Bài tập 16 trang 143 SGK Toán 11 NC

Bài tập 16 trang 143 SGK Toán 11 NC

a)

\(\begin{array}{l}
\lim \frac{{{n^2} + 4n - 5}}{{3{n^3} + {n^2} - 7}} = \lim \frac{{{n^3}\left( {\frac{1}{n} + \frac{4}{{{n^2}}} - \frac{5}{{{n^3}}}} \right)}}{{{n^3}\left( {3 + \frac{1}{n} + \frac{7}{{{n^3}}}} \right)}}\\
 = \lim \frac{{\frac{1}{n} + \frac{4}{{{n^2}}} - \frac{5}{{{n^3}}}}}{{3 + \frac{1}{n} + \frac{7}{{{n^3}}}}} = \frac{0}{3} = 0
\end{array}\)

b)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\lim \frac{{{n^5} + {n^4} - 3n - 2}}{{4{n^3} + 6{n^2} + 9}}\\
 = \lim {n^2}.\frac{{{n^3}\left( {1 + \frac{1}{n} - \frac{3}{{{n^4}}} - \frac{2}{{{n^5}}}} \right)}}{{{n^3}\left( {4 + \frac{6}{n} + \frac{9}{{{n^3}}}} \right)}}
\end{array}\\
{ = \lim {n^2}.\frac{{\left( {1 + \frac{1}{n} - \frac{3}{{{n^4}}} - \frac{2}{{{n^5}}}} \right)}}{{\left( {4 + \frac{6}{n} + \frac{9}{{{n^3}}}} \right)}} =  + \infty }
\end{array}\)

c)

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

d)

\(\lim \frac{{{3^n} - {{2.5}^n}}}{{7 + {{3.5}^n}}} = \lim \frac{{{{\left( {\frac{3}{5}} \right)}^n} - 2}}{{7.{{\left( {\frac{1}{5}} \right)}^n} + 3}} =  - \frac{2}{3}\)

(vì \(\lim {\left( {\frac{3}{5}} \right)^n} = \lim {\left( {\frac{1}{5}} \right)^n} = 0\))

-- Mod Toán 11