Bài 1 trang 69 SGK Toán 11 tập 1 - Chân trời sáng tạo

Đề bài

Tìm các giới hạn sau:

a) $\lim \frac{{ - 2n + 1}}{n}$                                        

b) $\lim \frac{{\sqrt {16{n^2} - 2} }}{n}$          

c) $\lim \frac{4}{{2n + 1}}$          

d) $\lim \frac{{{n^2} - 2n + 3}}{{2{n^2}}}$

Lời giải chi tiết

a) $\lim \frac{{ - 2n + 1}}{n} = \lim \frac{{n\left( { - 2 + \frac{1}{n}} \right)}}{n} = \lim \left( { - 2 + \frac{1}{n}} \right) =  - 2$

b) $\lim \frac{{\sqrt {16{n^2} - 2} }}{n} = \lim \frac{{\sqrt {{n^2}\left( {16 - \frac{2}{{{n^2}}}} \right)} }}{n} = \lim \frac{{n\sqrt {16 - \frac{2}{{{n^2}}}} }}{n} = \lim \sqrt {16 - \frac{2}{{{n^2}}}}  = 4$

c) $\lim \frac{4}{{2n + 1}} = \lim \frac{4}{{n\left( {2 + \frac{1}{n}} \right)}} = \lim \left( {\frac{4}{n}.\frac{1}{{2 + \frac{1}{n}}}} \right) = \lim \frac{4}{n}.\lim \frac{1}{{2 + \frac{1}{n}}} = 0$

d) $\lim \frac{{{n^2} - 2n + 3}}{{2{n^2}}} = \lim \frac{{{n^2}\left( {1 - \frac{2}{n} + \frac{3}{{{n^2}}}} \right)}}{{2{n^2}}} = \lim \frac{{1 - \frac{2}{n} + \frac{3}{{{n^2}}}}}{2} = \frac{1}{2}$