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Câu hỏi: Giới hạn $\underset{x\Rightarrow -\infty }{\mathop{\lim }} \dfrac{\sqrt{{{x}^{2}}+x+1}}{2x+1}$ là :
A. $\dfrac{1}{2}$.
B. $+\infty $.
C. $-\infty $.
D. $\dfrac{-1}{2}$.
A. $\dfrac{1}{2}$.
B. $+\infty $.
C. $-\infty $.
D. $\dfrac{-1}{2}$.
Ta có: $\underset{x\Rightarrow -\infty }{\mathop{\lim }} \dfrac{\sqrt{{{x}^{2}}+x+1}}{2x+1}=\underset{x\Rightarrow -\infty }{\mathop{\lim }} \dfrac{\sqrt{{{x}^{2}}\left( 1+\dfrac{1}{x}+\dfrac{1}{{{x}^{2}}} \right)}}{x\left( 2+\dfrac{1}{x} \right)}$
$=\underset{x\Rightarrow -\infty }{\mathop{\lim }} \dfrac{\left| x \right|\sqrt{1+\dfrac{1}{x}+\dfrac{1}{{{x}^{2}}}}}{x\left( 2+\dfrac{1}{x} \right)}$
$=\underset{x\Rightarrow -\infty }{\mathop{\lim }} \dfrac{-\sqrt{1+\dfrac{1}{x}+\dfrac{1}{{{x}^{2}}}}}{2+\dfrac{1}{x}}=-\dfrac{1}{2}$
$=\underset{x\Rightarrow -\infty }{\mathop{\lim }} \dfrac{\left| x \right|\sqrt{1+\dfrac{1}{x}+\dfrac{1}{{{x}^{2}}}}}{x\left( 2+\dfrac{1}{x} \right)}$
$=\underset{x\Rightarrow -\infty }{\mathop{\lim }} \dfrac{-\sqrt{1+\dfrac{1}{x}+\dfrac{1}{{{x}^{2}}}}}{2+\dfrac{1}{x}}=-\dfrac{1}{2}$
Đáp án D.