If limx→−∞ x2−x+1−ax−b=0, is

We have, limx→∞ x2−x+1−(ax+b)=0⇒limx→∞ x2−x+1−(ax+b)2x2−x+1+(ax+b)=0⇒limx→∞ x21−a2−x(1+2ab)+1−b2x2−x+1+ax+b=0⇒1−a2=0 and 1+2ab=0⇒a=±1 and b=∓1/2.For a=−1 and b=1/2 we observe that  limx→∞ x2−x+1−ax−b=limx→∞ x2−x+1+x−12→∞Hence, a=1 and b=−12