Tính các giới hạn sau: a) lim (2 + 4/3x)/(x^2 - 1); b) lim 1/(x - 2); c) lim (-5 + x)/(x + 3)

Giải SBT Toán 11 Bài tập cuối chương 3

Bài 44 trang 83 SBT Toán 11 Tập 1: Tính các giới hạn sau:

a) $\underset{x\to -\infty }{\lim }\frac{2+\frac{4}{3x}}{x^2-1}$ ;                         b) $\underset{x\to 2^{+}}{\lim }\frac{1}{x-2}$ ;                            c) $\underset{x\to -3^{+}}{\lim }\frac{-5+x}{x+3}$ ;

d) $\underset{x\to -\infty }{\lim }\frac{14x+2}{-7x+1}$ ;                        e) $\underset{x\to +\infty }{\lim }\frac{-2x^2}{3x+5}$ ;                          g) $\underset{x\to -\infty }{\lim }\frac{\sqrt{4x^2+1}}{x+2}$ ;

h) $\underset{x\to 1}{\lim }\frac{x-1}{x^2-1}$ ;                            i) $\underset{x\to 2}{\lim }\frac{x^2-5x+6}{x-2}$ ;                     k) $\underset{x\to 3}{\lim }\frac{-x^2+4x-3}{x^2+3x-18}$ .

Lời giải:

a) $\underset{x\to -\infty }{\lim }\frac{2+\frac{4}{3x}}{x^2-1}$ $=\underset{x\to -\infty }{\lim }\frac{\frac{2}{x^2}+\frac{4}{3x^3}}{1-\frac{1}{x^2}}$ $=\frac{\underset{x\to -\infty }{\lim }\left(\frac{2}{x^2}+\frac{4}{3x^3}\right)}{\underset{x\to -\infty }{\lim }\left(1-\frac{1}{x^2}\right)}=\frac{0}{1}=0$  .

b) $\underset{x\to 2^{+}}{\lim }\frac{1}{x-2}=+\infty$ .                            

c) Vì $\underset{x\to -3^{+}}{\lim }\left(-5+x\right)=-8<0$ ; $\underset{x\to -3^{+}}{\lim }\left(x+3\right)=0$  và x + 3 > 0 với mọi x > – 3.

Do đó, $\underset{x\to -3^{+}}{\lim }\frac{-5+x}{x+3}=-\infty$.

d) $\underset{x\to -\infty }{\lim }\frac{14x+2}{-7x+1}$ $=\underset{x\to -\infty }{\lim }\frac{14+\frac{2}{x}}{-7+\frac{1}{x}}=\frac{14}{-7}=-2$ .                         

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

g) 

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

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

i) $\underset{x\to 2}{\lim }\frac{x^2-5x+6}{x-2}$ $=\underset{x\to 2}{\lim }\frac{\left(x-2\right)\left(x-3\right)}{x-2}=\underset{x\to 2}{\lim }\left(x-3\right)=-1$ .           

k) $\underset{x\to 3}{\lim }\frac{-x^2+4x-3}{x^2+3x-18}$$=\underset{x\to 3}{\lim }\frac{\left(x-1\right)\left(3-x\right)}{\left(x+6\right)\left(x-3\right)}$ $=\underset{x\to 3}{\lim }\frac{-\left(x-1\right)}{x+6}=\frac{-2}{9}$ .