Let's take the limit n→∞lim(1−n5)n transform do replacement u=−5n then n→∞lim(1−n5)n = = u→∞lim(1+u1)−5u = u→∞lim(1+u1)−5u = ((u→∞lim(1+u1)u))−5 The limit u→∞lim(1+u1)u is second remarkable limit, is equal to e ~ 2.718281828459045 then ((u→∞lim(1+u1)u))−5=e−5
The final answer: n→∞lim(1−n5)n=e−5
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