Giải Bài 4 trang 79 SGK Toán 11 Chân trời sáng tạo tập 1 - CTST

Phương pháp giải - Xem chi tiết

Lời giải chi tiết

a) Áp dụng giới hạn một bên thường dùng,

Ta có : $\left\{ \begin{array}{l}1 > 0\\x - \left( { - 1} \right) > 0,x \to - {1^ + }\end{array} \right.$$\Rightarrow \mathop {\lim }\limits_{x \to - {1^ + }} \frac{1}{{x + 1}}$$= \mathop {\lim }\limits_{x \to - {1^ + }} \frac{1}{{x - \left( { - 1} \right)}}$$= + \infty$

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

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

$\Rightarrow \mathop {\lim }\limits_{x \to - \infty } \left( {1 - {x^2}} \right)$$= - \infty$

c) $\mathop {\lim }\limits_{x \to {3^ - }} \frac{x}{{3 - x}}$$= \mathop {\lim }\limits_{x \to {3^ - }} \frac{{ - x}}{{x - 3}}$$= \mathop {\lim }\limits_{x \to {3^ - }} \left( { - x} \right).\mathop {\lim }\limits_{x \to {3^ - }} \frac{1}{{x - 3}}$

Ta có: $\mathop {\lim }\limits_{x \to {3^ - }} \left( { - x} \right)$$= - \mathop {\lim }\limits_{x \to {3^ - }} x = - 3;$$\mathop {\lim }\limits_{x \to {3^ - }} \frac{1}{{x - 3}}$$= - \infty$

$\Rightarrow \mathop {\lim }\limits_{x \to {3^ - }} \frac{x}{{3 - x}}$$= + \infty$

-- Mod Toán 11 HỌC247