\(\dfrac{2x^2-3x+2}{x+1}-\left(ax+b\right)=\dfrac{2x^2-3x+2-\left(x+1\right)\left(ax+b\right)}{x+1}\)
\(=\dfrac{\left(2-a\right)x^2-\left(a+b+3\right)x+2-b}{x+1}\)
\(=\dfrac{\left(2-a\right)x-\left(a+b+3\right)+\dfrac{2-b}{x}}{1+\dfrac{1}{x}}\) (1)
Nếu \(a\ne2\Rightarrow\) giới hạn đã cho tiến tới vô cực (ktm)
\(\Rightarrow a=2\)
Khi đó \(\left(1\right)=\dfrac{-\left(a+b+3\right)}{1}=-\left(a+b+3\right)=0\)
\(\Rightarrow b+5=0\Rightarrow b=-5\)