If \(\lim _{x \rightarrow \infty}\left(\frac{x^2+x+1}{x+1}-a x-b\right)=4\) then

\(\begin{aligned} & \lim _{x \rightarrow \infty}\left(\frac{x^2+x+1}{x+1}-\mathrm{a} x-\mathrm{b}\right)=4 \\ & \Rightarrow \lim _{x \rightarrow \infty}\left(\frac{x^2+x+1-\mathrm{ax}(x+1)-\mathrm{b}(x+1)}{x+1}\right)=4 \\ & \Rightarrow \lim _{x \rightarrow \infty}\left[\frac{(1-\mathrm{a}) x^2+(1-\mathrm{a}-\mathrm{b}) x+(1-\mathrm{b})}{x+1}\right]=4\end{aligned}\)
Limit is finite, it exists when \((1-a)=0\)
\(\begin{array}{cc}
\therefore \quad & 1-a=0 \\
& \Rightarrow a=1
\end{array}\)
\(\begin{array}{ll}\therefore \quad & \lim _{x \rightarrow \infty}\left[\frac{1-a-b+\frac{(1-b)}{x}}{1+\frac{1}{x}}\right]=4 \\ & \Rightarrow 1-a-b=4 \\ & \Rightarrow \quad \mathrm{~b}=-4 \\ \therefore \quad & a=1, b=-4\end{array}\)