Mathematics - Limits Question with Solution | TestHub

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