Đáp án đúng: C khi $x\to -\infty \xrightarrow[{}]{}\frac{\sqrt{4{{x}^{2}}-x+1}}{x+1}\sim \frac{\sqrt{4{{x}^{2}}}}{x}=\frac{-2x}{x}=-2.$ Chọn C. Cụ thể: $\underset{x\to -\infty }{\mathop{\lim }}\,\frac{\sqrt{4{{x}^{2}}-x+1}}{x+1}=\underset{x\to -\infty }{\mathop{\lim }}\,\frac{-\sqrt{4-\frac{1}{x}+\frac{1}{{{x}^{2}}}}}{1+\frac{1}{x}}=\frac{-\sqrt{4}}{1}=-2.$
