ĐGNL ĐHQG Hà Nội - Tư duy định lượng - Giới hạn của dãy số

Giả sử\[\lim {u_n} = L,\lim {v_n} = M\]và c là một hằng số. Khi đó:

\[\lim \left( {{u_n} + {v_n}} \right) = L + M,\lim \left( {{u_n} - {v_n}} \right) = L - M,\lim \left( {{u_n}.{v_n}} \right) = L.M,\lim \left( {c.{u_n}} \right) = c.L\]

Công thức tính tổng cấp số nhân lùi vô hạn: \[S = {u_1} + {u_2} + ... + {u_n} + ... = \frac{{{u_1}}}{{1 - q}}\]

Ta có:\[\lim n = + \infty ,\lim \sqrt n = + \infty ,\lim \sqrt[3]{n} = + \infty ,\lim \frac{1}{n} = 0\]

Vậy chỉ có đáp án D là sai.

Ta có:\[\lim \left( {{u_n}.{v_n}} \right) = \lim \left( {\frac{1}{n}.{n^2}} \right) = \lim n = + \infty \]

\[\sqrt 2 >1 = >\lim {(\sqrt 2 )^n} = + \infty \]

\[\begin{array}{*{20}{l}}{S = {u_1} + {u_2} + ...}\\{ = {u_1}\left( {1 + q + {q^2} + ...} \right)}\\{ = \frac{{{u_1}}}{{1 - q}}}\end{array}\]

\[\lim {u_n} = \lim \frac{{1 - 4n}}{{5n}} = \lim \frac{{\frac{1}{n} - 4}}{5} = \frac{{ - 4}}{5} = - \frac{4}{5}.\]

\[\lim {u_n} = \lim \frac{{{n^2} - 3n}}{{1 - 4{n^3}}} = \lim \frac{{\frac{1}{n} - \frac{3}{{{n^2}}}}}{{\frac{1}{{{n^3}}} - 4}} = \frac{0}{{ - 4}} = 0.\]

\[\lim {u_n} = \lim \frac{{{n^2} - 3n}}{{1 - 4{n^3}}} = \lim \frac{{\frac{1}{n} - \frac{3}{{{n^2}}}}}{{\frac{1}{{{n^3}}} - 4}} = \frac{0}{{ - 4}} = 0.\]

\[\lim {u_n} = \lim \frac{{{3^n} + {5^n}}}{{{5^n}}} = \lim \frac{{{{\left( {\frac{3}{5}} \right)}^n} + 1}}{1} = \frac{1}{1} = 1.\]

\[\begin{array}{*{20}{l}}{\lim \frac{{2{n^2} - 3}}{{ - 2{n^3} - 4}} = \lim \frac{{\frac{2}{n} - \frac{3}{{{n^3}}}}}{{ - 2 - \frac{4}{{{n^3}}}}} = \frac{0}{{ - 2}} = 0.}\\{\lim \frac{{2{n^2} - 3}}{{ - 2{n^2} - 1}} = \lim \frac{{2 - \frac{3}{{{n^2}}}}}{{ - 2 - \frac{1}{{{n^2}}}}} = \frac{2}{{ - 2}} = - 1.}\\{\lim \frac{{2{n^2} - 3}}{{2{n^2} + 1}} = \lim \frac{{2 - \frac{3}{{{n^2}}}}}{{2 + \frac{1}{{{n^2}}}}} = \frac{2}{2} = 1.}\\{\lim \frac{{2{n^3} - 3}}{{2{n^2} - 1}} = \lim \frac{{2 - \frac{3}{{{n^3}}}}}{{\frac{2}{n} - \frac{1}{{{n^3}}}}} = + \infty .}\end{array}\]

Ta có:\[{n^3} - 2n + 1 = {n^3}\left( {1 - \frac{2}{{{n^2}}} + \frac{1}{{{n^3}}}} \right)\]

Vì\[\lim {n^3} = + \infty \] và\[\lim \left( {1 - \frac{2}{{{n^2}}} + \frac{1}{{{n^3}}}} \right) = 1 >0\] nên\[\lim \left( {{n^3} - 2n + 1} \right) = + \infty \]

Bước 1:

\[\lim \frac{{{2^{n + 1}} - {{3.5}^n} + 5}}{{{{3.2}^n} + {{9.5}^n}}} = \lim \frac{{{{2.2}^n} - {{3.5}^n} + 5}}{{{{3.2}^n} + {{9.5}^n}}}\]

\[ = \lim \frac{{2.{{\left( {\frac{2}{5}} \right)}^n} - 3 + 5.{{\left( {\frac{1}{5}} \right)}^n}}}{{3.{{\left( {\frac{2}{5}} \right)}^n} + 9}}\]

