Đáp án:
\[\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {2x - 1} - \sqrt[3]{{3{x^2} - 3x + 1}}}}{{{{\left( {x - 1} \right)}^2}}} = \frac{{ - 1}}{2}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {2x - 1} - \sqrt[3]{{3{x^2} - 3x + 1}}}}{{{{\left( {x - 1} \right)}^2}}}\\
= \mathop {\lim }\limits_{x \to 1} \left[ {\frac{{\sqrt {2x - 1} - x}}{{{{\left( {x - 1} \right)}^2}}} + \frac{{x - \sqrt[3]{{3{x^2} - 3x + 1}}}}{{{{\left( {x - 1} \right)}^2}}}} \right]\\
= \mathop {\lim }\limits_{x \to 1} \left[ {\frac{{\left( {\sqrt {2x - 1} - x} \right)\left( {\sqrt {2x - 1} + x} \right)}}{{{{\left( {x - 1} \right)}^2}\left( {\sqrt {2x - 1} + x} \right)}} + \frac{{\left( {x - \sqrt[3]{{3{x^2} - 3x + 1}}} \right)\left( {{x^2} + x.\sqrt[3]{{3{x^2} - 3x + 1}} + {{\sqrt[3]{{3{x^2} - 3x + 1}}}^2}} \right)}}{{{{\left( {x - 1} \right)}^2}.\left( {{x^2} + x.\sqrt[3]{{3{x^2} - 3x + 1}} + {{\sqrt[3]{{3{x^2} - 3x + 1}}}^2}} \right)}}} \right]\\
= \mathop {\lim }\limits_{x \to 1} \left[ {\frac{{2x - 1 - {x^2}}}{{{{\left( {x - 1} \right)}^2}\left( {\sqrt {2x - 1} + x} \right)}} + \frac{{{x^3} - \left( {3{x^2} - 3x + 1} \right)}}{{{{\left( {x - 1} \right)}^2}.\left( {{x^2} + x.\sqrt[3]{{3{x^2} - 3x + 1}} + {{\sqrt[3]{{3{x^2} - 3x + 1}}}^2}} \right)}}} \right]\\
= \mathop {\lim }\limits_{x \to 1} \left[ {\frac{{ - {{\left( {x - 1} \right)}^2}}}{{{{\left( {x - 1} \right)}^2}\left( {\sqrt {2x - 1} + x} \right)}} + \frac{{{{\left( {x - 1} \right)}^3}}}{{{{\left( {x - 1} \right)}^2}.\left( {{x^2} + x.\sqrt[3]{{3{x^2} - 3x + 1}} + {{\sqrt[3]{{3{x^2} - 3x + 1}}}^2}} \right)}}} \right]\\
= \mathop {\lim }\limits_{x \to 1} \left[ {\frac{{ - 1}}{{\sqrt {2x - 1} + x}} + \frac{{x - 1}}{{{x^2} + x.\sqrt[3]{{3{x^2} - 3x + 1}} + {{\sqrt[3]{{3{x^2} - 3x + 1}}}^2}}}} \right]\\
= \frac{{ - 1}}{{\sqrt {2.1 - 1} + 1}} + \frac{{1 - 1}}{{{1^2} + 1.\sqrt[3]{{{{3.1}^2} - 3.1 + 1}} + {{\sqrt[3]{{{{3.1}^2} - 3.1 + 1}}}^2}}}\\
= \frac{{ - 1}}{2}
\end{array}\)
