Let's take the limit x→∞lim(x3−3x) Let's divide numerator and denominator by x^3: x→∞lim(x3−3x) = x→∞lim(x311−x23) Do Replacement u=x1 then x→∞lim(x311−x23)=u→0+lim(u31−3u2) = 01−3⋅02=∞
The final answer: x→∞lim(x3−3x)=∞
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