Bài tập 4.47 trang 172 SBT Toán 11

a)

\(\lim \frac{{{{\left( { - 3} \right)}^n} + {{2.5}^n}}}{{1 - {5^n}}} = \lim \frac{{{{\left( { - \frac{3}{5}} \right)}^n} + 2}}{{{{\left( {\frac{1}{5}} \right)}^n} - 1}} =  - 2\)

b)

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

c) 

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

-- Mod Toán 11 HỌC247