Why is the limit of x^x (were x-->0) = 1 ?

Question

I know that the limit of x to the x power, where x approaches 0 is equal to 1. How do I show that?

Answer 1

If you simply plug in x=0, you get 0^0, which is an indeterminate form.

Take the natural log of x^x, and compute the limit of that. Since ln(a^b) = b ln a, that means computing

lim(x->0) x*ln(x).

If you plug in x=0, you get 0*-infinity, which is also an indeterminate form. Rewrite it as ln(x)/(1/x), which becomes -infinity / infinity as x->0. This is an indeterminate form that's suitable for l'Hospital's Rule.

Differentiate top and bottom, and you get

lim(x->0) 1/x / (-1/x^2) = lim(x->0) -x^2/x = lim(x->0) -x = 0.

That's the limit of x*ln(x), so exponentiate it to get the limit of x^x, which is e^0 = 1.