What's 0^0? I think its 1.

Proof:
0^0=(a-a)^0=0C0*a^0(-a)^0-0=1*1*1=1 where a's any real number(other than 0).

What's your view?

This is indeterminant form.
However, one can find the limit of x^x as x -> 0 from the right.

We begin like so: x^x = e^(ln(x^x)) = e^(x.ln(x))
Then we find the limit of x In x as x->0+
This is same as lim (x->0+) ln x / (1/x) an indeterminate 0/0 limit form.
By L'Hopital's Rule, this is the same as lim (x->0+) d/dx(ln x) / d/dx(1/x) providing the limit exists
Which is: lim (x->0+) d/dx(ln x) / d/dx(1/x) = lim(x->0+) (1/x) / (-1/x^2) = lim(x->0+) (-x) = -0 = 0
= lim (x->0+) ln x / (1/x)

So lim(x->0+) x^x = lim(x->0+) e^(x.ln(x)) = e^(lim(x->0+) x.ln(x)) = e^(0) = 1
So as x approaches 0 from the right hand side, x^x approaches 1.

Hello friend Raghavendran, no it is not 1.
Since a^0 means a/a, so 0^0 must be equated to 0/0 which is not 1
Hence 0^0 is in determinant.
Moreover 0C0 too is in determinant

it is an indeterminate..it was included in the first part of the indeterminate forms..

0^0 = @_@ = T.T = :) = (=D) = xD
So, 0^0 = xD
It does not exist..xD
haha..

x=0^0
log x =0log0
e^(0log0)=x
and we know 0 * any undefined number =0
therefore,
e^0=x
x=1.

Its indeterminate. The answer does not exist

Right, i didnt think you would need to prove it to us though, it should be common sense.

What if a is 0 then your proof falls apart. I think its indeterminate.