Cho \(m\) là một số thực. Biết \(\mathop {\lim }\limits_{x \to - \infty } [(m - x)(mx + 1)] = - \infty \). Xác định dấu của \(m\).

Chọn C

Ta có: + \(\mathop {\lim }\limits_{x \to  - 1} {\mkern 1mu} f(x) = 4 > 0\).

+ \(\mathop {\lim }\limits_{x \to  - 1} {\mkern 1mu} {\left( {x + 1} \right)^4} = 0\) và với \(\forall x \ne  - 1\) thì \({\left( {x + 1} \right)^4} > 0\).

Suy ra \(\mathop {\lim }\limits_{x \to  - 1} {\mkern 1mu} \frac{{f(x)}}{{{{\left( {x + 1} \right)}^4}}} =  + \infty \).

a) \(\mathop {\lim }\limits_{x \to - \infty } \left( {{x^2} - 10x} \right) = \mathop {\lim }\limits_{x \to - \infty } {x^2}\left( {1 - \frac{{10}}{x}} \right) = + \infty \).

b) \(\mathop {\lim }\limits_{x \to + \infty } \frac{{3{x^2} - 4x + 1}}{{2{x^2} + x + 1}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{{x^2}\left( {3 - \frac{4}{x} + \frac{1}{{{x^2}}}} \right)}}{{{x^2}\left( {2 + \frac{1}{x} + \frac{1}{{{x^2}}}} \right)}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{3 - \frac{4}{x} + \frac{1}{{{x^2}}}}}{{2 + \frac{1}{x} + \frac{1}{{{x^2}}}}} = \frac{3}{2}\)

c)

\(\begin{array}{*{20}{l}}{\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {{x^2} + x + 1} - 3x}}{{2 - 3x}}}&{ = \mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {{x^2}\left( {1 + \frac{1}{x} + \frac{1}{{{x^2}}}} \right)} - 3x}}{{x\left( {\frac{2}{x} - 3} \right)}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{ - x\sqrt {1 + \frac{1}{x} + \frac{1}{{{x^2}}}} - 3x}}{{x\left( {\frac{2}{x} - 3} \right)}}}\\{}&{ = \mathop {\lim }\limits_{x \to - \infty } \frac{{ - \sqrt {1 + \frac{1}{x} + \frac{1}{{{x^2}}}} - 3}}{{\frac{2}{x} - 3}} = \frac{{ - \sqrt 1 - 3}}{{ - 3}} = \frac{4}{3}}\end{array}\)

d)

\(\begin{array}{l}\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt[3]{{8{x^3} + 3{x^2} + 1}} - x}}{{\sqrt {4{x^2} - x + 2} + 3x}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt[3]{{{x^3}\left( {8 + \frac{3}{x} + \frac{1}{{{x^3}}}} \right)}} - x}}{{\sqrt {{x^2}\left( {4 - \frac{1}{x} + \frac{2}{{{x^2}}}} \right)} + 3x}}\\ = \mathop {\lim }\limits_{x \to - \infty } \frac{{x\sqrt[3]{{8 + \frac{3}{x} + \frac{1}{{{x^3}}}}} - x}}{{x\sqrt {4 - \frac{1}{x} + \frac{2}{{{x^2}}}} + 3x}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt[3]{{8 + \frac{3}{x} + \frac{1}{{{x^3}}}}} - 1}}{{\sqrt {4 - \frac{1}{x} + \frac{2}{{{x^2}}}} + 3}} = \frac{{\sqrt[3]{8} - 1}}{{ - \sqrt 4 + 3}} = 1\end{array}\)