Let's take the limit x→∞limx21 Let's divide numerator and denominator by x^2: x→∞limx21 = x→∞lim(x21) Do Replacement u=x1 then x→∞lim(x21)=u→0+limu2 = 02=0
The final answer: x→∞limx21=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