# Nike Dri Fit Leggings Ladies

﻿0 is an indeterminate form and has no specific solution

how you define it, since, as you point out, x^y should be zero if you approach (0,0) along the x=0 axis, and it should be one if you approach along the y=0 axis.0^0 is simply an expression that doesn't make sense. There isn't an answer, and there certainly isn't something we could agree to define it as. It is gibberish, nothing more, nothing less. One cannot assume just because there are mathematical symbols on paper that they make sense. Nike Dri Fit Leggings Ladies

Thanks, I was gonna say the same thing. :)

What is "indeterminate form"? What does it mean for expression to "have a specific solution"?You see, 0^0 = 1, and it's obvious to a mathematician. The only problem is that the function f: [0, \infty) x R R, f(x, y) = x^y is discontinuous in (0, 0) and that's what causes problems for instance, this is the source of the whole "indeterminate form" notion. If a function Nike Shorts Navy Blue

(a, b), then for every two sequences a_n, b_n, such that lim a_n = a, lim b_n = b, we have lim f(a_n, b_n) = f(a, b). That's why lim (a_n)^(b_n) = a^b if (a, b) != (0, 0), and this is "determinate form". But if (a, b) = (0, 0), then no matter how we define 0^0, it does not follow that lim (a^n)^(b^n) = a^b = 0^0, because in this case, lim (a_n)^(b_n) can be every positive value, and so mathematicians used to call it "indeterminate form" (it's not common today, though). The thing is, most people do not know, _why_ expressions like pi^e are supposed to make sense they just take exponentiation as given. Only then they need to make up some explanation why "exponentiation rules" are like this, and what exponentiation is about. . . we define 0^0 = 1, to be consistent with exponentiation rulesWell, you're going to be inconsistent with them no matter Pants Nike Hombre

True. That algorithm is not quite the definition of exponentiation, because that algorithm can't really be extended to work outside rational exponents. Exponentiation is defined across complex numbers (ignoring 0^0 for the moment). I think this is acceptable because I'm only shooting for an explanation, which doesn't need to be strict.

f is continuous in Nike Sports Bra Girls