Let's take the limit x→0+lim(1−2x)x1 transform do replacement u=(−2)x1 then x→0+lim(1−x12)x1 = = u→0+lim(1+u1)−2u = u→0+lim(1+u1)−2u = ((u→0+lim(1+u1)u))−2 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))−2=e−2
The final answer: x→0+lim(1−2x)x1=e−2
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