Simple concept about exponential

The answer is 1.

Limit x --> 0- of x^x = 1 and

Limit x --> 0+ of x^ x = 1 and so

Limit x --> 0 of x^x = 1

Therefore, 0^0 = 1.

Proof:

x^x

= e^(x Ln x)

= e^[Ln x/ (1/ x)]

--> e^[(1/ x)/ (-1/x^2)]

= e^(- x)

--> e^ 0

= 1

Ln (-1) doesn't matter for 0- is allowed. 0^0 = 0^(1 - 1) = 0/ 0 can be 1, as an indeterminate includes 1. (0) (n) = 0 => n = 0/ 0 for n can be some number. As x --> 3, (x - 3)(x+ 3)/ (x - 3) --> 6 as a form of 0/ 0 but NOT undefined.

0^(N - N) = 0^ N/ 0^ N becomes 0^0 because of same number of zeroes which is therefore gives a sense of 1 in limit. 0/ 0 does not emphasize the number of zeroes in numerator or denominator. This makes 0/ 0 to differ from 0^0.

For x^x, the index has a stronger influence to make 1 than the influence of base to make 0. This is obvious in e^0. As 0^0 is reasonable to be 1, we should set or change the answer into 1 instead of undefined. (Mathematics are reasonable things which should subject to correcting acts with only reason we can posses. Otherwise, we are not following the truth.)

I remembered there was one case when I met 0^0 which ought to be 1 particularly but I can't state as I don't remember exactly. If you find any other form of 0^0 not reasonable to be 1, then please write here to tell me. Thanks!