Bài tập 56 trang 177 SGK Toán 11 NC

a)

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

(vì \(\lim \sqrt n  =  + \infty ,lim\left( {\sqrt {3 - \frac{1}{n}}  + \sqrt {2 - \frac{1}{n}} } \right) = \sqrt 3  + \sqrt 2  > 0\))

b)

\(\begin{array}{l}
\lim {u_n} = \lim \frac{{{4^n} - {5^n}}}{{{2^n} + {{3.5}^n}}}\\
 = \lim \frac{{{{\left( {\frac{4}{5}} \right)}^n} - 1}}{{{{\left( {\frac{2}{5}} \right)}^n} + 3}} =  - \frac{1}{3}
\end{array}\)

(vì \(\lim {\left( {\frac{4}{5}} \right)^n} = 0,\lim {\left( {\frac{2}{5}} \right)^n} = 0\))

-- Mod Toán 11 HỌC247