Let's take the limit x→∞lim(x2+2x) Let's divide numerator and denominator by x^2: x→∞lim(x2+2x) = x→∞lim(x211+x2) Do Replacement u=x1 then x→∞lim(x211+x2)=u→0+lim(u22u+1) = 02⋅0+1=∞
The final answer: x→∞lim(x2+2x)=∞
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