Can L'Hopital's Rule be applied to limits with multiple zeros?

[rman144]

I have been working with a limit for a while now but cannot for the life of me seem to solve it. Any ideas:


lim[x appr. 0] of (x^x)/((e^x)-1)


I've tried turning x^x into e^(x ln(x)), but the root of my problem is that I'm unsure of whether or not I can use L'Hopital's because technically, (0^0)/0 is not necessarily of the form 0/0.

 

[mathman]

e^x - 1 ~ x. Therefore your expression is ~ x^(1+x) -> 0^1 = 0.

 

[rman144]

You make a good point, but, assuming that is correct, wouldn't it go to:

x^(x-1) >>> oo

 

[lurflurf]

You make a good point, but, assuming that is correct, wouldn't it go to:

x^(x-1) >>> oo

That is right.
This is straight forward
x^x->1
exp(x)-1->0
1/0->infinity

 

[mathman]

You make a good point, but, assuming that is correct, wouldn't it go to:

x^(x-1) >>> oo

You're right. I misread your original expression - I missed the / .