If $\underset{x\to \infty }{\lim }\left(\frac{x^2+x+1}{x+1}-ax-b\right)=4$then a - b =

Given $\underset{x\to \infty }{\lim }\left(\frac{x^2+x+1}{x+1}-ax-b\right)=4$ $\Rightarrow \underset{x\to \infty }{\lim }\frac{x^2+x+1-a{x}^2-ax-bx-b}{(x+1)}=4$ $\Rightarrow \underset{x\to \infty }{\lim }\frac{(1-a)x^2+(1-a-b)x+(1-b)}{(x+1)}=4$$\Rightarrow 1-a=0and1-a-b=4\Rightarrow b=-4,a=1$