mathop lim limitsx to pm giới hạn d căn x^6 - x + 1 căn [3]x^3 + x + 1
![mathop lim limitsx to pm giới hạn d căn x^6 - x + 1 căn [3]x^3 + x + 1](https://tuhoc365.vn/wp-content/uploads/2020/03/qa-238x145.png "mathop lim limitsx to pm giới hạn d căn x^6 - x + 1 căn [3]x^3 + x + 1")
Câu hỏi
Nhận biết\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{\sqrt {{x^6} - x + 1} }}{{\sqrt[3]{{{x^3} + x + 1}}}}\)
Đáp án đúng: B
Lời giải của Tự Học 365
Giải chi tiết:
\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{\sqrt {{x^6} - x + 1} }}{{\sqrt[3]{{{x^3} + x + 1}}}}\)
\(\begin{array}{l} + )\,\,\,\mathop {\lim }\limits_{x \to + \infty } \dfrac{{\sqrt {{x^6} - x + 1} }}{{\sqrt[3]{{{x^3} + x + 1}}}} = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{\sqrt {{x^6}} \sqrt {1 - \dfrac{1}{{{x^5}}} + \dfrac{1}{{{x^6}}}} }}{{\sqrt[3]{{{x^3}}}.\sqrt[3]{{1 + \dfrac{1}{{{x^2}}} + \dfrac{1}{{{x^3}}}}}}}\\ = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{{x^3}\sqrt {1 - \dfrac{1}{{{x^5}}} + \dfrac{1}{{{x^6}}}} }}{{x.\sqrt[3]{{1 + \dfrac{1}{{{x^2}}} + \dfrac{1}{{{x^3}}}}}}} = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{{x^2}\sqrt {1 - \dfrac{1}{{{x^5}}} + \dfrac{1}{{{x^6}}}} }}{{\sqrt[3]{{1 + \dfrac{1}{{{x^2}}} + \dfrac{1}{{{x^3}}}}}}}\end{array}\)
Ta có: \(\left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to + \infty } {x^2} = + \infty \\\mathop {\lim }\limits_{x \to + \infty } \dfrac{{\sqrt {1 - \dfrac{1}{{{x^5}}} + \dfrac{1}{{{x^6}}}} }}{{\sqrt[3]{{1 + \dfrac{1}{{{x^2}}} + \dfrac{1}{{{x^3}}}}}}} = 1 > 0\end{array} \right. \Rightarrow \mathop {\lim }\limits_{x \to + \infty } \dfrac{{\sqrt {{x^6} - x + 1} }}{{\sqrt[3]{{{x^3} + x + 1}}}} = + \infty \)
\(\begin{array}{l} + )\,\,\,\mathop {\lim }\limits_{x \to - \infty } \dfrac{{\sqrt {{x^6} - x + 1} }}{{\sqrt[3]{{{x^3} + x + 1}}}} = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\sqrt {{x^6}} \sqrt {1 - \dfrac{1}{{{x^5}}} + \dfrac{1}{{{x^6}}}} }}{{\sqrt[3]{{{x^3}}}.\sqrt[3]{{1 + \dfrac{1}{{{x^2}}} + \dfrac{1}{{{x^3}}}}}}}\\ = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{ - {x^3}\sqrt {1 - \dfrac{1}{{{x^5}}} + \dfrac{1}{{{x^6}}}} }}{{x.\sqrt[3]{{1 + \dfrac{1}{{{x^2}}} + \dfrac{1}{{{x^3}}}}}}} = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{ - {x^2}\sqrt {1 - \dfrac{1}{{{x^5}}} + \dfrac{1}{{{x^6}}}} }}{{\sqrt[3]{{1 + \dfrac{1}{{{x^2}}} + \dfrac{1}{{{x^3}}}}}}}\end{array}\)
Ta có: \(\left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to - \infty } \left( { - {x^2}} \right) = - \infty \\\mathop {\lim }\limits_{x \to - \infty } \dfrac{{\sqrt {1 - \dfrac{1}{{{x^5}}} + \dfrac{1}{{{x^6}}}} }}{{\sqrt[3]{{1 + \dfrac{1}{{{x^2}}} + \dfrac{1}{{{x^3}}}}}}} = 1 > 0\end{array} \right. \Rightarrow \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\sqrt {{x^6} - x + 1} }}{{\sqrt[3]{{{x^3} + x + 1}}}} = - \infty \).
Vậy \(\mathop {\lim }\limits_{x \to + \infty } \dfrac{{\sqrt {{x^6} - x + 1} }}{{\sqrt[3]{{{x^3} + x + 1}}}} = + \infty \), \(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{\sqrt {{x^6} - x + 1} }}{{\sqrt[3]{{{x^3} + x + 1}}}} = - \infty \).
Chọn B.
