Let's take the limit x→∞lim1−x1 Let's divide numerator and denominator by x: x→∞lim1−x1 = x→∞lim(x(−1+x1)1) Do Replacement u=x1 then x→∞lim(x(−1+x1)1)=u→0+lim(u−1u) = −10=0
The final answer: x→∞lim1−x1=0
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