Tính các giới hạn sau: a) lim (-5x + 2)/(3x + 1);  b) lim (-2x + 3)/(3x^2 + 2x + 5)

Giải SBT Toán 11 Bài 2: Giới hạn của hàm số

Bài 21 trang 76 SBT Toán 11 Tập 1: Tính các giới hạn sau:

a) $\underset{x\to -\infty }{\lim }\frac{-5x+2}{3x+1}$ ;                                 b) $\underset{x\to -\infty }{\lim }\frac{-2x+3}{3x^2+2x+5}$ ;

c) $\underset{x\to +\infty }{\lim }\frac{\sqrt{9x^2+3}}{x+1}$ ;                               d) $\underset{x\to -\infty }{\lim }\frac{\sqrt{9x^2+3}}{x+1}$ ;

e) $\underset{x\to 1}{\lim }\frac{2x^2-8x+6}{x^2-1}$ ;                               g) $\underset{x\to -3}{\lim }\frac{-x^2+2x+15}{x^2+4x+3}$ .

Lời giải:

a) $\underset{x\to -\infty }{\lim }\frac{-5x+2}{3x+1}=\underset{x\to -\infty }{\lim }\frac{-5+\frac{2}{x}}{3+\frac{1}{x}}=\frac{-5}{3}$ .                      

b) $\underset{x\to -\infty }{\lim }\frac{-2x+3}{3x^2+2x+5}$ $=\underset{x\to -\infty }{\lim }\frac{\frac{-2}{x}+\frac{3}{x^2}}{3+\frac{2}{x}+\frac{5}{x^2}}=\frac{0}{3}=0$ .

c)  

d)  

$=\underset{x\to -\infty }{\lim }\frac{\left(-x\right).\sqrt{9+\frac{3}{x^2}}}{x\left(1+\frac{1}{x}\right)}=\underset{x\to -\infty }{\lim }\frac{-\sqrt{9+\frac{3}{x^2}}}{1+\frac{1}{x}}=\frac{-\sqrt{9}}{1}=-3$.         

e) $\underset{x\to 1}{\lim }\frac{2x^2-8x+6}{x^2-1}$ $=\underset{x\to 1}{\lim }\frac{\left(2x-6\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=\underset{x\to 1}{\lim }\frac{2x-6}{x+1}=-2$.                          

g) $\underset{x\to -3}{\lim }\frac{-x^2+2x+15}{x^2+4x+3}$ $=\underset{x\to -3}{\lim }\frac{\left(5-x\right)\left(x+3\right)}{\left(x+1\right)\left(x+3\right)}=\underset{x\to -3}{\lim }\frac{5-x}{x+1}=-4$.