mathop lim limitsx to 0 căn [3]1 + 3x - 1 - x căn 1 - x x^3 + x^2
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Câu hỏi
Nhận biết\(\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[3]{{1 + 3x}} - 1 - x\sqrt {1 - x} }}{{{x^3} + {x^2}}}\)
Đáp án đúng: A
Lời giải của Tự Học 365
Giải chi tiết:
\(\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[3]{{1 + 3x}} - 1 - x\sqrt {1 - x} }}{{{x^3} + {x^2}}}\)
\(\begin{array}{l} = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[3]{{1 + 3x}} - 1}}{{{x^3} + {x^2}}} - \mathop {\lim }\limits_{x \to 0} \frac{{x\sqrt {1 - x} }}{{{x^3} + {x^2}}}\\ = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sqrt[3]{{1 + 3x}} - 1} \right)\left( {{{\sqrt[3]{{1 + 3x}}}^2} + \sqrt[3]{{1 + 3x}} + 1} \right)}}{{\left( {{x^3} + {x^2}} \right)\left( {{{\sqrt[3]{{1 + 3x}}}^2} + \sqrt[3]{{1 + 3x}} + 1} \right)}} - \mathop {\lim }\limits_{x \to 0} \frac{{x\sqrt {1 - x} }}{{x\left( {{x^2} + 1} \right)}}\\ = \mathop {\lim }\limits_{x \to 0} \frac{{3x}}{{x\left( {{x^2} + 1} \right)\left( {{{\sqrt[3]{{1 + 3x}}}^2} + \sqrt[3]{{1 + 3x}} + 1} \right)}} - \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 - x} }}{{{x^2} + 1}}\\ = \mathop {\lim }\limits_{x \to 0} \frac{3}{{\left( {{x^2} + 1} \right)\left( {{{\sqrt[3]{{1 + 3x}}}^2} + \sqrt[3]{{1 + 3x}} + 1} \right)}} - \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 - x} }}{{{x^2} + 1}}\\ = \frac{3}{{1\left( {{1^2} + 1 + 1} \right)}} - \frac{1}{{{0^2} + 1}} = 0 \end{array}\)
Chọn A.
