Let's take the limit x→∞lim(2(−1)x) Let's divide numerator and denominator by x: x→∞lim(2(−1)x) = x→∞lim(−1)2x11 Do Replacement u=x1 then x→∞lim(−1)2x11=u→0+lim(−2u1) = −0⋅21=−∞
The final answer: x→∞lim(2(−1)x)=−∞
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