Bài 6 trang 86

Đề bài

Tìm các giới hạn sau:

a) $lim\frac{3n-1}{n}$                                                   

b) $lim\frac{\sqrt{n^2+2}}{n}$      

c) $lim\frac{2}{3n+1}$                                                     

d) $lim\frac{\left(n+1\right)\left(2n+2\right)}{n^2}$

Lời giải chi tiết

a) $lim\frac{3n-1}{n}=lim\frac{n\left(3-\frac{1}{n}\right)}{n}=lim\left(3-\frac{1}{n}\right)=3-0=3$

b) $lim\frac{\sqrt{n^2+2}}{n}=lim\frac{\sqrt{n^2\left(1+\frac{2}{n^2}\right)}}{n}=lim\frac{n\sqrt{1+\frac{2}{n^2}}}{n}=lim\sqrt{1+\frac{2}{n^2}}=1+0=1$

c) $lim\frac{2}{3n+1}=lim\frac{2}{n\left(3+\frac{1}{n}\right)}=lim\left(\frac{2}{n}.\frac{1}{3+\frac{1}{n}}\right)=lim\frac{2}{n}.lim\frac{1}{3+\frac{1}{n}}=0.\frac{1}{3+0}=0$

d) $lim\frac{\left(n+1\right)\left(2n+2\right)}{n^2}=lim\frac{n\left(1+\frac{1}{n}\right).2n\left(1+\frac{1}{n}\right)}{n^2}=lim\frac{2n^2{\left(1+\frac{1}{n}\right)}^2}{n^2}$

                                                      $=lim2{\left(1+\frac{1}{n}\right)}^2=2.{\left(1+0\right)}^2=2$