Math question about degrees/ exponents

Math question about degrees/ exponents?
what is 0 to the zero power/ 0^0....HELP!!!




its not zero!!!


use calculus plz

It's a matter of definition. The function f(x,y) = x^y is not continuous at (0,0). The conventional choice is to leave it undefined, much like 0/0, as an indeterminate form.

In some treatments of discrete math, particularly combinatorics, you'll see 0^0 defined to be 1. That simplifies many proofs without costing a lot. You don't worry too much about limits or continuity in discrete math. It's not a bad choice. Approach (0,0) on any straight-line path *except* along the y-axis and the limit of x^y is 1. (First and 4th quadrants, anyway. Fractional powers of negative x values are problematical in the reals.)

Still, you can find non-straight paths to make x^y approach any non-negative limit you like as (x,y)->(0,0). It's fun to work that out. (Hint: Consider logarithms...)

I believe from my knowledge this is known as "indeterminant."
Assuming you try to solve this with limits...

The limit as x approaches zero from the positive range of x^0 is what?
10^0, 9^0, etc.-->1

And...

The limit as x approaches zero from the positive range of 0^x is what?
0^10, 0^9, etc.-->0

Thus, in general, knowing that the limit of f as x approaches a c is positive, and the limit of g as x approaches c is 0. is not sufficient to calculate the limit of f^g as x approaches c.

If the functions f and g are analytic and f is not identically zero in a neighbourhood of c on the complex plane, then the limit of f(z) g(z) will always be 1.

Sorry if its unclear, difficult to explain without a visual.

Since,
. . . . . x^0 = x/x = 1,
SO,
. . . . 0^0 = 0/0 = undetermined(Infinity) >=====< ANSWER

It depends on what it approaches, I suppose. So we'd need to know the original function. Otherwise, we won't know.