mathop lim limitsx to 0 căn 1 + alpha x . căn [3]1 + beta x. căn [4]1 + gamma x - 1x( alpha beta g
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Câu hỏi
Nhận biết\(\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + \alpha x} .\sqrt[3]{{1 + \beta x}}.\sqrt[4]{{1 + \gamma x}} - 1}}{x}\,\,\,\,\left( {\alpha \beta \gamma \ne 0} \right)\)
Đáp án đúng: A
Lời giải của Tự Học 365
Giải chi tiết:
\(\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + \alpha x} .\sqrt[3]{{1 + \beta x}}.\sqrt[4]{{1 + \gamma x}} - 1}}{x}\,\,\,\,\left( {\alpha \beta \gamma \ne 0} \right)\)
\(\begin{array}{l} = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sqrt {1 + \alpha x} - 1} \right)\sqrt[3]{{1 + \beta x}}.\sqrt[4]{{1 + \gamma x}}}}{x} + \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sqrt[3]{{1 + \beta x}} - 1} \right).\sqrt[4]{{1 + \gamma x}}}}{x} + \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[4]{{1 + \gamma x}} - 1}}{x}\\ = {L_1} + {L_2} + {L_3}\end{array}\)
\(\begin{array}{l}{L_1} = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sqrt {1 + \alpha x} - 1} \right)\left( {\sqrt {1 + \alpha x} + 1} \right)\sqrt[3]{{1 + \beta x}}.\sqrt[4]{{1 + \gamma x}}}}{{x\left( {\sqrt {1 + \alpha x} + 1} \right)}}\\\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to 0} \frac{{\alpha x\sqrt[3]{{1 + \beta x}}.\sqrt[4]{{1 + \gamma x}}}}{{x\left( {\sqrt {1 + \alpha x} + 1} \right)}}\\\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to 0} \frac{{\alpha \sqrt[3]{{1 + \beta x}}.\sqrt[4]{{1 + \gamma x}}}}{{\sqrt {1 + \alpha x} + 1}} = \frac{\alpha }{2}\end{array}\)
\(\begin{array}{l}{L_2} = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sqrt[3]{{1 + \beta x}} - 1} \right).\sqrt[4]{{1 + \gamma x}}}}{x}\\\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sqrt[3]{{1 + \beta x}} - 1} \right)\left( {{{\sqrt[3]{{1 + \beta x}}}^2} + \sqrt[3]{{1 + \beta x}} + 1} \right).\sqrt[4]{{1 + \gamma x}}}}{{x\left( {{{\sqrt[3]{{1 + \beta x}}}^2} + \sqrt[3]{{1 + \beta x}} + 1} \right)}}\\\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to 0} \frac{{\beta x.\sqrt[4]{{1 + \gamma x}}}}{{x\left( {{{\sqrt[3]{{1 + \beta x}}}^2} + \sqrt[3]{{1 + \beta x}} + 1} \right)}}\\\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to 0} \frac{{\beta .\sqrt[4]{{1 + \gamma x}}}}{{{{\sqrt[3]{{1 + \beta x}}}^2} + \sqrt[3]{{1 + \beta x}} + 1}} = \frac{\beta }{3}\end{array}\)
\(\begin{array}{l}{L_3} = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[4]{{1 + \gamma x}} - 1}}{x}\\\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sqrt[4]{{1 + \gamma x}} - 1} \right)\left( {{{\sqrt[4]{{1 + \gamma x}}}^3} + {{\sqrt[4]{{1 + \gamma x}}}^2} + \sqrt[4]{{1 + \gamma x}} + 1} \right)}}{{x\left( {{{\sqrt[4]{{1 + \gamma x}}}^3} + {{\sqrt[4]{{1 + \gamma x}}}^2} + \sqrt[4]{{1 + \gamma x}} + 1} \right)}}\\\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to 0} \frac{{\gamma x}}{{x\left( {{{\sqrt[4]{{1 + \gamma x}}}^3} + {{\sqrt[4]{{1 + \gamma x}}}^2} + \sqrt[4]{{1 + \gamma x}} + 1} \right)}}\\\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to 0} \frac{\gamma }{{{{\sqrt[4]{{1 + \gamma x}}}^3} + {{\sqrt[4]{{1 + \gamma x}}}^2} + \sqrt[4]{{1 + \gamma x}} + 1}} = \frac{\gamma }{4}\end{array}\)
Vậy \(\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + \alpha x} .\sqrt[3]{{1 + \beta x}}.\sqrt[4]{{1 + \gamma x}} - 1}}{x} = \frac{\alpha }{2} + \frac{\beta }{3} + \frac{\gamma }{4}\).
Chọn A.
