Bài tập 21 trang 76 SBT Toán 11 Tập 1 Cánh diều - CD

a) Ta có: \(\mathop {\lim }\limits_{x \to - \infty } \frac{{ - 5x + 2}}{{3x + 1}}\)

\(= \mathop {\lim }\limits_{x \to - \infty } \frac{{x\left( { - 5 + \frac{2}{x}} \right)}}{{x\left( {3 + \frac{1}{x}} \right)}} \\= \mathop {\lim }\limits_{x \to - \infty } \frac{{ - 5 + \frac{2}{x}}}{{3 + \frac{1}{x}}} \\= \frac{{\mathop {\lim }\limits_{x \to - \infty } \left( { - 5} \right) + \mathop {\lim }\limits_{x \to - \infty } \frac{2}{x}}}{{\mathop {\lim }\limits_{x \to - \infty } 3 + \mathop {\lim }\limits_{x \to - \infty } \frac{1}{x}}}\)

\( = \frac{{ - 5 + 0}}{{3 + 0}} = \frac{{ - 5}}{3}\)

b) Ta có: \(\mathop {\lim }\limits_{x \to - \infty } \frac{{ - 2x + 3}}{{3{x^2} + 2x + 5}}\)

\(= \mathop {\lim }\limits_{x \to - \infty } \frac{{{x^2}\left( {\frac{{ - 2}}{x} + \frac{3}{{{x^2}}}} \right)}}{{{x^2}\left( {3 + \frac{2}{x} + \frac{5}{{{x^2}}}} \right)}} \\= \mathop {\lim }\limits_{x \to - \infty } \frac{{\frac{{ - 2}}{x} + \frac{3}{{{x^2}}}}}{{3 + \frac{2}{x} + \frac{5}{{{x^2}}}}}\)

\( = \frac{{\mathop {\lim }\limits_{x \to - \infty } \frac{{ - 2}}{x} + \mathop {\lim }\limits_{x \to - \infty } \frac{3}{{{x^2}}}}}{{\mathop {\lim }\limits_{x \to - \infty } 3 + \mathop {\lim }\limits_{x \to - \infty } \frac{2}{x} + \mathop {\lim }\limits_{x \to - \infty } \frac{5}{{{x^2}}}}} \\= \frac{{0 + 0}}{{3 + 0 + 0}} = 0\)

c) Ta có: \(\mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {9{x^2} + 3} }}{{x + 1}}\)

\(= \mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {{x^2}\left( {9 + \frac{3}{{{x^2}}}} \right)} }}{{x\left( {1 + \frac{1}{x}} \right)}} \\= \mathop {\lim }\limits_{x \to + \infty } \frac{{x\sqrt {9 + \frac{3}{{{x^2}}}} }}{{x\left( {1 + \frac{1}{x}} \right)}} \\= \mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {9 + \frac{3}{{{x^2}}}} }}{{1 + \frac{1}{x}}} \\= \frac{{\mathop {\lim }\limits_{x \to + \infty } \sqrt {9 + \frac{3}{{{x^2}}}} }}{{\mathop {\lim }\limits_{x \to + \infty } \left( {1 + \frac{1}{x}} \right)}}\)

Do \(\mathop {\lim }\limits_{x \to + \infty } \left( {9 + \frac{3}{{{x^2}}}} \right) = \mathop {\lim }\limits_{x \to + \infty } 9 + \mathop {\lim }\limits_{x \to + \infty } \frac{3}{{{x^2}}} = 9 + 0 = 9\).

Nên \(\mathop {\lim }\limits_{x \to + \infty } \sqrt {9 + \frac{3}{{{x^2}}}} = \sqrt 9 = 3\).

Mặt khác, \(\mathop {\lim }\limits_{x \to + \infty } \left( {1 + \frac{1}{x}} \right) = \mathop {\lim }\limits_{x \to + \infty } 1 + \mathop {\lim }\limits_{x \to + \infty } \frac{1}{x} = 1 + 0 = 1\).

Suy ra \(\mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {9{x^2} + 3} }}{{x + 1}} = \frac{{\mathop {\lim }\limits_{x \to + \infty } \sqrt {9 + \frac{3}{{{x^2}}}} }}{{\mathop {\lim }\limits_{x \to + \infty } \left( {1 + \frac{1}{x}} \right)}} = \frac{3}{1} = 3\).

d) Ta có: \(\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {9{x^2} + 3} }}{{x + 1}}\)

\(= \mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {{x^2}\left( {9 + \frac{3}{{{x^2}}}} \right)} }}{{x\left( {1 + \frac{1}{x}} \right)}} \\= \mathop {\lim }\limits_{x \to - \infty } \frac{{\left( { - x} \right)\sqrt {9 + \frac{3}{{{x^2}}}} }}{{x\left( {1 + \frac{1}{x}} \right)}}\)

-- Mod Toán 11 HỌC247