mathop lim limitsx to pm giới hạn d( 3x - 1 ) căn x^6 + x + 1 căn x^8 - x + 2
Câu hỏi
Nhận biết\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{\left( {3x - 1} \right)\sqrt {{x^6} + x + 1} }}{{\sqrt {{x^8} - x + 2} }}\)
Đáp án đúng: C
Lời giải của Tự Học 365
Giải chi tiết:
\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{\left( {3x - 1} \right)\sqrt {{x^6} + x + 1} }}{{\sqrt {{x^8} - x + 2} }}\)
\(\begin{array}{l}\,\,\,\,\mathop {\lim }\limits_{x \to + \infty } \dfrac{{\left( {3x - 1} \right)\sqrt {{x^6} + x + 1} }}{{\sqrt {{x^8} - x + 2} }}\\ = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{x\left( {3 - \dfrac{1}{x}} \right){x^3}\sqrt {1 + \dfrac{1}{{{x^5}}} + \dfrac{1}{{{x^6}}}} }}{{{x^4}\sqrt {1 - \dfrac{1}{{{x^7}}} + \dfrac{2}{{{x^8}}}} }}\\ = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{\left( {3 - \dfrac{1}{x}} \right)\sqrt {1 + \dfrac{1}{{{x^5}}} + \dfrac{1}{{{x^6}}}} }}{{\sqrt {1 - \dfrac{1}{{{x^7}}} + \dfrac{2}{{{x^8}}}} }} = \dfrac{{3.1}}{1} = 3\end{array}\)
\(\begin{array}{l}\,\,\,\,\mathop {\lim }\limits_{x \to - \infty } \dfrac{{\left( {3x - 1} \right)\sqrt {{x^6} + x + 1} }}{{\sqrt {{x^8} - x + 2} }}\\ = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{ - x\left( {3 - \dfrac{1}{x}} \right){x^3}\sqrt {1 + \dfrac{1}{{{x^5}}} + \dfrac{1}{{{x^6}}}} }}{{{x^4}\sqrt {1 - \dfrac{1}{{{x^7}}} + \dfrac{2}{{{x^8}}}} }}\\ = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{ - \left( {3 - \dfrac{1}{x}} \right)\sqrt {1 + \dfrac{1}{{{x^5}}} + \dfrac{1}{{{x^6}}}} }}{{\sqrt {1 - \dfrac{1}{{{x^7}}} + \dfrac{2}{{{x^8}}}} }} = \dfrac{{ - 3.1}}{1} = - 3\end{array}\)
Vậy \(\mathop {\lim }\limits_{x \to + \infty } \dfrac{{\left( {3x - 1} \right)\sqrt {{x^6} + x + 1} }}{{\sqrt {{x^8} - x + 2} }} = 3\), \(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{\left( {3x - 1} \right)\sqrt {{x^6} + x + 1} }}{{\sqrt {{x^8} - x + 2} }} = - 3\).
Chọn C.
