Google
Câu hỏi: Trong các mệnh đề sau, mệnh đề nào sai?
A. ${\underset{x\to +\infty }{\mathop{\lim }} \left( \sqrt{{{x}^{2}}-x+1}+x-2 \right)=+\infty }$.
B. ${\underset{x\to {{1}^{-}}}{\mathop{\lim }} \dfrac{3x-2}{x-1}=+\infty }$.
C. ${\underset{x\to 1}{\mathop{\lim }} \dfrac{3x-2}{x+1}=\dfrac{1}{2}}$.
D. ${\underset{x\to -\infty }{\mathop{\lim }} \left( \sqrt{{{x}^{2}}-x+1}+x-2 \right)=-\dfrac{3}{2}}$.
A. ${\underset{x\to +\infty }{\mathop{\lim }} \left( \sqrt{{{x}^{2}}-x+1}+x-2 \right)=+\infty }$.
B. ${\underset{x\to {{1}^{-}}}{\mathop{\lim }} \dfrac{3x-2}{x-1}=+\infty }$.
C. ${\underset{x\to 1}{\mathop{\lim }} \dfrac{3x-2}{x+1}=\dfrac{1}{2}}$.
D. ${\underset{x\to -\infty }{\mathop{\lim }} \left( \sqrt{{{x}^{2}}-x+1}+x-2 \right)=-\dfrac{3}{2}}$.
+) Ta có:
${{\lim }_{x\to +\infty }}\left( \sqrt{{{x}^{2}}-x+1}+x-2 \right)={{\lim }_{x\to +\infty }}\left( \sqrt[x]{1-\dfrac{1}{x}+\dfrac{1}{{{x}^{2}}}}+x-2 \right)={{\lim }_{x\to +\infty }}x\left( \sqrt{1-\dfrac{1}{x}+\dfrac{1}{{{x}^{2}}}}+1-\dfrac{2}{x} \right)=+\infty $
Vì ${{\lim }_{x\to +\infty }}\left( \sqrt{1-\dfrac{1}{x}+\dfrac{1}{{{x}^{2}}}}+1-\dfrac{2}{x} \right)=2>0;{{\lim }_{x\to +\infty }}x=+\infty $
+ Có ${{\lim }_{x\to {{1}^{-}}}}\left( 3x-2 \right)=1>0;{{\lim }_{x\to {{1}^{-}}}}\left( x-1 \right)=0$ và $x-1<0,\forall x<1$ nên ${{\lim }_{x\to {{1}^{-}}}}\left( \dfrac{3x-2}{x-1} \right)=-\infty $
+ Có ${{\lim }_{x\to 1}}\dfrac{3x-2}{x+1}=\dfrac{3.1-2}{1+1}=\dfrac{1}{2}$
+ Có ${{\lim }_{x\to -\infty }}\left( \sqrt{{{x}^{2}}-x+1}+x-2 \right)={{\lim }_{x\to -\infty }}\left( \dfrac{{{\left( \sqrt{{{x}^{2}}-x+1} \right)}^{2}}-{{\left( x-2 \right)}^{2}}}{\sqrt{{{x}^{2}}-x+1}-x+2} \right)$
$={{\lim }_{x\to -\infty }}\left( \dfrac{3x-3}{\sqrt{{{x}^{2}}-x+1}-x+2} \right)={{\lim }_{x\to -\infty }}\left( \dfrac{3-\dfrac{3}{x}}{-\sqrt{1-\dfrac{1}{x}+\dfrac{1}{{{x}^{2}}}}-1+\dfrac{2}{x}} \right)=-\dfrac{3}{2}$
${{\lim }_{x\to +\infty }}\left( \sqrt{{{x}^{2}}-x+1}+x-2 \right)={{\lim }_{x\to +\infty }}\left( \sqrt[x]{1-\dfrac{1}{x}+\dfrac{1}{{{x}^{2}}}}+x-2 \right)={{\lim }_{x\to +\infty }}x\left( \sqrt{1-\dfrac{1}{x}+\dfrac{1}{{{x}^{2}}}}+1-\dfrac{2}{x} \right)=+\infty $
Vì ${{\lim }_{x\to +\infty }}\left( \sqrt{1-\dfrac{1}{x}+\dfrac{1}{{{x}^{2}}}}+1-\dfrac{2}{x} \right)=2>0;{{\lim }_{x\to +\infty }}x=+\infty $
+ Có ${{\lim }_{x\to {{1}^{-}}}}\left( 3x-2 \right)=1>0;{{\lim }_{x\to {{1}^{-}}}}\left( x-1 \right)=0$ và $x-1<0,\forall x<1$ nên ${{\lim }_{x\to {{1}^{-}}}}\left( \dfrac{3x-2}{x-1} \right)=-\infty $
+ Có ${{\lim }_{x\to 1}}\dfrac{3x-2}{x+1}=\dfrac{3.1-2}{1+1}=\dfrac{1}{2}$
+ Có ${{\lim }_{x\to -\infty }}\left( \sqrt{{{x}^{2}}-x+1}+x-2 \right)={{\lim }_{x\to -\infty }}\left( \dfrac{{{\left( \sqrt{{{x}^{2}}-x+1} \right)}^{2}}-{{\left( x-2 \right)}^{2}}}{\sqrt{{{x}^{2}}-x+1}-x+2} \right)$
$={{\lim }_{x\to -\infty }}\left( \dfrac{3x-3}{\sqrt{{{x}^{2}}-x+1}-x+2} \right)={{\lim }_{x\to -\infty }}\left( \dfrac{3-\dfrac{3}{x}}{-\sqrt{1-\dfrac{1}{x}+\dfrac{1}{{{x}^{2}}}}-1+\dfrac{2}{x}} \right)=-\dfrac{3}{2}$
Đáp án B.