Let's take the limit x→0+lim(2x+1)x2 transform do replacement u=2x1 then x→0+lim(1+x12)x2 = = u→0+lim(1+u1)4u = u→0+lim(1+u1)4u = ((u→0+lim(1+u1)u))4 The limit u→0+lim(1+u1)u is second remarkable limit, is equal to e ~ 2.718281828459045 then ((u→0+lim(1+u1)u))4=e4
The final answer: x→0+lim(2x+1)x2=e4
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