Question

lim x->o+ (x^x - 1)/ (lnx +x -1) find limit using l'Hospital's Rule

Answer 1

Do you have to use L'Hospital? I don't need you need and I am unsure that you can. Bottom goes to infinity, top seems to go for some constant value, either 0 or 1.

Answer 2

yea i have to use L'Hospital

Answer 3

Is that:\[\lim_{x -> 0+} \frac{x^x - 1}{lnx + x -1}\]?

Answer 4

yes

Answer 5

i know its supposed to be in a 0/0 form but the lnx gives me a decimal when i plug in 0

Answer 6

Well, let's just plow through the L'Hospital. I will think this over in a bit. But the bottom derivative is more or less straightforward, right? It's D{lnx} + D{x} + D{-1} = 1/x + 1 + 0. The hardest part is the top, so remember the chain rule?\[D(x^{x}) = x^xD(x) + x^{x}\ln{x}D(x) = x^{x}(1 + lnx)\]So we are left with:\[\frac{x^x(1+lnx)}{\frac{1}{x} + 1}\]And that does not seem really simplified. That's my problem with this, D(x^x) = x^x*f(x). That never simplifies. Don't know, maybe try L'Hospital again?