Bài tập 9 trang 85 SBT Toán 11 Tập 1 Chân trời sáng tạo - CTST

a) Ta có: $\underset{x\to +\infty }{\lim }\frac{x}{x+4}=\underset{x\to +\infty }{\lim }\frac{1}{1+\frac{4}{x}}=\frac{1}{1+4\cdot 0}=1.$

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

c) Ta có: Với x < 0 thì $\sqrt{x^2}=\mathrm{|x|}=-x,$ nên ta có:

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

$=\underset{x\to -\infty }{\lim }\frac{3+\frac{1}{x}}{-\sqrt{1-\frac{2}{x}}}$

$=-\frac{3+0}{\sqrt{1-2\cdot 0}}=-3.$

d) Ta có: $\underset{x\to +\infty }{\lim }\left(x-\sqrt{x^2+2x}\right)$

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

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

$=\underset{x\to +\infty }{\lim }\frac{-2}{1+\sqrt{1+\frac{2}{x}}}$

$=\frac{-2}{1+1}=-1.$

-- Mod Toán 11 HỌC247