Câu hỏi 10 - Mục Bài tập trang 85

1. Nội dung câu hỏi

Tính các giới hạn sau:

a) \(\mathop {\lim }\limits_{x \to  - \infty } \left( {{x^3} + 2{x^2} - 1} \right)\);

b) \(\mathop {\lim }\limits_{x \to  + \infty } \frac{{{x^3} + 2{x^2}}}{{3{x^2} + 1}}\);

c) \(\mathop {\lim }\limits_{x \to  - \infty } \sqrt {{x^2} - 2x + 3} \).

3. Lời giải chi tiết 

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

Vì \(\mathop {\lim }\limits_{x \to  - \infty } {x^3} \) \( =  - \infty ;\mathop {\lim }\limits_{x \to  - \infty } \left( {1 + \frac{2}{x} - \frac{1}{{{x^3}}}} \right) \) \( = \mathop {\lim }\limits_{x \to  - \infty } 1 + \mathop {\lim }\limits_{x \to  - \infty } \frac{2}{x} - \mathop {\lim }\limits_{x \to  - \infty } \frac{1}{{{x^3}}} \) \( = 1 > 0\)

Do đó, \(\mathop {\lim }\limits_{x \to  - \infty } \left( {{x^3} + 2{x^2} - 1} \right) \) \( = \mathop {\lim }\limits_{x \to  - \infty } \left[ {{x^3}\left( {1 + \frac{2}{x} - \frac{1}{{{x^3}}}} \right)} \right] \) \( =  - \infty \)

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

Ta có: \(\mathop {\lim }\limits_{x \to  + \infty } x \) \( =  + \infty ,\mathop {\lim }\limits_{x \to  + \infty } \frac{{1 + \frac{2}{x}}}{{3 + \frac{1}{{{x^2}}}}} \) \( = \frac{{1 + \mathop {\lim }\limits_{x \to  + \infty } \frac{2}{x}}}{{3 + \mathop {\lim }\limits_{x \to  + \infty } \frac{1}{{{x^2}}}}} \) \( = \frac{1}{3} > 0\)

Do đó, \(\mathop {\lim }\limits_{x \to  + \infty } \frac{{{x^3} + 2{x^2}}}{{3{x^2} + 1}} \) \( = \mathop {\lim }\limits_{x \to  + \infty } \left[ {x.\frac{{1 + \frac{2}{x}}}{{3 + \frac{1}{{{x^2}}}}}} \right] \) \( =  + \infty \)

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

Ta có: \(\mathop {\lim }\limits_{x \to  - \infty } \left( { - x} \right) \) \( =  + \infty ;\mathop {\lim }\limits_{x \to  - \infty } \sqrt {1 - \frac{2}{x} + \frac{3}{{{x^2}}}}  \) \( = \sqrt {1 - \mathop {\lim }\limits_{x \to  - \infty } \frac{2}{x} + \mathop {\lim }\limits_{x \to  - \infty } \frac{3}{{{x^2}}}}  \) \( = 1 > 0\)

Do đó, \(\mathop {\lim }\limits_{x \to  - \infty } \sqrt {{x^2} - 2x + 3}  \) \( = \mathop {\lim }\limits_{x \to  - \infty } \left[ { - x\sqrt {1 - \frac{2}{x} + \frac{3}{{{x^2}}}} } \right] \) \( =  + \infty \).