Bài 4.4 trang 156 SBT đại số và giải tích 11

Đề bài

Tính giới hạn của các dãy số có số hạng tổng quát sau đây, khi \(\displaystyle n \to  + \infty \)

a) \(\displaystyle {a_n} = {{2n - 3{n^3} + 1} \over {{n^3} + {n^2}}}\) ;

b) \(\displaystyle {b_n} = {{3{n^3} - 5n + 1} \over {{n^2} + 4}}\) ;

c) \(\displaystyle {c_n} = {{2n\sqrt n } \over {{n^2} + 2n - 1}}\) ;

d) \(\displaystyle {u_n} = {2^n} + {1 \over n}\) ;

e) \(\displaystyle {v_n} = {\left( { - {{\sqrt 2 } \over \pi }} \right)^n} + {{{3^n}} \over {{4^n}}}\) ;

f) \(\displaystyle {u_n} = {{{3^n} - {4^n} + 1} \over {{{2.4}^n} + {2^n}}}\) ;

g) \(\displaystyle {v_n} = {{\sqrt {{n^2} + n - 1}  - \sqrt {4{n^2} - 2} } \over {n + 3}}\) ;

a) \(\lim {a_n} = \lim \dfrac{{2n - 3{n^3} + 1}}{{{n^3} + {n^2}}}\) \( = \lim \dfrac{{{n^3}\left( {\dfrac{2}{{{n^2}}} - 3 + \dfrac{1}{{{n^3}}}} \right)}}{{{n^3}\left( {1 + \dfrac{1}{n}} \right)}}\) \( = \lim \dfrac{{\dfrac{2}{{{n^2}}} - 3 + \dfrac{1}{{{n^3}}}}}{{1 + \dfrac{1}{n}}}\) \( = \dfrac{{0 - 3 + 0}}{{1 + 0}} = \dfrac{{ - 3}}{1} =  - 3\)

b) \(\lim {b_n} = \lim \dfrac{{3{n^3} - 5n + 1}}{{{n^2} + 4}}\) \( = \lim \dfrac{{{n^3}\left( {3 - \dfrac{5}{{{n^2}}} + \dfrac{1}{{{n^3}}}} \right)}}{{{n^3}\left( {\dfrac{1}{n} + \dfrac{4}{{{n^3}}}} \right)}}\) \( = \lim \dfrac{{3 - \dfrac{5}{{{n^2}}} + \dfrac{1}{{{n^3}}}}}{{\dfrac{1}{n} + \dfrac{4}{{{n^3}}}}}\) \( =  + \infty \)

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

c) \(\lim {c_n} = \lim \dfrac{{2n\sqrt n }}{{{n^2} + 2n - 1}}\) \( = \lim \dfrac{{2{n^2}.\dfrac{1}{{\sqrt n }}}}{{{n^2}\left( {1 + \dfrac{2}{n} - \dfrac{1}{{{n^2}}}} \right)}}\) \( = \lim \dfrac{{\dfrac{2}{{\sqrt n }}}}{{1 + \dfrac{2}{n} - \dfrac{1}{{{n^2}}}}}\) \( = \dfrac{0}{{1 + 0 - 0}} = 0\)

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

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

(vì \(\left| { - \dfrac{{\sqrt 2 }}{\pi }} \right| < 1\) và \(\dfrac{3}{4} < 1\) nên \(\lim {\left( { - \dfrac{{\sqrt 2 }}{\pi }} \right)^n} = \lim {\left( {\dfrac{3}{4}} \right)^n} = 0\))

f) \(\lim {u_n} = \lim \dfrac{{{3^n} - {4^n} + 1}}{{{{2.4}^n} + {2^n}}}\) \( = \lim \dfrac{{{{\left( {\dfrac{3}{4}} \right)}^n} - 1 + \dfrac{1}{{{4^n}}}}}{{2 + {{\left( {\dfrac{2}{4}} \right)}^n}}}\) \( = \dfrac{{0 - 1 + 0}}{{2 + 0}} =  - \dfrac{1}{2}\)

g) \(\lim {v_n} = \lim \dfrac{{\sqrt {{n^2} + n - 1}  - \sqrt {4{n^2} - 2} }}{{n + 3}}\) \( = \lim \dfrac{{n\sqrt {1 + \dfrac{1}{n} - \dfrac{1}{{{n^2}}}}  - n\sqrt {4 - \dfrac{2}{{{n^2}}}} }}{{n\left( {1 + \dfrac{3}{n}} \right)}}\) \( = \lim \dfrac{{\sqrt {1 + \dfrac{1}{n} - \dfrac{1}{{{n^2}}}}  - \sqrt {4 - \dfrac{2}{{{n^2}}}} }}{{1 + \dfrac{3}{n}}}\) \( = \dfrac{{1 - 2}}{1} =  - 1\).