Bài tập 63 trang 179 SGK Toán 11 NC

a. \({\lim \frac{{n - 2\sqrt n \sin 2n}}{{2n}} = \lim \left( {\frac{1}{2} - \frac{{\sin 2n}}{{\sqrt n }}} \right) = \frac{1}{2}}\)

vì \({\left| {\frac{{\sin 2n}}{{\sqrt n }}} \right| \le \frac{1}{{\sqrt n }},\lim \frac{1}{{\sqrt n }} = 0}\)

Chọn B

b. \(\lim \frac{{{n^2} - 3{n^3}}}{{2{n^3} + 5n - 2}} = \lim \frac{{\frac{1}{n} - 3}}{{2 + \frac{5}{{{n^2}}} - \frac{2}{{{n^3}}}}} =  - \frac{3}{2}.\)

Chọn C

c. \(\lim \frac{{{3^n} - 1}}{{{2^n} - {{2.3}^n} + 1}} = \lim \frac{{1 - {{\left( {\frac{1}{3}} \right)}^n}}}{{{{\left( {\frac{2}{3}} \right)}^n} - 2 + {{\left( {\frac{1}{3}} \right)}^n}}} =  - \frac{1}{2}\)

Chọn A

d. \(\lim \left( {2n - 3{n^3}} \right) = \lim {n^3}\left( {\frac{2}{{{n^2}}} - 3} \right) =  - \infty \)

Chọn B.

-- Mod Toán 11 HỌC247