Trả lời câu hỏi 44 - Mục câu hỏi trắc nghiệm trang 83

1. Nội dung câu hỏi

Tính các giới hạn sau:

a) \(\mathop {\lim }\limits_{x \to  - \infty } \frac{{2 + \frac{4}{{3x}}}}{{{x^2} - 1}}\)                           

b) \(\mathop {\lim }\limits_{x \to {2^ + }} \frac{1}{{x - 2}}\)                  

c) \(\mathop {\lim }\limits_{x \to  - {3^ + }} \frac{{ - 5 + x}}{{x + 3}}\)

d) \(\mathop {\lim }\limits_{x \to  - \infty } \frac{{14x + 2}}{{ - 7x + 1}}\)                          

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

g) \(\mathop {\lim }\limits_{x \to  - \infty } \frac{{\sqrt {4{x^2} + 1} }}{{x + 2}}\)

h) \(\mathop {\lim }\limits_{x \to 1} \frac{{x - 1}}{{{x^2} - 1}}\)                     

i) \(\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} - 5x + 6}}{{x - 2}}\)               

k) \(\mathop {\lim }\limits_{x \to 3} \frac{{ - {x^2} + 4x - 3}}{{{x^2} + 3x - 18}}\)

3. Lời giải chi tiết

a) Ta có \(\mathop {\lim }\limits_{x \to  - \infty } \left( {2 + \frac{4}{{3x}}} \right) = \mathop {\lim }\limits_{x \to  - \infty } 2 + \mathop {\lim }\limits_{x \to  - \infty } \frac{4}{{3x}} = 2 + 0 = 2\).

Mặt khác, \(\mathop {\lim }\limits_{x \to  - \infty } \left( {{x^2} - 1} \right) = \mathop {\lim }\limits_{x \to  - \infty } \left[ {{x^2}\left( {1 - \frac{1}{{{x^2}}}} \right)} \right] = \mathop {\lim }\limits_{x \to  - \infty } {x^2}.\mathop {\lim }\limits_{x \to  - \infty } \left( {1 - \frac{1}{{{x^2}}}} \right) =  + \infty \)

Suy ra \(\mathop {\lim }\limits_{x \to  - \infty } \frac{{2 + \frac{4}{{3x}}}}{{{x^2} - 1}} = 0\).

b) Ta có \(\mathop {\lim }\limits_{x \to {2^ + }} \frac{1}{{x - 2}} =  + \infty \).

c) Ta có \(\mathop {\lim }\limits_{x \to  - {3^ + }} \left( { - 5 + x} \right) = \left( { - 5} \right) + \left( { - 3} \right) =  - 2 < 0\).

Suy ra \(\mathop {\lim }\limits_{x \to  - {3^ + }} \frac{{ - 5 + x}}{{x + 3}} =  - \infty \).

d) Ta có:\(\mathop {\lim }\limits_{x \to  - \infty } \frac{{14x + 2}}{{ - 7x + 1}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{x\left( {14 + \frac{2}{x}} \right)}}{{x\left( { - 7 + \frac{1}{x}} \right)}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{14 + \frac{2}{x}}}{{ - 7 + \frac{1}{x}}} = \frac{{\mathop {\lim }\limits_{x \to  - \infty } 14 + \mathop {\lim }\limits_{x \to  - \infty } \frac{2}{x}}}{{\mathop {\lim }\limits_{x \to  - \infty } \left( { - 7} \right) + \mathop {\lim }\limits_{x \to  - \infty } \frac{1}{x}}}\)

\( = \frac{{14 + 0}}{{ - 7 + 0}} =  - 2\).

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

Ta thấy \(\mathop {\lim }\limits_{x \to  + \infty } \left( { - 2x} \right) =  - \infty \) và \(\mathop {\lim }\limits_{x \to  + \infty } \left( {3 + \frac{5}{x}} \right) = \mathop {\lim }\limits_{x \to  + \infty } 3 + \mathop {\lim }\limits_{x \to  + \infty } \frac{5}{x} = 3 + 0 = 3\).

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

g) Ta có: \(\mathop {\lim }\limits_{x \to  - \infty } \frac{{\sqrt {4{x^2} + 1} }}{{x + 2}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{\sqrt {{x^2}\left( {4 + \frac{1}{{{x^2}}}} \right)} }}{{x\left( {1 + \frac{2}{x}} \right)}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{\left| x \right|\sqrt {4 + \frac{1}{{{x^2}}}} }}{{x\left( {1 + \frac{2}{x}} \right)}}\)

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

Vì \(\mathop {\lim }\limits_{x \to  - \infty } \left( {4 + \frac{1}{{{x^2}}}} \right) = \mathop {\lim }\limits_{x \to  - \infty } 4 + \mathop {\lim }\limits_{x \to  - \infty } \frac{1}{{{x^2}}} = 4 + 0 = 4\) nên \(\mathop {\lim }\limits_{x \to  - \infty } \sqrt {4 + \frac{1}{{{x^2}}}}  = \sqrt 4  = 2\).

Mặt khác, \(\mathop {\lim }\limits_{x \to  - \infty } \left( {1 + \frac{2}{x}} \right) = \mathop {\lim }\limits_{x \to  - \infty } 1 + \mathop {\lim }\limits_{x \to  - \infty } \frac{2}{x} = 1 + 0 = 1\).

Như vậy \(\mathop {\lim }\limits_{x \to  - \infty } \frac{{\sqrt {4{x^2} + 1} }}{{x + 2}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{ - \sqrt {4 + \frac{1}{{{x^2}}}} }}{{1 + \frac{2}{x}}} = \frac{{ - 2}}{1} =  - 2\).

h) Ta có \(\mathop {\lim }\limits_{x \to 1} \frac{{x - 1}}{{{x^2} - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{x - 1}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{1}{{x + 1}} = \frac{{\mathop {\lim }\limits_{x \to 1} 1}}{{\mathop {\lim }\limits_{x \to 1} x + \mathop {\lim }\limits_{x \to 1} 1}} = \frac{1}{{1 + 1}} = \frac{1}{2}\).

i) \(\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} - 5x + 6}}{{x - 2}} = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {x - 2} \right)\left( {x - 3} \right)}}{{x - 2}} = \mathop {\lim }\limits_{x \to 2} \left( {x - 3} \right) = \mathop {\lim }\limits_{x \to 2} x + \mathop {\lim }\limits_{x \to 2} 3 = 2 + 3 = 5\).

k) \(\mathop {\lim }\limits_{x \to 3} \frac{{ - {x^2} + 4x - 3}}{{{x^2} + 3x - 18}} = \mathop {\lim }\limits_{x \to 3} \frac{{\left( {x - 3} \right)\left( {1 - x} \right)}}{{\left( {x - 3} \right)\left( {x + 6} \right)}} = \mathop {\lim }\limits_{x \to 3} \frac{{1 - x}}{{x + 6}} = \frac{{\mathop {\lim }\limits_{x \to 3} 1 - \mathop {\lim }\limits_{x \to 3} x}}{{\mathop {\lim }\limits_{x \to 3} x + \mathop {\lim }\limits_{x \to 3} 6}} = \frac{{1 - 3}}{{3 + 6}} = \frac{{ - 2}}{9}\).