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