Giải bài 21 trang 76 sách bài tập toán 11 – Cánh diều

Đề bài

Tính các giới hạn sau:

a) \(\mathop {\lim }\limits_{x \to  – \infty } \frac{{ – 5x + 2}}{{3x + 1}}\)                   

b) \(\mathop {\lim }\limits_{x \to  – \infty } \frac{{ – 2x + 3}}{{3{x^2} + 2x + 5}}\)

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

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

e) \(\mathop {\lim }\limits_{x \to 1} \frac{{2{x^2} – 8x + 6}}{{{x^2} – 1}}\)               

g) \(\mathop {\lim }\limits_{x \to  – 3} \frac{{ – {x^2} + 2x + 15}}{{{x^2} + 4x + 3}}\)

Lời giải chi tiết

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

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

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

e) Ta có: \(\mathop {\lim }\limits_{x \to 1} \frac{{2{x^2} – 8x + 6}}{{{x^2} – 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x – 1} \right)\left( {2x – 6} \right)}}{{\left( {x – 1} \right)\left( {x + 1} \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{{2x – 6}}{{x + 1}} = \frac{{\mathop {\lim }\limits_{x \to 1} 2x – \mathop {\lim }\limits_{x \to 1} 6}}{{\mathop {\lim }\limits_{x \to 1} x + \mathop {\lim }\limits_{x \to 1} 1}}\)

\( = \frac{{2.1 – 6}}{{1 + 1}} =  – 2\).

f) Ta có: \(\mathop {\lim }\limits_{x \to  – 3} \frac{{ – {x^2} + 2x + 15}}{{{x^2} + 4x + 3}} = \mathop {\lim }\limits_{x \to  – 3} \frac{{\left( {x + 3} \right)\left( {5 – x} \right)}}{{\left( {x + 3} \right)\left( {x + 1} \right)}} = \mathop {\lim }\limits_{x \to  – 3} \frac{{5 – x}}{{x + 1}} = \frac{{\mathop {\lim }\limits_{x \to  – 3} 5 – \mathop {\lim }\limits_{x \to  – 3} x}}{{\mathop {\lim }\limits_{x \to  – 3} x + \mathop {\lim }\limits_{x \to  – 3} 1}}\)

\( = \frac{{5 – \left( { – 3} \right)}}{{\left( { – 3} \right) + 1}} =  – 4\).