Bài tập 55 trang 177 SGK Toán 11 NC

a)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\lim {u_n} = \lim \frac{{2{n^3} - n - 3}}{{5n - 1}}\\
 = \lim \frac{{{n^3}\left( {2 - \frac{1}{{{n^2}}} - \frac{3}{{{n^3}}}} \right)}}{{{n^3}\left( {\frac{5}{{{n^2}}} - \frac{1}{{{n^3}}}} \right)}}
\end{array}\\
{ = \lim \frac{{2 - \frac{1}{{{n^2}}} - \frac{3}{{{n^3}}}}}{{\frac{5}{{{n^2}}} - \frac{1}{{{n^3}}}}} =  + \infty }
\end{array}\)

vì \(\lim \left( {2 - \frac{1}{{{n^2}}} - \frac{3}{{{n^3}}}} \right) = 2,\)

\(\lim \left( {\frac{5}{{{n^2}}} - \frac{1}{{{n^3}}}} \right) = 0;5n - 1 > 0\)

b)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\lim {u_n} = \lim \frac{{\sqrt {{n^4} - 2n + 3} }}{{ - 2{n^2} + 3}}\\
 = \lim \frac{{{n^2}\sqrt {1 - \frac{2}{{{n^3}}} + \frac{3}{{{n^4}}}} }}{{{n^2}\left( { - 2 + \frac{3}{{{n^2}}}} \right)}}
\end{array}\\
{ = \lim \frac{{\sqrt {1 - \frac{2}{{{n^3}}} + \frac{3}{{{n^4}}}} }}{{ - 2 + \frac{3}{{{n^2}}}}} =  - \frac{1}{2}}
\end{array}\)

c)

\(\begin{array}{l}
\lim {u_n} = \lim \left( { - 2{n^2} + 3n - 7} \right)\\
 = \lim {n^2}\left( { - 2 + \frac{3}{n} - \frac{7}{{{n^2}}}} \right) =  - \infty 
\end{array}\)

(vì \(\lim {n^2} =  + \infty ,\lim \left( { - 2 + \frac{3}{n} - \frac{7}{{{n^2}}}} \right) =  - 2 < 0\))

d)

\(\begin{array}{l}
\lim {u_n} = \lim \sqrt[3]{{{n^9} + 8{n^2} - 7}}\\
 = \lim {n^3}.\sqrt[3]{{1 + \frac{8}{{{n^7}}} - \frac{7}{{{n^9}}}}} =  + \infty 
\end{array}\)

(vì \(\lim {n^3} =  + \infty ,\lim \sqrt[3]{{1 + \frac{8}{{{n^7}}} - \frac{7}{{{n^9}}}}} = 1 > 0\))

-- Mod Toán 11 HỌC247