mathop lim limitsx to 1 căn 5 - x^3 - căn [3]x^2 + 7x^2 - 1
![mathop lim limitsx to 1 căn 5 - x^3 - căn [3]x^2 + 7x^2 - 1](https://tuhoc365.vn/wp-content/uploads/2020/03/qa-238x145.png "mathop lim limitsx to 1 căn 5 - x^3 - căn [3]x^2 + 7x^2 - 1")
Câu hỏi
Nhận biết\(\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {5 - {x^3}} - \sqrt[3]{{{x^2} + 7}}}}{{{x^2} - 1}}\)
Đáp án đúng: B
Lời giải của Tự Học 365
Giải chi tiết:
\(\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {5 - {x^3}} - \sqrt[3]{{{x^2} + 7}}}}{{{x^2} - 1}}\)
\(\begin{array}{l} = \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {5 - {x^3}} - 2 + 2 - \sqrt[3]{{{x^2} + 7}}}}{{{x^2} - 1}}\\ = \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {5 - {x^3}} - 2}}{{{x^2} - 1}} + \mathop {\lim }\limits_{x \to 1} \frac{{2 - \sqrt[3]{{{x^2} + 7}}}}{{{x^2} - 1}}\\ = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {\sqrt {5 - {x^3}} - 2} \right)\left( {\sqrt {5 - {x^3}} + 2} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)\left( {\sqrt {5 - {x^3}} + 2} \right)}} + \mathop {\lim }\limits_{x \to 1} \frac{{\left( {2 - \sqrt[3]{{{x^2} + 7}}} \right)\left( {4 + 2\sqrt[3]{{{x^2} + 7}} + {{\sqrt[3]{{{x^2} + 7}}}^2}} \right)}}{{\left( {{x^2} - 1} \right)\left( {4 + 2\sqrt[3]{{{x^2} + 7}} + {{\sqrt[3]{{{x^2} + 7}}}^2}} \right)}}\\ = \mathop {\lim }\limits_{x \to 1} \frac{{1 - {x^3}}}{{\left( {x - 1} \right)\left( {x + 1} \right)\left( {\sqrt {5 - {x^3}} + 2} \right)}} + \mathop {\lim }\limits_{x \to 1} \frac{{1 - {x^2}}}{{\left( {x - 1} \right)\left( {x + 1} \right)\left( {4 + 2\sqrt[3]{{{x^2} + 7}} + {{\sqrt[3]{{{x^2} + 7}}}^2}} \right)}}\\ = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {1 - x} \right)\left( {{x^2} + x + 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)\left( {\sqrt {5 - {x^3}} + 2} \right)}} - \mathop {\lim }\limits_{x \to 1} \frac{1}{{4 + 2\sqrt[3]{{{x^2} + 7}} + {{\sqrt[3]{{{x^2} + 7}}}^2}}}\\ = - \mathop {\lim }\limits_{x \to 1} \frac{{{x^2} + x + 1}}{{\left( {x + 1} \right)\left( {\sqrt {5 - {x^3}} + 2} \right)}} - \mathop {\lim }\limits_{x \to 1} \frac{1}{{4 + 2\sqrt[3]{{{x^2} + 7}} + {{\sqrt[3]{{{x^2} + 7}}}^2}}}\\ = - \frac{3}{{2.\left( {2 + 2} \right)}} - \frac{1}{{4 + 2.2 + {2^2}}} = - \frac{{11}}{{24}}\end{array}\)
Chọn B.
