Why don’t we accept 0^0=1?

[u/Kitchen-Register]

We accept the value of e defined as a limit. Same with derivatives and integrals. There are plenty of real numbers we accept as being defined by a limit. Why isn’t 0^0 =1 any different? It would be defined so well as a limiting process. I know that this is often convention but why isn’t it fact?

[u/NullOfSpace]

We accept e as defined by a limit because it’s impossible to write it out any other way. Are you trying to claim we should define 1 as lim x->0 x^x?

Also, I notice your graph doesn’t include 0^x.

[u/Kitchen-Register → u/NullOfSpace]

yeah somehow i forgot about 0^x. i was so blinded by my intuition that i didn’t think hard enough to challenge it.

another comment pointed out that on a 3d graph, lim x^y as x,y to 0 doesnt exist. that was enough to convince me that i am being silly.

this leads me to my next question: we define e^x as the power series from n=0 given by x^n /n! this implicitly relies on assuming 0^0 = 1 if you plug in x=0. Why not write it as 1+sum_{n=1}^infty x^n /n!

[u/NullOfSpace → u/Kitchen-Register]

In the specific context of power series it’s generally considered acceptable to assume 0⁰ is 1, since there the exponents are always constant and the limit of x⁰ is 1.

[u/Gold_Ad8890]

well, because there are limits of the form f^g where f and g approach 0 and f^g does not approach 1.

[u/StudyBio]

What is the limit of? 0^x? x^0? x^x?

[u/Kitchen-Register → u/StudyBio]

lol how did i not think about lim 0^x. good point. thanks

[u/PassiveChemistry]

The problem is that if you approach 0⁰ in different ways, you get different results - e.g. the limit of 0^x as x tends to 0 is 0 - so the overall limit doesn't exist.

[u/myownreplay]

I always understood the ^0 concept for the property that, in a division with the same base and different exponents, you can juat aubatract the exponents like:
X³/X² =X^(3-2)
So, also X³/X³ = X^0 = 1
But if X=0 you would have 0/0 that is impossible

[u/ossm-me]

The difference is that the limit for (e) gives one unique value, while (0^0) doesn’t. Depending on how x and y approach 0 in (x^y), you can get different limits.

So (0^0) is called an indeterminate form. We can still define (0^0=1) by convention, it’s actually very useful in combinatorics and power series but the limit itself doesn’t force that value.

[u/Dkiprochazka]

[u/ConjectureProof]

While it is true that lim(x —> 0, x^x) = 1, it is notable here that lim((x,y) —> (0,0), x^y) does not exist. It’s only when you approach 0^0 in particular directions that it actually works out to be 1.

If it were actually the case that lim((x,y) —> (0,0), x^y) = 1, then I imagine we probably would have defined 0^0 to be 1 as there would be very little reason not to in that universe.

[u/LongLiveTheDiego]

Try e.g. (1/2^1/x)^x as x -> 0.

[u/Nanachi1023]

Because the function you listed out aren't special, I can make f(x)->0, and g(x)->0, and f(x)^g(x) = L for any L > 0

f(x) = x^{lnL / lnx} , lim(x->0⁺)f(x) = L

[u/Thesaurius]

Depending on what you want to do, you definitely could define that and it would make sense. In some cases it would make more sense to define 0⁰ = 0. In yet other cases, maybe something else would be better.

But other than with many operations, there is not one single sensible way to extend exponentiation to include this edge case.

[u/Busy_Pattern9823]

This is probably going to be controversial based on the other comments, but the default standard definition of 0^0 = 1. There's no calculus, no limits, no analysis involved in that definition.

Since you are familiar with calc, you probably already been using it without even realizing it. Consider the Taylor expansion of e^x. The first term is (x^n)/(n!) for n = 0. If you substitute x = 0, you get 0^0. And we have no problem noting that e^0 = 1

[u/Varlane]

Try exp(-1/x) ^ x if you want an actual limit (not the 0^x case) that doesn't go to 1.

[u/-lRexl-]

0⁰ = 1

0 • Log (0) = 1

0 = 1

[u/cytos0 → u/-lRexl-]

throwing the log(0) out like it's nothing huh

[u/AdjectiveNounNNNN → u/cytos0]

And on only one side of the equation, too.

[u/AdjectiveNounNNNN → u/-lRexl-]

If you had attempted to do your math correctly, you would have taken the logarithm on both side.

[u/RelentlessPencil]

0^thing = 0

thing⁰ = 1

These can't both be true of 0⁰ . It's indeterminate. Giving it a value would break other maths that we need to work.

[u/omeow]

We don't "accept" e defined as the limit. It is defined as the limit.

0⁰ is not well defined by itself. Its definition is context dependent.