Using Wolfram Alpha to prove 1=0

Unfortunately the author seems uninformed. He claims that 0^0 = 1, not indeterminate which is what Wolfram Alpha claims (and as anyone knows, if you use indeterminates you can prove any number equals any other number, which the author does in uncached screenshots). The problem though is that 0^0 changes its meaning depending on the context. If you have an function f(x,y) = x^y, it's not continuous at (0,0) because the limit of x -> 0 is 0 and the limit y -> 0 is 1, making the value undefined! (Which is still different from 1, 0, or indeterminate.) I used to know of a better blog post on the issue, but here's http://mathforum.org/dr.math/faq/faq.0.to.0.power.html

Oh wait, I found it: http://www.askamathematician.com/?p=4524 "Zero raised to the zero power is one. Why? Because mathematicians said so."

Sort of in the same character of dispute is the fact that 0 and 1 aren't probabilities: http://lesswrong.com/lw/mp/0_and_1_are_not_probabilities/

I'm also unsure what the author is entering since the screenshot isn't cached. If I try http://www.wolframalpha.com/input/?i=summation+x^n%2Fn!+from... wolfram gives the correct e^0 = 1.

Edit: Ah, he's using http://www.wolframalpha.com/input/?i=summation+0^n%2Fn!+from... (substituting 0 in explicitly instead of instructing Wolfram to substitute on x) which is quite an interesting quirk to give 0.

I thought so because for x → 0,

0^x → 0

x^0 → 1

x^x → 1

and I couldn't find any other expressions of the type "0^0" that have a limit different from 0 and 1. I even wrote a "proof" for it, which of course was flawed, but it took others (and me) quite some time to find the exact mistake in my "proof".

0^0 is completely indefinite in the sense that you can find "0^0" expressions for any limit a you want it to be. Here's an example for that kind of expressions:

x ^ (ln a / ln x) → a

Although I wasn't able to come up with those as a schoolboy, this is an easy exercise for any math student.

http://webcache.googleusercontent.com/search?q=cache:nY9J5bU...

From my (fuzzy) math memories, I thought that:

exp(x)= 1 + (x^1/1!) + (x^2/2!) + ... + (x^n/n!) + o(x^n) _near_ x=0