Let's take the limit x→∞lim(x6) Let's divide numerator and denominator by x: x→∞lim(x6) = x→∞lim(16x1) Do Replacement u=x1 then x→∞lim(16x1)=u→0+lim(6u) = 0⋅6=0
The final answer: x→∞lim(x6)=0
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