Let's take the limit x→∞lim(1−3x1)6x transform do replacement u=−31x then x→∞lim(1−3x1)6x = = u→∞lim(1+u1)−2u = u→∞lim(1+u1)−2u = ((u→∞lim(1+u1)u))−2 The limit u→∞lim(1+u1)u is second remarkable limit, is equal to e ~ 2.718281828459045 then ((u→∞lim(1+u1)u))−2=e−2
The final answer: x→∞lim(1−3x1)6x=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