Bước 2:

\[ = \frac{{2.0 - 3 + 5.0}}{{3.0 + 9}} = \frac{{ - 3}}{9} = - \frac{1}{3}.\]

\[\lim \frac{{{{(2 - 5n)}^3}{{(n + 1)}^2}}}{{2 - 25{n^5}}} = \lim \frac{{\frac{{{{(2 - 5n)}^3}}}{{{n^3}}}.\frac{{{{(n + 1)}^2}}}{{{n^2}}}}}{{\frac{{2 - 25{n^5}}}{{{n^5}}}}} = \frac{{{{\left( {\frac{{2 - 5n}}{n}} \right)}^3}.{{\left( {\frac{{n + 1}}{n}} \right)}^2}}}{{\frac{2}{{{n^5}}} - 25}}\]

\[ = \lim \frac{{{{\left( {\frac{2}{n} - 5} \right)}^3}.{{\left( {1 + \frac{1}{n}} \right)}^2}}}{{\frac{2}{{{n^5}}} - 25}} = \frac{{{{\left( {0 - 5} \right)}^3}{{\left( {1 + 0} \right)}^2}}}{{0 - 25}} = \frac{{{{( - 5)}^3}{{.1}^2}}}{{ - 25}} = 5\]

Cách 1:

\[\begin{array}{l}lim\frac{{\sqrt {{n^2} - 3n - 5} - \sqrt {9{n^2} + 3} }}{{2n - 1}}\\ = lim\frac{{\left( {\sqrt {{n^2} - 3n - 5} - \sqrt {9{n^2} + 3} } \right).\left( {\sqrt {{n^2} - 3n - 5} + \sqrt {9{n^2} + 3} } \right)}}{{\left( {\sqrt {{n^2} - 3n - 5} + \sqrt {9{n^2} + 3} } \right).\left( {2n - 1} \right)}}\\ = lim\frac{{({n^2} - 3n - 5) - (9{n^2} + 3)}}{{\left( {\sqrt {{n^2} - 3n - 5} + \sqrt {9{n^2} + 3} } \right).\left( {2n - 1} \right)}}\\ = lim\frac{{ - 8{n^2} - 3n - 8}}{{\left( {\sqrt {{n^2} - 3n - 5} + \sqrt {9{n^2} + 3} } \right).\left( {2n - 1} \right)}}\\ = lim\frac{{ - 8 - \frac{3}{n} - \frac{8}{{{n^2}}}}}{{\left( {\sqrt {1 - \frac{3}{n} - \frac{5}{{{n^2}}}} + \sqrt {9 + \frac{3}{{{n^2}}}} } \right)\left( {2 - \frac{1}{n}} \right)}} = \frac{{ - 8}}{{4.2}} = - 1\end{array}\]

Cách 2: Chia cả tử và mẫu cho n.

\[\lim \frac{{\sqrt {{n^2} - 3n - 5} - \sqrt {9{n^2} + 3} }}{{2n - 1}} = \lim \frac{{\sqrt {1 - \frac{3}{n} - \frac{5}{{{n^2}}}} - \sqrt {9 + \frac{3}{{{n^2}}}} }}{{2 - \frac{1}{n}}} = \lim \frac{{1 - 3}}{2} = - 1\]

\[\lim \frac{{2{n^2} - n + 4}}{{\sqrt {2{n^4} - {n^2} + 1} }} = \lim \frac{{2 - \frac{1}{n} + \frac{4}{{{n^2}}}}}{{\sqrt {2 - \frac{1}{{{n^2}}} + \frac{1}{{{n^4}}}} }} = \frac{2}{{\sqrt 2 }} = \sqrt 2 .\]

\[\lim \left( {\sqrt {{n^2} - n} - n} \right) = \lim \frac{{\left( {\sqrt {{n^2} - n} - n} \right).\left( {\sqrt {{n^2} - n} + n} \right)}}{{\sqrt {{n^2} - n} + n}}\]

\[ = \lim \frac{{{n^2} - n - {n^2}}}{{\sqrt {{n^2} - n} + n}} = \lim \frac{{ - n}}{{\sqrt {{n^2} - n} + n}} = \lim \frac{{ - 1}}{{\sqrt {1 - \frac{1}{n}} + 1}} = \frac{{ - 1}}{2} = - \frac{1}{2}.\]

\[\begin{array}{*{20}{l}}{\lim (\sqrt {{n^2} - n + 1} - \sqrt {{n^2} + 1} )}\\{ = \lim \frac{{(\sqrt {{n^2} - n + 1} - \sqrt {{n^2} + 1} )(\sqrt {{n^2} - n + 1} + \sqrt {{n^2} + 1} )}}{{\sqrt {{n^2} - n + 1} + \sqrt {{n^2} + 1} }}}\\{ = \lim \frac{{{n^2} - n + 1 - {n^2} - 1}}{{\sqrt {{n^2} - n + 1} + \sqrt {{n^2} + 1} }}}\\{ = \lim \frac{{ - n}}{{\sqrt {{n^2} - n + 1} + \sqrt {{n^2} + 1} }}}\\{ = \lim \frac{{ - 1}}{{\sqrt {1 - \frac{1}{n} + \frac{1}{{{n^2}}}} + \sqrt {1 + \frac{1}{{{n^2}}}} }}}\\{ = - \frac{1}{2}}\end{array}\]

