Bài 9 trang 86

Đề bài

Tìm các giới hạn sau:

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

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

Lời giải chi tiết

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

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

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