Giới hạn của dãy số

$\lim {u_n} = \lim \dfrac{{1 - 4n}}{{5n}} = \lim \dfrac{{\dfrac{1}{n} - 4}}{5} = \dfrac{{ - 4}}{5} =  - \dfrac{4}{5}.$

Ta tính giới hạn của các dãy số trong từng đáp án:

+) Đáp án A:           

\(\mathop {\lim }\limits_{} {u_n} = \mathop {\lim }\limits_{} \dfrac{{{{\left( {2017 - n} \right)}^{2018}}}}{{n{{\left( {2018 - n} \right)}^{2017}}}} = \mathop {\lim }\limits_{} \left[ {\dfrac{{2017 - n}}{n}.{{\left( {\dfrac{{2017 - n}}{{2018 - n}}} \right)}^{2017}}} \right]\)

\( = \mathop {\lim }\limits_{} \left[ {\left( {\dfrac{{2017}}{n} - 1} \right){{\left( {\dfrac{{\dfrac{{2017}}{n} - 1}}{{\dfrac{{2018}}{n} - 1}}} \right)}^{2017}}} \right] =  - 1\).

+) Đáp án B:

\(\mathop {\lim }\limits_{} {u_n} = \mathop {\lim }\limits_{} n\left( {\sqrt {{n^2} + 2018}  - \sqrt {{n^2} + 2016} } \right) = \mathop {\lim }\limits_{} \dfrac{{n\left( {{n^2} + 2018 - {n^2} - 2016} \right)}}{{\sqrt {{n^2} + 2018}  + \sqrt {{n^2} + 2016} }}\)

\( = \mathop {\lim }\limits_{} \dfrac{{2n}}{{\sqrt {{n^2} + 2018}  + \sqrt {{n^2} + 2016} }} = \mathop {\lim }\limits_{} \dfrac{2}{{\sqrt {1 + \dfrac{{2018}}{{{n^2}}}}  + \sqrt {1 + \dfrac{{2016}}{{{n^2}}}} }} = 1\).

+) Đáp án C:           

Ta có \({u_{n + 1}} - 1 = \dfrac{1}{2}\left( {{u_n} - 1} \right)\)

\(\begin{array}{l}{u_n} - 1 = \dfrac{1}{2}\left( {{u_{n - 1}} - 1} \right)\\{u_{n - 1}} - 1 = \dfrac{1}{2}\left( {{u_{n - 2}} - 1} \right)\\{u_{n - 2}} - 1 = \dfrac{1}{2}\left( {{u_{n - 3}} - 1} \right)\\...\\{u_2} - 1 = \dfrac{1}{2}\left( {{u_1} - 1} \right)\end{array}\)

\( \Rightarrow \left( {{u_n} - 1} \right)\left( {{u_{n - 1}} - 1} \right)\left( {{u_{n - 2}} - 1} \right)...\left( {{u_2} - 1} \right)\) \( = \dfrac{1}{2}\left( {{u_{n - 1}} - 1} \right).\dfrac{1}{2}\left( {{u_{n - 2}} - 1} \right).\dfrac{1}{2}\left( {{u_{n - 3}} - 1} \right)...\dfrac{1}{2}\left( {{u_1} - 1} \right)\)

\( = \dfrac{1}{{{2^{n - 1}}}}\left( {{u_{n - 1}} - 1} \right)\left( {{u_{n - 2}} - 1} \right)...\left( {{u_2} - 1} \right)\left( {{u_1} - 1} \right)\)

\( \Rightarrow {u_n} - 1 = \dfrac{1}{{{2^{n - 1}}}}\left( {{u_1} - 1} \right)\)

\( \Rightarrow {u_n} = \dfrac{{2016}}{{{2^{n - 1}}}} + 1 \Leftrightarrow {u_n} = 4032.{\left( {\dfrac{1}{2}} \right)^n} + 1\)\( \Rightarrow \mathop {\lim }\limits_{} {u_n} = 1\).

Ở câu C, các em cũng có thể giả sử \(\lim {u_n} = a\) thì \({u_{n + 1}} \to a\) khi \(n \to \infty \), do đó ta có \(a = \dfrac{1}{2}\left( {a + 1} \right) \Leftrightarrow a = 1\)

+) Đáp án D:

Ta có \({u_n} = \dfrac{1}{{1.2}} + \dfrac{1}{{2.3}} + \dfrac{1}{{3.4}} + ... + \dfrac{1}{{n\left( {n + 1} \right)}} = 1 - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + ... + \dfrac{1}{n} - \dfrac{1}{{n + 1}} = 1 - \dfrac{1}{{n + 1}} = \dfrac{n}{{n + 1}}\)

\( \Rightarrow \mathop {\lim }\limits_{} {u_n} = \mathop {\lim }\limits_{} \dfrac{n}{{n + 1}} = 1\).