\[{u_n} = \frac{1}{{1.2}} + \frac{1}{{2.3}} + \frac{1}{{3.4}} + ... + \frac{1}{{n.\left( {n + 1} \right)}}\]

\[ = \frac{{2 - 1}}{{1.2}} + \frac{{3 - 2}}{{2.3}} + ... + \frac{{n + 1 - n}}{{n.\left( {n + 1} \right)}}\]

\[ = 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + .... + \frac{1}{n} - \frac{1}{{n + 1}}\]

\[ = 1 - \frac{1}{{n + 1}}\]

\[ \Rightarrow \lim {u_n} = \lim \left( {1 - \frac{1}{{n + 1}}} \right) = 1.\]

\[\begin{array}{*{20}{l}}{{u_n} = \frac{1}{{1.3}} + \frac{1}{{3.5}} + ... + \frac{1}{{\left( {2n - 1} \right).\left( {2n + 1} \right)}}}\\{ = \frac{1}{2}.\left( {1 - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + ... + \frac{1}{{2n - 1}} - \frac{1}{{2n + 1}}} \right)}\\{ = \frac{1}{2}.\left( {1 - \frac{1}{{2n + 1}}} \right)}\\{ \Rightarrow \lim {u_n} = \lim \frac{1}{2}\left( {1 - \frac{1}{{2n + 1}}} \right) = \frac{1}{2}.}\end{array}\]

Ta có\[\left| {\frac{{\sin \left( {n!} \right)}}{{{n^2} + 1}}} \right| \le \frac{1}{{{n^2} + 1}}\]  mà\[\lim \frac{1}{{{n^2} + 1}} = 0\] nên chọn đáp án A.

\[\lim {u_n} = \lim \frac{{\left( {2n + 1} \right)\left( {1 - 3n} \right)}}{{\sqrt[3]{{{n^3} + 5n - 1}}}} = \lim \frac{{ - 6{n^2} - n + 1}}{{\sqrt[3]{{{n^3} + 5n - 1}}}}\]

\[ = \lim \frac{{\frac{{ - 6{n^2} - n + 1}}{{{n^2}}}}}{{\sqrt[3]{{\frac{{{n^3} + 5n - 1}}{{{n^6}}}}}}} = \lim \frac{{ - 6 - \frac{1}{n} + \frac{1}{{{n^2}}}}}{{\sqrt[3]{{\frac{1}{{{n^3}}} + \frac{5}{{{n^5}}} - \frac{1}{{{n^6}}}}}}} = - \infty .\]

Ta có\[1,\;a,\;{a^2},\;...,\;{a^n}\] là một cấp số nhân có công bội a

\[ \Rightarrow 1 + a + {a^2} + ... + {a^n} = \frac{{1 - {a^{n + 1}}}}{{1 - a}}.\]

 Tương tự:  \[1 + b + {b^2} + ... + {b^n} = \frac{{1 - {b^{n + 1}}}}{{1 - b}}\]

\[ \Rightarrow \lim I = \lim \frac{{\frac{{1 - {a^{n + 1}}}}{{1 - a}}}}{{\frac{{1 - {b^{n + 1}}}}{{1 - b}}}} = \lim \left( {\frac{{1 - {a^{n + 1}}}}{{1 - a}}.\frac{{1 - b}}{{1 - {b^{n + 1}}}}} \right) = \lim \left( {\frac{{1 - {a^{n + 1}}}}{{1 - {b^{n + 1}}}}.\frac{{1 - b}}{{1 - a}}} \right) = \frac{{1 - b}}{{1 - a}}.\]

(Vì\[\left| a \right| < 1,\;\;\left| b \right| < 1 \Rightarrow \lim {a^{n + 1}} = \lim {b^{n + 1}} = 0\])

\[{u_2} = \sqrt {1.2.3.4 + 1} = 5,{u_n} >0,\forall n = 1;2;...\]

Ta có:

\[{u_{n + 1}} = \sqrt {{u_n}({u_n} + 1)({u_n} + 2)({u_n} + 3) + 1} \]

\[ = \sqrt {(u_n^2 + 3{u_n})(u_n^2 + 3{u_n} + 2) + 1} \]

\[ = \sqrt {{{(u_n^2 + 3{u_n})}^2} + 2(u_n^2 + 3{u_n}) + 1} \]

\[ = \sqrt {{{(u_n^2 + 3{u_n} + 1)}^2}} = u_n^2 + 3{u_n} + 1\]

