Let's take the limit x→∞limx6 Let's divide numerator and denominator by x^6: x→∞limx6 = x→∞limx611 Do Replacement u=x1 then x→∞limx611=u→0+limu61 = 01=∞
The final answer: x→∞limx6=∞
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