# what is limit of x^x as x tends 0

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Answer is 1.

Consider the right-hand limit:

lim x^x

x -> 0+

Use the property y = e^(ln(y)). Since e and natural log are inverses of each other, they cancel each other out.

lim e^(ln(x^x))

x -> 0+

Use the log property to move the x outside of the log.

lim e^[x ln(x)]

x -> 0+

Move the limit inside of the exponent.

e^[ lim (x ln(x)) ]

. . x -> 0+

Now the limit is of the form [0*infinity], which is indeterminate. Let's move x to the denominator as (1/x).

e^[ lim ( ln(x)/(1/x) ) ]

. . x -> 0+

How the form is [infinity/infinity]. Use L'Hospital's rule.

The derivative of ln(x) is (1/x). The derivative of (1/x) is (-1/x^2).

e^[ lim ( (1/x)/(-1/x^2) ) ]

. . x -> 0+

Simplify.

e^[ lim ( (-x^2)/x ]

. . x -> 0+

e^[ lim (-x ) ]

. . x -> 0+

And evaluate the limit

e^0

Which is just

1

As for the left-hand limit, I believe a similar argument could work, since the imaginary part approaches zero, but I'm not sure of the details.

**Sources:**

**limit of x^x as xtends to 0 is indeterminate **