Let's take the limit x→∞limx7 Let's divide numerator and denominator by x^7: x→∞limx7 = x→∞limx711 Do Replacement u=x1 then x→∞limx711=u→0+limu71 = 01=∞
The final answer: x→∞limx7=∞
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