Bài tập 2 trang 84 SBT Toán 11 Tập 1 Chân trời sáng tạo - CTST

a) Ta có: $\underset{x\to -3}{\lim }\left(8+3x-x^2\right)$

= $8+3\underset{x\to -3}{\lim }x-\underset{x\to -3}{\lim }{x}^2$

= 8 + 3.(-3) - (-3)² = - 10.

b) Ta có: $\underset{x\to 2}{\lim }\left[\left(5x-1\right)\left(2-4x\right)\right]$

$=\underset{x\to 2}{\lim }\left(5x-1\right)\cdot \underset{x\to 2}{\lim }\left(2-4x\right)$

$=\left(5\underset{x\to 2}{\lim }x-1\right)\cdot \left(2-4\underset{x\to 2}{\lim }x\right)$

= (5.2 ‒ 1)(2 ‒ 4.2) = ‒ 54.

c) Ta có: $\underset{x\to -2}{\lim }\frac{x^2-x}{{\left(2x+1\right)}^2}$

$=\frac{\underset{x\to -2}{\lim }\left(x^2-x\right)}{\underset{x\to -2}{\lim }\left(4x^2+4x+1\right)}$

$=\frac{\underset{x\to -2}{\lim }{x}^2-\underset{x\to -2}{\lim }x}{4\underset{x\to -2}{\lim }{x}^2+4\underset{x\to -2}{\lim }x+1}$

$=\frac{{\left(-2\right)}^2-\left(-2\right)}{4\cdot {\left(-2\right)}^2+4\cdot \left(-2\right)+1}=\frac{6}{9}=\frac{2}{3}.$

d) Ta có: $\underset{x\to -1}{\lim }\sqrt{10-2x^2}$

$=\sqrt{\underset{x\to -1}{\lim }\left(10-2x^2\right)}$

$=\sqrt{10-\underset{x\to -1}{\lim }2x^2}$

$=\sqrt{10-2\underset{x\to -1}{\lim }{x}^2}$

$=\sqrt{10-2.{\left(-1\right)}^2}=\sqrt{8}=2\sqrt{2}.$

-- Mod Toán 11 HỌC247