Let's take the limit x→∞lim(−3x) Let's divide numerator and denominator by x: x→∞lim(−3x) = x→∞lim(−1)31x11 Do Replacement u=x1 then x→∞lim(−1)31x11=u→0+lim(−u3) = −03=−∞
The final answer: x→∞lim(−3x)=−∞
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