Let's take the limit n→∞limn1 Let's divide numerator and denominator by n: n→∞limn1 = n→∞lim(n1) Do Replacement u=n1 then n→∞lim(n1)=u→0+limu = 0=0
The final answer: n→∞limn1=0
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