Bài tập 10 trang 85 SBT Toán 11 Tập 1 Chân trời sáng tạo - CTST

a) $\underset{x\to -\infty }{\lim }\left(x^3+2x^2-1\right)=\underset{x\to -\infty }{\lim }\left[x^3\left(1+\frac{2}{x}-\frac{1}{x^3}\right)\right]$

Ta có: $\underset{x\to -\infty }{\lim }{x}^3=-\infty$ và $\underset{x\to -\infty }{\lim }\left(1+\frac{2}{x}-\frac{1}{x^3}\right)=1+0-0=1.$

Suy ra $\underset{x\to -\infty }{\lim }\left(x^3+2x^2-1\right)=\underset{x\to -\infty }{\lim }\left[x^3\left(1+\frac{2}{x}-\frac{1}{x^3}\right)\right]=-\infty .$

b) $\underset{x\to +\infty }{\lim }\frac{x^3+2x^2}{3x^2+1}=\underset{x\to +\infty }{\lim }\left[x\cdot \frac{x^2+2x}{3x^2+1}\right]$

Ta có: $\underset{x\to +\infty }{\lim }x=+\infty$ và $\underset{x\to +\infty }{\lim }\frac{x^2+2x}{3x^2+1}=\underset{x\to +\infty }{\lim }\frac{1+\frac{2}{x}}{3+\frac{1}{x^2}}=\frac{1}{3}$

Suy ra $\underset{x\to +\infty }{\lim }\frac{x^3+2x^2}{3x^2+1}=\underset{x\to +\infty }{\lim }\left[x\cdot \frac{x^2+2x}{3x^2+1}\right]=+\infty .$

c) $\underset{x\to -\infty }{\lim }\sqrt{x^2-2x+3}=\underset{x\to -\infty }{\lim }\sqrt{x^2\left(1-\frac{2}{x}+\frac{3}{x^2}\right)}$

$=\underset{x\to -\infty }{\lim }\left(\left|x\right|\sqrt{1-\frac{2}{x}+\frac{3}{x^2}}\right)$

Ta có: $\underset{x\to -\infty }{\lim }\left|x\right|=\underset{x\to -\infty }{\lim }\left(-x\right)=+\infty$ và $\underset{x\to -\infty }{\lim }\sqrt{1-\frac{2}{x}+\frac{3}{x^2}}=1$

Suy ra $\underset{x\to -\infty }{\lim }\sqrt{x^2-2x+3}=\underset{x\to -\infty }{\lim }\left(\left|x\right|\sqrt{1-\frac{2}{x}+\frac{3}{x^2}}\right)=+\infty .$

-- Mod Toán 11 HỌC247