Bài tập 66 trang 180 SGK Toán 11 NC

Bài tập 66 trang 180 SGK Toán 11 NC

a.

\(\begin{array}{l}
\lim \frac{{2n + 3}}{{2 - 3n}} = \lim \frac{{2 + \frac{3}{n}}}{{\frac{2}{n} - 3}} =  - 23\\
\lim \frac{{{n^2} - {n^3}}}{{2{n^3} + 1}} = \lim \frac{{\frac{1}{n} - 1}}{{2 + \frac{1}{{{n^3}}}}} =  - \frac{1}{2}\\
\lim \frac{{{n^2} + n}}{{ - 2n - {n^2}}} = \lim \frac{{1 + \frac{1}{n}}}{{ - \frac{2}{n} - 1}} =  - 1\\
\lim \frac{{{n^3}}}{{{n^2} + 3}} =  + \infty 
\end{array}\)

Chọn C

b.

\(\begin{array}{l}
\lim \frac{{{n^2} - 3n + 2}}{{{n^2} + n}} = \lim \frac{{1 - \frac{3}{n} + \frac{2}{{{n^2}}}}}{{1 + \frac{1}{n}}} = 1\\
\lim \frac{{{n^3} + 2n - 1}}{{n - 2{n^3}}} = \lim \frac{{1 + \frac{2}{{{n^2}}} - \frac{1}{{{n^3}}}}}{{\frac{1}{{{n^3}}} - 2}} =  - \frac{1}{2}\\
\lim \frac{{2{n^2} - 3n}}{{{n^3} + 3n}} = \lim \frac{{\frac{2}{n} - \frac{3}{{{n^2}}}}}{{1 + \frac{3}{{{n^2}}}}} = 0\\
\lim \frac{{{n^2} - n + 1}}{{2n - 1}} = \lim \frac{{1 - \frac{1}{n} + \frac{1}{{{n^2}}}}}{{\frac{2}{n} - \frac{1}{{{n^2}}}}} =  + \infty 
\end{array}\)

Chọn D

c.

\(\begin{array}{l}
\lim \frac{{{2^n} + 1}}{{{{3.2}^n} - {3^n}}} = \lim \frac{{{{\left( {\frac{2}{3}} \right)}^n} + {{\left( {\frac{1}{3}} \right)}^n}}}{{3.{{\left( {\frac{2}{3}} \right)}^n} - 1}} = 0\\
\lim \frac{{{2^n} + 3}}{{1 - {2^n}}} = \lim \frac{{1 + \frac{3}{{{2^n}}}}}{{{{\left( {\frac{1}{2}} \right)}^n} - 1}} =  - 1\\
\lim \frac{{1 - {n^3}}}{{{n^2} + 2n}} =  - \infty \\
\lim \frac{{\left( {2n + 1} \right){{\left( {n - 3} \right)}^2}}}{{n - 2{n^3}}} =  - 1
\end{array}\)

Chọn A.

-- Mod Toán 11