Let's take the limit x→0+lim(x+1)x1 transform do replacement u=x1 then x→0+lim(1+x11)x1 = = u→0+lim(1+u1)u = u→0+lim(1+u1)u = ((u→0+lim(1+u1)u)) 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))=e
The final answer: x→0+lim(x+1)x1=e
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