Chọn phương án C.

\(\begin{aligned}
\lim\limits_{x\to-\infty}\dfrac{\sqrt{4x^2-x+1}}{x+1}&=\lim\limits_{x\to-\infty}\dfrac{\sqrt{x^2\left(4-\dfrac{1}{x}+\dfrac{1}{x^2}\right)}}{x+1}\\
&=\lim\limits_{x\to-\infty}\dfrac{|x|\sqrt{4-\dfrac{1}{x}+\dfrac{1}{x^2}}}{x+1}\\
&=\lim\limits_{x\to-\infty}\dfrac{-x\sqrt{4-\dfrac{1}{x}+\dfrac{1}{x^2}}}{x\left(1+\dfrac{1}{x}\right)}\\
&=\lim\limits_{x\to-\infty}\dfrac{-\sqrt{4-\dfrac{1}{x}+\dfrac{1}{x^2}}}{1+\dfrac{1}{x}}\\
&=\dfrac{-\sqrt{4-0+0}}{1+0}=-2.
\end{aligned}\)