Ta có: \(\lim \dfrac{{3{u_n} - 1}}{{{u_n} + 1}} = \dfrac{{3\lim {u_n} - 1}}{{\lim {u_n} + 1}}\) \( = \dfrac{{3.3 - 1}}{{3 + 1}} = \dfrac{8}{4} = 2\)

$\lim {u_n} = \lim \dfrac{{{n^2} - 3n}}{{1 - 4{n^2}}} = \lim \dfrac{{1 - \dfrac{3}{n}}}{{\dfrac{1}{{{n^2}}} - 4}} = \dfrac{1}{{ - 4}} =  - \dfrac{1}{4}.$

Xét \(g\left( n \right) = \dfrac{{f\left( {2n - 1} \right)}}{{f\left( {2n} \right)}} \Rightarrow g\left( n \right) = \dfrac{{{{\left( {4{n^2} - 2n + 1} \right)}^2} + 1}}{{{{\left( {4{n^2} + 2n + 1} \right)}^2} + 1}}\).

\(g\left( n \right) = \dfrac{{{{\left( {4{n^2} + 1} \right)}^2} - 4n\left( {4{n^2} + 1} \right) + \left( {4{n^2} + 1} \right)}}{{{{\left( {4{n^2} + 1} \right)}^2} + 4n\left( {4{n^2} + 1} \right) + \left( {4{n^2} + 1} \right)}} = \dfrac{{4{n^2} + 1 - 4n + 1}}{{4{n^2} + 1 + 4n + 1}} = \dfrac{{{{\left( {2n - 1} \right)}^2} + 1}}{{{{\left( {2n + 1} \right)}^2} + 1}}\)

$ \Rightarrow {u_n} = \dfrac{2}{{10}}.\dfrac{{10}}{{26}}.\dfrac{{26}}{{50}}....\dfrac{{{{\left( {2n - 3} \right)}^2} + 1}}{{{{\left( {2n - 1} \right)}^2} + 1}}.\dfrac{{{{\left( {2n - 1} \right)}^2} + 1}}{{{{\left( {2n + 1} \right)}^2} + 1}} = \dfrac{2}{{{{\left( {2n + 1} \right)}^2} + 1}}$

\( \Rightarrow \lim n\sqrt {{u_n}}  = \lim \sqrt {\dfrac{{2{n^2}}}{{4{n^2} + 4n + 2}}}  = \dfrac{1}{{\sqrt 2 }}.\)

Định lý: Cho hai dãy số \(\left( {{u_n}} \right)\) và \(\left( {{v_n}} \right)\). Nếu \(\left| {{u_n}} \right| \le {v_n}\) với mọi \(n\) và \(\lim {v_n} = 0\) thì \(\lim {u_n} = 0\).

$\lim {u_n} = \lim \dfrac{{{3^n} + {5^n}}}{{{5^n}}} = \lim \dfrac{{{{\left( {\dfrac{3}{5}} \right)}^n} + 1}}{1} = \dfrac{1}{1} = 1.$

Ta có: $\lim n\left( {\sqrt {4{n^2} + 3}  - \sqrt[3]{{8{n^3} + n}}} \right)$$ = \lim n\left[ {\left( {\sqrt {4{n^2} + 3}  - 2n} \right) + \left( {2n - \sqrt[3]{{8{n^3} + n}}} \right)} \right]$

$ = \lim \left[ {n\left( {\sqrt {4{n^2} + 3}  - 2n} \right) + n\left( {2n - \sqrt[3]{{8{n^3} + n}}} \right)} \right]$.

Ta có: $\lim n\left( {\sqrt {4{n^2} + 3}  - 2n} \right)$$ = \lim \dfrac{{3n}}{{\left( {\sqrt {4{n^2} + 3}  + 2n} \right)}}$$ = \lim \dfrac{3}{{\left( {\sqrt {4 + \dfrac{3}{{{n^2}}}}  + 2} \right)}} = \dfrac{3}{4}$.

Ta có: $\lim n\left( {2n - \sqrt[3]{{8{n^3} + n}}} \right)$$ = \lim \dfrac{{ - {n^2}}}{{\left( {4{n^2} + 2n\sqrt[3]{{8{n^3} + n}} + \sqrt[3]{{{{\left( {8{n^3} + n} \right)}^2}}}} \right)}}$

$ = \lim \dfrac{{ - 1}}{{\left( {4 + 2\sqrt[3]{{8 + \dfrac{1}{{{n^2}}}}} + \sqrt[3]{{{{\left( {8 + \dfrac{1}{{{n^2}}}} \right)}^2}}}} \right)}} =  - \dfrac{1}{{12}}$.

