Đáp án:
\[\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[3]{{1 + 3x}} - 1 - x\sqrt {1 - x} }}{{{x^3} + {x^2}}} = - \frac{1}{2}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[3]{{1 + 3x}} - 1 - x\sqrt {1 - x} }}{{{x^3} + {x^2}}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sqrt[3]{{1 + 3x}} - \left( {x + 1} \right)} \right) + x - x\sqrt {1 - x} }}{{{x^3} + {x^2}}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{\left( {1 + 3x} \right) - {{\left( {x + 1} \right)}^3}}}{{{{\sqrt[3]{{1 + 3x}}}^2} + \sqrt[3]{{1 + 3x}}.\left( {x + 1} \right) + {{\left( {x + 1} \right)}^2}}} + x.\frac{{1 - \left( {1 - x} \right)}}{{1 + \sqrt {1 - x} }}}}{{{x^3} + {x^2}}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{ - {x^3} - 3{x^2}}}{{{{\sqrt[3]{{1 + 3x}}}^2} + \sqrt[3]{{1 + 3x}}.\left( {x + 1} \right) + {{\left( {x + 1} \right)}^2}}} + \frac{{{x^2}}}{{1 + \sqrt {1 - x} }}}}{{{x^2}\left( {x + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 0} \left( { - \frac{{x + 3}}{{\left( {{{\sqrt[3]{{1 + 3x}}}^2} + \sqrt[3]{{1 + 3x}}.\left( {x + 1} \right) + {{\left( {x + 1} \right)}^2}} \right).\left( {x + 1} \right)}} + \frac{1}{{\left( {x + 1} \right).\left( {1 + \sqrt {1 - x} } \right)}}} \right)\\
= - \frac{{0 + 3}}{{\left( {{{\sqrt[3]{{1 + 3.0}}}^2} + \sqrt[3]{{1 + 3.0}}.\left( {0 + 1} \right) + {{\left( {0 + 1} \right)}^2}} \right)\left( {0 + 1} \right)}} + \frac{1}{{\left( {0 + 1} \right)\left( {1 + \sqrt {1 - 0} } \right)}}\\
= \frac{{ - 3}}{{3.1}} + \frac{1}{{1.2}} = - 1 + \frac{1}{2} = - \frac{1}{2}
\end{array}\)
