Nếu \(a\ne2\Rightarrow\lim\limits_{x\rightarrow+\infty}\left(\sqrt{4x^2-3x+1}-\left(ax+b\right)\right)=\lim\limits_{x\rightarrow+\infty}x\left(\sqrt{4-\dfrac{3}{x}+\dfrac{1}{x^2}}-a-\dfrac{b}{x}\right)\)
\(=\lim\limits_{x\rightarrow+\infty}\left(2-a\right)x=\infty\) (ktm)
\(\Rightarrow a=2\)
Khi đó:
\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{4x^2-3x+1}-\left(2x+b\right)\right)=\lim\limits_{x\rightarrow+\infty}\dfrac{-\left(4b+3\right)x+1-b^2}{\sqrt{4x^2-3x+1}+2x+b}\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{-\left(4b+3\right)+\dfrac{1-b^2}{x}}{\sqrt{4-\dfrac{3}{x}+\dfrac{1}{x^2}}+2+\dfrac{b}{x}}=\dfrac{-\left(4b+3\right)}{4}\)
\(\Rightarrow-\left(4b+3\right)=0\Rightarrow-4b=3\)
\(\Rightarrow T=2+3=5\)
