Bài tập 4.4 trang 156 SBT Toán 11

a) \(\lim \frac{{2n - 3{n^3} + 1}}{{{n^3} + {n^2}}} = \lim \frac{{ - 3 + \frac{2}{{{n^2}}} + \frac{1}{{{n^3}}}}}{{1 + \frac{1}{n}}} =  - 3\)

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

b) \(\lim \frac{{3{n^3} - 5n + 1}}{{{n^2} + 4}} = \lim \frac{{3 - \frac{5}{{{n^2}}} + \frac{1}{{{n^3}}}}}{{\frac{1}{n} + \frac{4}{{{n^3}}}}} =  + \infty \)

Vì \(\lim \left( {3 - \frac{5}{{{n^2}}} + \frac{1}{{{n^3}}}} \right) = 3 > 0\) và \(\lim \left( {\frac{1}{n} + \frac{4}{{{n^3}}}} \right) = 0\)

c) \(\lim \frac{{2n\sqrt n }}{{{n^2} + 2n - 1}} = \lim \frac{{\frac{{\sqrt 2 }}{{\sqrt n }}}}{{1 + \frac{2}{n} - \frac{1}{{{n^2}}}}} = 0\)

Vì \(\lim \frac{2}{{\sqrt n }} = 0,\lim \left( {1 + \frac{2}{n} - \frac{1}{{{n^2}}}} \right) = 1 > 0\)

d) \(\lim \left( {{2^n} + \frac{1}{n}} \right) = \lim {2^n} + \lim \frac{1}{n} =  + \infty \)

e) \(\lim \left[ {{{\left( { - \frac{{\sqrt 2 }}{\pi }} \right)}^n} + \frac{{{3^n}}}{{{4^n}}}} \right] = \lim {\left( { - \frac{{\sqrt 2 }}{\pi }} \right)^n} + \lim {\left( {\frac{3}{4}} \right)^n} = 0\)

Vì \(\left| { - \frac{{\sqrt 2 }}{\pi }} \right| < 1 \Rightarrow \lim {\left( { - \frac{{\sqrt 2 }}{\pi }} \right)^n} = 0\) và \(\frac{3}{4} < 1 \Rightarrow \lim {\left( {\frac{3}{4}} \right)^n} = 0\)

f) \(\lim \frac{{{3^n} - {4^n} + 1}}{{{{2.4}^n} + {2^n}}} = \lim \frac{{{{\left( {\frac{3}{4}} \right)}^n} + {{\left( {\frac{1}{4}} \right)}^n} - 1}}{{2 + {{\left( {\frac{2}{4}} \right)}^n}}} =  - \frac{1}{2}\)

g)

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

-- Mod Toán 11 HỌC247