Giải bài 9 trang 85 sách bài tập toán 11 - Chân trời sáng tạo tập 1

Đề bài

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

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

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