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

\( \Rightarrow \mathop {\lim }\limits_{x \to {3^ – }} \frac{x}{{3 – x}} =  + \infty \)