Question

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

i know that i have to use L'Hopitals Rule but im not sure how to move on from there. any help will be greatly appreciated

Answer 1

do you know this property?\[\lim_{x\to a}f(x)=\lim_{x\to a}e^{\ln f(x)}\]?

Answer 2

yes i did

Answer 3

ok, so in a more convenient notation...\[\large\lim_{x\to 0}(1-x)^{\frac2 x}=\exp\left(\lim_{x\to 0}\ln(1-x)^{\frac2 x}\right)\]use log properties to simplify

Answer 4

in case you haven't seen it, \[\exp(...)=e^{...}\]it's just a notation thing

Answer 5

ok so now i would take the 2/x down so that i would get 2(1-x)/x?

Answer 6

don't forget the ln

Answer 7

so ln2(1-x)/x?

Answer 8

the 2 goes in front of the ln

Answer 9

\[\ln x^a=a\ln x\]

Answer 10

2ln(1-x)/x?

Answer 11

yes, that is what is in the exponent
now use l'hospital

Answer 12

i use l'hopitals rule just on the (1-x)/x?

Answer 13

leave the 2 in there, but yes\[\large\lim_{x\to 0}(1-x)^{\frac2 x}=\exp\left(\lim_{x\to 0}\ln(1-x)^{\frac2 x}\right)=\exp\left(\lim_{x\to 0}{2\ln(1-x)\over x}\right)\]

Answer 14

so the derivative will be -1/1 = -1

so i have e^2(-1)= 1/e^2?

Answer 15

bingo :)

Answer 16

thank you very much :D

Answer 17

welcome!