mathop lim limitsx to pm infty ( căn 9x^2 - x + 1 - 3x + 1 )
Câu hỏi
Nhận biết\( \mathop { \lim } \limits_{x \to \pm \infty } \left( { \sqrt {9{x^2} - x + 1} - 3x + 1} \right) \)
Đáp án đúng: A
Lời giải của Tự Học 365
Giải chi tiết:
Ta có:
\(\begin{array}{l}\,\,\,\,\,\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {9{x^2} - x + 1} - 3x + 1} \right)\\ = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{\left( {\sqrt {9{x^2} - x + 1} - 3x + 1} \right)\left( {\sqrt {9{x^2} - x + 1} + 3x - 1} \right)}}{{\sqrt {9{x^2} - x + 1} + 3x - 1}}\\ = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{9{x^2} - x + 1 - 9{x^2} + 6x - 1}}{{\sqrt {9{x^2} - x + 1} + 3x - 1}}\\ = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{5x}}{{\sqrt {9{x^2} - x + 1} + 3x - 1}}\\ = \mathop {\lim }\limits_{x \to + \infty } \dfrac{5}{{\sqrt {9 - \dfrac{1}{x} + \dfrac{1}{{{x^2}}}} + 3 - \dfrac{1}{x}}}\\ = \dfrac{5}{{3 + 3}} = \dfrac{5}{6}\end{array}\)
\(\begin{array}{l}\,\,\,\,\,\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {9{x^2} - x + 1} - 3x + 1} \right)\\ = \mathop {\lim }\limits_{x \to - \infty } x\left( { - \sqrt {9 - \dfrac{1}{x} + \dfrac{1}{{{x^2}}}} - 3 + \dfrac{1}{x}} \right) = + \infty \end{array}\)
Vì \(\mathop {\lim }\limits_{x \to - \infty } x = - \infty ;\,\,\mathop {\lim }\limits_{x \to - \infty } \left( { - \sqrt {9 - \dfrac{1}{x} + \dfrac{1}{{{x^2}}}} - 3 + \dfrac{1}{x}} \right) = - 6 < 0\).
Chọn A.
