Let's take the limit x→∞lim(x26) Let's divide numerator and denominator by x^2: x→∞lim(x26) = x→∞lim(16x21) Do Replacement u=x1 then x→∞lim(16x21)=u→0+lim(6u2) = 6⋅02=0
The final answer: x→∞lim(x26)=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