Bài 8 trang 86

Đề bài

Tìm các giới hạn sau:

a) $\underset{x\to -1}{lim}\left(3x^2-x+2\right)$  

b) $\underset{x\to 4}{lim}\frac{x^2-16}{x-4}$                                          

c) $\underset{x\to 2}{lim}\frac{3-\sqrt{x+7}}{x-2}$

Lời giải chi tiết

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

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

b) $\underset{x\to 4}{lim}\frac{x^2-16}{x-4}=\underset{x\to 4}{lim}\frac{\left(x-4\right)\left(x+4\right)}{x-4}=\underset{x\to 4}{lim}\left(x+4\right)=\underset{x\to 4}{lim}x+\underset{x\to 4}{lim}4=4+4=8$

c) $\underset{x\to 2}{lim}\frac{3-\sqrt{x+7}}{x-2}=\underset{x\to 2}{lim}\frac{\left(3-\sqrt{x+7}\right)\left(3+\sqrt{x+7}\right)}{\left(x-2\right)\left(3+\sqrt{x+7}\right)}=\underset{x\to 2}{lim}\frac{3^2-\left(x+7\right)}{\left(x-2\right)\left(3+\sqrt{x+7}\right)}$

                                         $=\underset{x\to 2}{lim}\frac{2-x}{\left(x-2\right)\left(3+\sqrt{x+7}\right)}=\underset{x\to 2}{lim}\frac{-\left(x-2\right)}{\left(x-2\right)\left(3+\sqrt{x+7}\right)}=\underset{x\to 2}{lim}\frac{-1}{3+\sqrt{x+7}}$

                                         $=\frac{\underset{x\to 2}{lim}\left(-1\right)}{\underset{x\to 2}{lim}3+\sqrt{\underset{x\to 2}{lim}x+\underset{x\to 2}{lim}7}}=\frac{-1}{3+\sqrt{2+7}}=-\frac{1}{6}$