Let's take the limit t→∞lim(1⋅t1) Let's divide numerator and denominator by t: t→∞lim(1⋅t1) = t→∞lim(1t1) Do Replacement u=t1 then t→∞lim(1t1)=u→0+limu = 0=0
The final answer: t→∞lim(1⋅t1)=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