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

Đề bài

Tìm các giới hạn sau:

a) $\mathop {\lim }\limits_{x \to  - {1^ + }} \frac{1}{{x + 1}}$;              

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

c) $\mathop {\lim }\limits_{x \to {3^ - }} \frac{x}{{3 - x}}$.

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

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