Let's take the limit x→0+lim(3x+1)x5 transform do replacement u=3x1 then x→0+lim(1+x13)x5 = = u→0+lim(1+u1)15u = u→0+lim(1+u1)15u = ((u→0+lim(1+u1)u))15 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))15=e15
The final answer: x→0+lim(3x+1)x5=e15
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