Let's take the limit x→∞lim(xx+2)3x transform x→∞lim(xx+2)3x = x→∞lim(xx+2)3x = x→∞lim(xx+x2)3x = x→∞lim(1+x2)3x = do replacement u=2x then x→∞lim(1+x2)3x = = u→∞lim(1+u1)6u = u→∞lim(1+u1)6u = ((u→∞lim(1+u1)u))6 The limit u→∞lim(1+u1)u is second remarkable limit, is equal to e ~ 2.718281828459045 then ((u→∞lim(1+u1)u))6=e6
The final answer: x→∞lim(xx+2)3x=e6
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