Bài 9 trang 86 SGK Toán 11 tập 1 - Chân trời sáng tạo

Đề bài

Tìm các giới hạn sau:

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

b) \(\mathop {\lim }\limits_{x \to  - \infty } \frac{{x - 2}}{{{x^2}}}\).

Lời giải chi tiết

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

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

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