\(\begin{array}{l} \Rightarrow {u_{n + 1}} + 1 = u_n^2 + 3{u_n} + 2 = ({u_n} + 1)({u_n} + 2)\\ \Rightarrow \frac{1}{{{u_{n + 1}} + 1}} = \frac{1}{{({u_n} + 1)({u_n} + 2)}} = \frac{1}{{{u_n} + 1}} - \frac{1}{{{u_n} + 2}}\\ \Rightarrow \frac{1}{{{u_n} + 2}} = \frac{1}{{{u_n} + 1}} - \frac{1}{{{u_{n + 1}} + 1}}\end{array}\)

Do đó: 

\(\begin{array}{l}\\{v_n} = \mathop \sum \limits_{i = 1}^n \frac{1}{{{u_i} + 2}} = \mathop \sum \limits_{i = 1}^n \left( {\frac{1}{{{u_i} + 1}} - \frac{1}{{{u_{i + 1}} + 1}}} \right)\end{array}\)

\[ = \frac{1}{{{u_1} + 1}} - \frac{1}{{{u_{n + 1}} + 1}} = \frac{1}{2} - \frac{1}{{{u_{n + 1}} + 1}}\]

Xét hiệu\[{u_{n + 1}} - {u_n} = u_n^2 + 3{u_n} + 1 - {u_n} = {\left( {{u_n} + 1} \right)^2} >0\]

\[ \Rightarrow \left( {{u_n}} \right)\]là dãy tăng.

Giả sử

\[\lim {u_{n + 1}} = \lim {u_n} = a >0 \Rightarrow a = {a^2} + 3a + 1 \Rightarrow {a^2} + 2a + 1 = 0 \Leftrightarrow a = - 1\,\,\left( {ktm} \right) \Rightarrow \lim {u_n} = + \infty \]\[ \Rightarrow \lim {v_n} = \frac{1}{2} - \frac{1}{{{u_{n + 1}} + 1}} = \frac{1}{2} - 0 = \frac{1}{2}.\]

Bước 1:

Vì \[\lim \frac{1}{n} = 0;\lim \frac{1}{{{n^2}}} = 0\]

Bước 2:

Nên\[\lim \left( {\frac{2}{n} + \frac{3}{{{n^2}}}} \right) = 2.0 + 3.0 = 0\]

Bước 1:

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

Bước 2:

\[ = \lim \frac{{ - 3 + \frac{1}{n}}}{{2 + \frac{5}{n} - \frac{2}{{{n^3}}}}} = \frac{{ - 3 + 0}}{{2 + 0 - 0}} = \frac{{ - 3}}{2}\]

Bước 1:

\[\lim \frac{{n + 1}}{{2n - 3}} = \lim \frac{{n\left( {1 + \frac{1}{n}} \right)}}{{n\left( {2 - \frac{3}{n}} \right)}} = \lim \frac{{1 + \frac{1}{n}}}{{2 - \frac{3}{n}}}\]

Bước 2:

\[ = \frac{{1 + 0}}{{2 - 0}} = \frac{1}{2}\]

Dãy số \[\left( {{u_n}} \right)\]có giới hạn \[ - \infty \] nếu mọi số hạng của dãy số đều nhỏ hơn một số âm tùy ý cho trước kể từ một số hạng nào đó trở đi.

Khi đó, ta viết\[\mathop {\lim }\limits_{n \to + \infty } \left( {{u_n}} \right) = - \infty \]viết tắt là\[\lim \left( {{u_n}} \right) = - \infty \] hoặc\[\lim {u_n} = - \infty \]

Bước 1: Tìm cấp số nhân

Ta có:

\[\begin{array}{l}{{\rm{S}}_1} = {a^2}\\{{\rm{S}}_2} = {\left( {\frac{{a\sqrt 2 }}{2}} \right)^2} = {a^2} \cdot \frac{1}{2}\\{{\rm{S}}_3} = {\left( {\frac{{a\sqrt 2 }}{2} \cdot \frac{{\sqrt 2 }}{2}} \right)^2}\\ \cdots \\{{\rm{S}}_{\rm{n}}} = {a^2} \cdot {\left( {\frac{1}{2}} \right)^{n - 1}}\end{array}\]

Có\[{S_1};{S_2};{S_3}; \ldots \] là một cấp số nhân lùi vô hạn với:

- Số hạng đầu:\[{S_1} = {a^2}\]

- Công bội:\[q = \frac{1}{2}\]

Bước 2: Sử dụng công thức tổng cấp số nhân lùi vô hạn

Do đó:\[S = {S_1} + {S_2} + {S_3} + \ldots = \frac{{{S_1}}}{{1 - q}} = \frac{{{a^2}}}{{1 - \frac{1}{2}}} = 2{a^2}\]