What is 0^0?

Question

On desmos, if f(x)=x^x, f(0)=1. The function's domain is also restricted to x>=0. I do not know why since -1^-1 is -1 and -2^-2 is -1/4.
In school, I was taught that x^0=1 and 0^x=0 so why is one overriding the other?
So why is x^x restricted to x>=0 and why does 0^0=1?

Answer 1

what is 1 to the power 2..1*1 ...so...1^2 =1
what is 2 to the power 3... 2*2*2 so 8
then what is 0 to the power 0?
it is 0
Physically speaking, it is nothing..i mean..it is saying like nothing has happened...i.e 0(nothing) raised to the power 0(nothing).
some call it indeterminate form also..

Answer 2

It is indeterminate (you cannot determine the value).
The rule for limits is that the limit must be the same, regardless of the "direction" you take to get there.
Bluntly, a limit must be unique.
As soon as you can show more than one limit value, then the value does not exist and the operation cannot be determined.

Let's look at n^0
and allow n to get closer and closer to zero (for example, n = 1/2, 1/4, 1/8...)
The value of n^0 will be 1 for all values of n, therefore the limit is "1"
If this is valid, then 0^0 = 1

Now consider 0^n
and allow n to get closer and closer to zero.
The value of 0^n will be 0 for all values of n, therefore the limit is "0"
If this is valid, then 0^0 = 0

Both operations are valid, yet they reach a different limit. Therefore the limit does not exist.

Answer 3

While the actual answer is "indefinite", I would think it would be agreed upon that the value sould be '1'

Answer 4

Indeterminate form

Answer 5

ITS INDIFINITE

Answer 6

1 = (y^x)/(y^x) = y^(x-x) = y^0
But I don't think y can be 0, cause 0^0 is undetermined.
If you calculate, for instance, .0001^.0001, you'll see the result is .99, but not 1.

Answer 7

In school you were taught that 0^(-1) = 1/0 = 0 ?

We know that 1^1 = 1, so by binomial expansion
1 = 1^1
1 = (1+0)^1
1 = 1 * 1^1 * 0^0 + 1 * 1^0 * 0^1
1 = 1 * 1 * 0^0 + 1 * 1 * 0
1 = 0^0 + 0
then 0^0 = 1

This may not be strictly true, but it is at least a useful value. Like 0! = 1.

Answer 8

0^0 = 0...but limit as x ---> 0+ of f(x) = x^x = 1...thus f(x) is not continuous at x = 0...and the proper definition of x^x is e^(x ln x )

Answer 9

You must find the limit as x approaches 0 of f(x) = x^x. Use can use l'hopitals rule or just put in a small number for x and see what happens. If x=.001 then x^x is close to 1.

X can not be negative because of fractional negative values. If x = -1/2 then x^x = 1 / sqrt(-1/2) and you can not take the square root of a negative number.

Answer 10

Any number to the zeroth power is 1. At least, that's what I remember.