If _ x arrow ∞ ( x^ 2 +x+1 x+1 -a x-b)=4 , then

Given: $\lim _{x \rightarrow \infty}$\left(\frac{x^{2}+x+1}{x+1}-a x-b\right)=4 $$\begin{array}{l} \Rightarrow \lim _{x \rightarrow \infty} \frac{x^{2}+x+1-a x^{2}-a x-b x-b}{x+1}=4 \\ \Rightarrow \lim _{x \rightarrow \infty} \frac{(1-a) x^{2}+(1-a-b) x+(1-b)}{x+1}=4 \end{array}$$$For this limit to be finite$1-a=0 \Rightarrow a=1$then given limit reduces to$\lim _{x \rightarrow \infty} \frac{-b x+(1-b)}{x+1}=4 \Rightarrow \lim _{x \rightarrow \infty} \frac{-b+\frac{(1-b)}{x}}{1+\frac{1}{x}}=4 \Rightarrow-b=4$or$b=-4, \therefore a=1, b=-4