Câu hỏi 9 - Mục Bài tập trang 85

1. Nội dung câu hỏi

Tìm các giới hạn sau:

a) \(\mathop {\lim }\limits_{x \to  + \infty } \frac{x}{{x + 4}}\);

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

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

d) \(\mathop {\lim }\limits_{x \to  + \infty } \left( {x - \sqrt {{x^2} + 2x} } \right)\).

3. Lời giải chi tiết 

a) \(\mathop {\lim }\limits_{x \to  + \infty } \frac{x}{{x + 4}} \) \( = \mathop {\lim }\limits_{x \to  + \infty } \frac{1}{{1 + \frac{4}{x}}} \) \( = \frac{1}{{1 + 4\mathop {\lim }\limits_{x \to  + \infty } \frac{1}{x}}} \) \( = \frac{1}{{1 + 4.0}} \) \( = 1\);

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

c) Với \(x < 0\) thì \(\sqrt {{x^2}}  \) \( = \left| x \right| \) \( =  - x\).

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

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

\( \) \( = \mathop {\lim }\limits_{x \to  + \infty } \frac{{ - 2}}{{1 + \sqrt {1 + \frac{2}{x}} }} \) \( = \frac{{ - 2}}{{1 + \sqrt {1 + \mathop {\lim }\limits_{x \to  + \infty } \frac{2}{x}} }} \) \( = \frac{{ - 2}}{{1 + \sqrt {1 + 0} }} \) \( =  - 1\).