`lim_(x->0^+) (x^x - 1)/(ln(x) + x - 1)` Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule doesn’t apply, explain why.
The denominator has limit `-oo+0-1=-oo.`
The numerator has limit 0 (I'll prove this below).
Therefore there is no indeterminacy, `0/oo=0.`
(zero is the answer).
Now prove that `x^x->1` when `x->0^+:`
`x^x = (e^lnx)^x = e^(x*lnx).`
`x*lnx = lnx/(1/x),` limit of this exists by l'Hospital's Rule: `oo/oo,` everything is differentiable for x>0, `f'/g' = (1/x)/(-1/x^2) = -x -> 0.`