Yas asked in Science & MathematicsMathematics · 11 months ago

# 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
• 11 months ago

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.

• 11 months ago

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..

• 11 months ago

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.

• 11 months ago

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

• 11 months ago

Indeterminate form

• 11 months ago

ITS INDIFINITE

• 11 months ago

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.

• 11 months ago

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
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9 months agoReport

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• ted s
Lv 7
11 months ago

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 )