Let's take the limit x→0+lim(1−3x)x1 transform do replacement u=(−3)x1 then x→0+lim(1−x13)x1 = = u→0+lim(1+u1)−3u = u→0+lim(1+u1)−3u = ((u→0+lim(1+u1)u))−3 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))−3=e−3
The final answer: x→0+lim(1−3x)x1=e−3
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