If limx→∞ x3+1x2+1−(ax+b)=2 ,then

Given that, limx→∞ x3+1x2+1−(ax+b)=2⇒ limx→∞ x3(1−a)−bx2−ax+(1−b)x2+1=2⇒ limx→∞ x(1−a)−b−ax+(1−b)x21+1x2=2This limit will exist, if1−a=0 and b=−2⇒a=1 and b=−2