Let's take the limit x→0+lim(1−x)x1 transform do replacement u=(−1)x1 then x→0+lim(1−x11)x1 = = u→0+lim(1+u1)−u = u→0+lim(1+u1)−u = ((u→0+lim(1+u1)u))−1 The limit u→0+lim(1+u1)u is second remarkable limit, is equal to e ~ 2.718281828459045 then ((u→0+lim(1+u1)u))−1=e−1
The final answer: x→0+lim(1−x)x1=e−1
Lopital's rule
There is no sense to apply Lopital's rule to this function since there is no indeterminateness of 0/0 or oo/oo type