Let's take the limit x→∞lim(x32) Let's divide numerator and denominator by x^3: x→∞lim(x32) = x→∞lim(12x31) Do Replacement u=x1 then x→∞lim(12x31)=u→0+lim(2u3) = 2⋅03=0
The final answer: x→∞lim(x32)=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