Let's take the limit x→∞lim(16x) Let's divide numerator and denominator by x: x→∞lim(16x) = x→∞lim161x11 Do Replacement u=x1 then x→∞lim161x11=u→0+lim(u16) = 016=∞
The final answer: x→∞lim(16x)=∞
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