mathop lim limitsx to 1 d căn 3x + 6 . căn [3]x + 7 - 6x - 1
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Câu hỏi
Nhận biết\(\mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {3x + 6} .\sqrt[3]{{x + 7}} - 6}}{{x - 1}}\)
Đáp án đúng: D
Lời giải của Tự Học 365
Giải chi tiết:
\(\begin{array}{l}\,\,\,\,\mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {3x + 6} .\sqrt[3]{{x + 7}} - 6}}{{x - 1}}\\ = \mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {3x + 6} \left( {\sqrt[3]{{x + 7}} - 2} \right) + 2\left( {\sqrt {3x + 6} - 3} \right)}}{{x - 1}}\\ = \mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {3x + 6} \left( {\sqrt[3]{{x + 7}} - 2} \right)}}{{x - 1}} + 2\mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {3x + 6} - 3}}{{x - 1}}\\ = \mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {3x + 6} \left( {\sqrt[3]{{x + 7}} - 2} \right)\left( {{{\left( {\sqrt[3]{{x + 7}}} \right)}^2} + \sqrt[3]{{x + 7}} + 2} \right)}}{{\left( {x - 1} \right)\left( {{{\left( {\sqrt[3]{{x + 7}}} \right)}^2} + \sqrt[3]{{x + 7}} + 2} \right)}} + 2\mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {\sqrt {3x + 6} - 3} \right)\left( {\sqrt {3x + 6} + 3} \right)}}{{\left( {x - 1} \right)\left( {\sqrt {3x + 6} + 3} \right)}}\\ = \mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {3x + 6} \left( {x + 7 - 8} \right)}}{{\left( {x - 1} \right)\left( {{{\left( {\sqrt[3]{{x + 7}}} \right)}^2} + \sqrt[3]{{x + 7}} + 2} \right)}} + 2\mathop {\lim }\limits_{x \to 1} \dfrac{{3x + 6 - 9}}{{\left( {x - 1} \right)\left( {\sqrt {3x + 6} + 3} \right)}}\\ = \mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {3x + 6} }}{{{{\left( {\sqrt[3]{{x + 7}}} \right)}^2} + \sqrt[3]{{x + 7}} + 2}} + 2\mathop {\lim }\limits_{x \to 1} \dfrac{3}{{\sqrt {3x + 6} + 3}}\\ = \dfrac{3}{{4 + 2 + 2}} + 2.\dfrac{3}{{3 + 3}} = \dfrac{3}{8} + 2.\dfrac{1}{2} = \dfrac{{11}}{8}\end{array}\)
