Iflimx→∞(√x2−x+1−ax−b)=0then the values of a and b are given by
We have,limx→∞(√x2−x+1−ax−b)=0⇒limy→∞(√y2+y+1+ay−b)=0 [putting x=-y; x→,∞,y→∞]⇒limy→∞[y(1+1y+1y2)1/2+ay−b]=0⇒limy→∞[y{1+12(1y+1y2+.........)}+ay−b]=0⇒limy→∞[y(1+a)+(12−b)+12y+........]=0⇒1+a=0 and (1/2)−b=0 ⇒ a=−1 and b=1/2