x for negative values of x is not defined (0/0 for all negative values). So we can't really say that 0^x is not defined for x = 0, though that would have been a really good idea. It's not a discontinuity per say, because it's just a one-sided test, and we can't test it from the other side.

it would make more sense to define 0^0 using the first limit you provided, since it is continuous at all points except for that one hole at 0. In other words, both x^-0 is 1 and x^+0 is 1, whereas only 0^+x is 0. For a limit to be defined at a single point, both sides of the limit must be satisfied.

The main idea behind the choosing of those limits is just to show that for some limits, it goes to zero, as for others, it goes toward one.

 Spoiler

I assume you already know this, but I'll say it just in case: You cannot really define a point through limits, so it doesn't really matter whether I can show that the limit is defined for a single point or not. I can make my own function:



And if I take the limits from both the positive and negative side, you see that



Even though the function is equal to pi when x = 1. Same applies to this function:



But we know that it is indeterminate at that point.

Otherwise, the step function f(x) = floor(x) would equal both 1 and 0 at x=1, when it can only actually exist at one of the two points or the function would not be a function.

True enough. What we need to find though, is the actual value of 0^0. If we knew that answer as a rational number, it would be easy to find floor(0^0).

Furthermore:


If x = 0, there is just no possible selection of y that can make k = 0.

Another great example that 0^0 is a weird thing!

This Schrödinger-number is funny, as we cannot look inside the theoretical box.

Noah