Bài tập 6 trang 134 SGK Toán 11 NC

a) Ta có:

\(\begin{array}{l}
\lim {u_n} = \lim \frac{{{n^2}\left( {1 - \frac{3}{n} + \frac{5}{{{n^2}}}} \right)}}{{{n^2}\left( {2 - \frac{1}{{{n^2}}}} \right)}} = \lim \frac{{1 - \frac{3}{n} + \frac{5}{{{n^2}}}}}{{2 - \frac{1}{{{n^2}}}}}\\
 = \frac{{\lim 1 - \lim \frac{3}{n} + \lim \frac{5}{{{n^2}}}}}{{\lim 2 - \lim \frac{1}{{{n^2}}}}} = \frac{{1 - 0 + 0}}{{2 - 0}} = \frac{1}{2}
\end{array}\)

b) Ta có

\(\begin{array}{l}
\lim {u_n} = \lim \frac{{{n^4}\left( {\frac{{ - 2}}{{{n^2}}} + \frac{1}{{{n^3}}} + \frac{2}{{{n^4}}}} \right)}}{{{n^4}\left( {3 + \frac{5}{{{n^4}}}} \right)}}\\
 = \lim \frac{{\frac{{ - 2}}{{{n^2}}} + \frac{1}{{{n^3}}} + \frac{2}{{{n^4}}}}}{{3 + \frac{5}{{{n^4}}}}} = \frac{0}{3} = 0
\end{array}\)

c) Ta có

\(\begin{array}{l}
\lim {u_n} = \lim \frac{{{n^2}\sqrt {\frac{2}{{{n^2}}} - \frac{1}{{{n^3}}}} }}{{{n^2}\left( {\frac{1}{{{n^2}}} - 3} \right)}}\\
 = \lim \frac{{\sqrt {\frac{2}{{{n^2}}} - \frac{1}{{{n^3}}}} }}{{\left( {\frac{1}{{{n^2}}} - 3} \right)}} = \frac{0}{{ - 3}} = 0
\end{array}\)

d) Chia cả tử và mẫu u_n cho 4ⁿ ta được:

\(\lim {u_n} = \lim \frac{1}{{2.{{\left( {\frac{3}{4}} \right)}^n} + 1}} = 1\)

(vì \(\lim {\left( {\frac{3}{4}} \right)^n} = 0\))

-- Mod Toán 11 HỌC247