# What is 0^0?

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?

Relevance
• 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..

• 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.

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

• ITS INDIFINITE

• 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.

• 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.

• Φ² = Φ+1
Lv 7
7 months agoReport

Thank you for asking this question. Now that over a month has passed, can you please award the Best Answer to whichever answer you find most deserving?

• 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 )

• 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.

• Anonymous
9 months ago

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