Đáp án:
\[\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {2x - 1} + {x^2} - 3x + 1}}{{\sqrt[3]{{x - 2}} + {x^2} - x + 1}} = 0\]
Giải thích các bước giải:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {2x - 1} + {x^2} - 3x + 1}}{{\sqrt[3]{{x - 2}} + {x^2} - x + 1}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{\left( {\sqrt {2x - 1} - 1} \right) + {x^2} - 3x + 2}}{{\left( {\sqrt[3]{{x - 2}} + 1} \right) + \left( {{x^2} - x} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{\frac{{2x - 1 - 1}}{{\sqrt {2x - 1} + 1}} + \left( {x - 1} \right)\left( {x - 2} \right)}}{{\frac{{x - 2 + 1}}{{{{\sqrt[3]{{x - 2}}}^2} - \sqrt[3]{{x - 2}} + 1}} + x\left( {x - 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{\frac{{2\left( {x - 1} \right)}}{{\sqrt {2x - 1} + 1}} + \left( {x - 1} \right)\left( {x - 2} \right)}}{{\frac{{x - 1}}{{{{\sqrt[3]{{x - 2}}}^2} - \sqrt[3]{{x - 2}} + 1}} + x\left( {x - 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{\frac{2}{{\sqrt {2x - 1} + 1}} + x - 2}}{{\frac{1}{{{{\sqrt[3]{{x - 2}}}^2} - \sqrt[3]{{x - 2}} + 1}} + x}}\\
= \frac{{\frac{2}{{\sqrt {2.1 - 1} + 1}} + 1 - 2}}{{\frac{1}{{{{\sqrt[3]{{1 - 2}}}^2} - \sqrt[3]{{1 - 2}} + 1}} + 1}}\\
= \frac{{\frac{2}{2} - 1}}{{\frac{1}{3} + 1}} = 0
\end{array}\)
