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: $\mathop {\lim }\limits_{x \to  - 3} \frac{{ - {x^2} + 2x + 15}}{{{x^2} + 4x + 3}} = \mathop {\lim }\limits_{x \to  - 3} \frac{{\left( {x + 3} \right)\left( {5 - x} \right)}}{{\left( {x + 3} \right)\left( {x + 1} \right)}} = \mathop {\lim }\limits_{x \to  - 3} \frac{{5 - x}}{{x + 1}} = \frac{{\mathop {\lim }\limits_{x \to  - 3} 5 - \mathop {\lim }\limits_{x \to  - 3} x}}{{\mathop {\lim }\limits_{x \to  - 3} x + \mathop {\lim }\limits_{x \to  - 3} 1}}$

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