find limx^(1/(1-e^-x) when x -- 0

Question

find limx^(1/(1-e^-x) when x --> 0

gg · Answer

$$\lim_{x \rightarrow 0}x ^{1/(1-e ^{-x})}$$

gg · Answer

$$\lim_{x \rightarrow 0}x ^{1/\ln (1-e ^{-x})}$$

gg · Answer

the last one is correct

anonymous · Answer

Take it piece by piece$$e ^{0}$$is what?

gg · Answer

it's 1

anonymous · Answer

$$\ln (1-1)$$=What is $$\ln 0$$It is a trick question.

gg · Answer

$$-\infty$$

gg · Answer

?

anonymous · Answer

Great. I never got that right each time my teacher asked. You are ahead of the game. So put it together, what is the answer?

gg · Answer

I got x^0, but x goes to 0, so I got 0^0

anonymous · Answer

Actually 1 over 0 to the power of infinity, I think. Either way our little game ended in a indeterminate form. This is one where you set y equal to your original thingy. Take ln y and take ln of your thingy. You ever did any of those?

gg · Answer

I don't understand what do you want to say, I am not so good with English. 
Can you just write solution?

anonymous · Answer

$$\ln y =\ln x ^{1/\ln(1-e ^{-x)}}$$

gg · Answer

and now I have to find limes of this? That's all? :)

anonymous · Answer

Take lim of each side

gg · Answer

ok :) thank you very much :)

anonymous · Answer

I forgot: L'hopital

gg · Answer

what L'hopital ? :)

anonymous · Answer

L'hopital = 1) derivative, 2) limit. Does not work here.