How do you solve for the limit of x^x as x-->0+

I don't need the numerical answer, I know it's 1. What I don't know is how to get there. I know perfectly well how to use Lhopital's rule for fractions, but again, I'm stumped on this.

Use logarithms. (This is a useful maneuver for indeterminate forms of exponential type.)

Let L = lim(x→0+) x^x.
==> ln L = lim(x→0+) ln(x^x) = lim(x→0+) x ln x.

We can rewrite this in the form "∞/∞" to use L'Hopital's Rule:
ln L = lim(x→0+) ln x / x^(-1)
......= lim(x→0+) x^(-1) / (-x^(-2))
......= lim(x→0+) -x
......= 0.

Since ln L = 0, we have L = e^0 = 1.

I hope this helps!

I think you could take the natural logarithm, find the limit of that, and then exponentiate. It sounds like the limit of the log will be 0.
lim x^x x--->0
lim x ln x
now as a fraction:
lim lnx/*=(1/x)
now this is negative infinity over infinity, so you can use L'Hopital.

Let L = x^x
Then ln L = x ln x

lim ln L {as x --> 0+} = lim x ln x = lim ln x/(1/x) --> -∞/∞

Use l'hospital's Rule: lim 1/x/(-1/x^2) = lim (-x) = 0

Since ln L --> 0, L --> e^0 = 1.

So the limit is 1.