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