Let's take the limit x→∞lim(x+2x+1)x+1 transform x→∞lim(x+2x+1)x+1 = x→∞lim(x+2(x+2)−1)x+1 = x→∞lim(−x+21+x+2x+2)x+1 = x→∞lim(1−x+21)x+1 = do replacement u=−1x+2 then x→∞lim(1−x+21)x+1 = = u→∞lim(1+u1)−u−1 = u→∞lim(1+u1(1+u1)−u) = u→∞lim1+u11u→∞lim(1+u1)−u = u→∞lim(1+u1)−u = ((u→∞lim(1+u1)u))−1 The limit u→∞lim(1+u1)u is second remarkable limit, is equal to e ~ 2.718281828459045 then ((u→∞lim(1+u1)u))−1=e−1
The final answer: x→∞lim(x+2x+1)x+1=e−1
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