Vậy $\lim n\left( {\sqrt {4{n^2} + 3}  - \sqrt[3]{{8{n^3} + n}}} \right) = \dfrac{3}{4} - \dfrac{1}{{12}}$$ = \dfrac{2}{3}$.

Định lý: Nếu \(\left| q \right| < 1\) thì \(\lim {q^n} = 0\).

Ta có: \({1^2} + {2^2} + {3^2} + ... + {n^2} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\).

Khi đó: \(\lim \sqrt {\dfrac{{{1^2} + {2^2} + {3^2} + ... + {n^2}}}{{2n\left( {n + 7} \right)\left( {6n + 5} \right)}}}  = \lim \sqrt {\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{{12n\left( {n + 7} \right)\left( {6n + 5} \right)}}} \)\( = \lim \sqrt {\dfrac{{\left( {1 + \dfrac{1}{n}} \right)\left( {2 + \dfrac{1}{n}} \right)}}{{12\left( {1 + \dfrac{7}{n}} \right)\left( {6 + \dfrac{5}{n}} \right)}}} \)\( = \dfrac{1}{6}\).

Ta có: \({n^3} - 2n + 1 = {n^3}\left( {1 - \dfrac{2}{{{n^2}}} + \dfrac{1}{{{n^3}}}} \right)\).

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

Đặt \(f\left( x \right) = {x^3} - 3{x^2} + \left( {2m - 2} \right)x + m - 3\). Ta thấy hàm số liên tục trên \(\mathbb{R}\).

Dễ thấy nếu \(x \to  - \infty \)thì \(f\left( x \right) \to  - \infty \) hay \(f\left( x \right) < 0\)

Suy ra điều kiện cần để \(f\left( x \right) = 0\) có \(3\) nghiệm thỏa \({x_1} <  - 1 < {x_2} < {x_3}\) là $f\left( { - 1} \right) > 0 \Leftrightarrow  - m - 5 > 0 \Leftrightarrow m <  - 5$.

Điều kiện đủ: với \(m <  - 5\) ta có

*) \(\mathop {\lim }\limits_{x \to  - \infty } f\left( x \right) =  - \infty \) nên tồn tại \(a <  - 1\) sao cho \(f\left( a \right) < 0\)

Mặt khác \(f\left( { - 1} \right) =  - m - 5 > 0\). Suy ra \(f\left( a \right).f\left( { - 1} \right) < 0\).

Do đó tồn tại \({x_1} \in \left( {a; - 1} \right)\) sao cho \(f\left( {{x_1}} \right) = 0\).

*) \(f\left( 0 \right) = m - 3 < 0\), \(f\left( { - 1} \right) > 0\). Suy ra \(f\left( 0 \right).f\left( { - 1} \right) < 0\).

Do đó tồn tại \({x_2} \in \left( { - 1;0} \right)\) sao cho \(f\left( {{x_2}} \right) = 0\).

*) \(\mathop {\lim }\limits_{x \to  + \infty } f\left( x \right) =  + \infty \) nên tồn tại \(b > 0\) sao cho \(f\left( b \right) > 0\)

Mặt khác \(f\left( 0 \right) < 0\). Suy ra \(f\left( 0 \right).f\left( b \right) < 0\).

Do đó tồn tại \({x_3} \in \left( {0;b} \right)\) sao cho \(f\left( {{x_3}} \right) = 0\).

Vậy \(m <  - 5\) thỏa mãn yêu cầu bài toán.

Bước 1:

$\lim \dfrac{{{2^{n + 1}} - {{3.5}^n} + 5}}{{{{3.2}^n} + {{9.5}^n}}}$ $ = \lim \dfrac{{{{2.2}^n} - {{3.5}^n} + 5}}{{{{3.2}^n} + {{9.5}^n}}} $

$= \lim \dfrac{{2.{{\left( {\dfrac{2}{5}} \right)}^n} - 3 + 5.{{\left( {\dfrac{1}{5}} \right)}^n}}}{{3.{{\left( {\dfrac{2}{5}} \right)}^n} + 9}}$

Bước 2:

$ =\dfrac{2.0-3+5.0}{3.0+9}= \dfrac{{ - 3}}{9} =  - \dfrac{1}{3}.$

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

Ta có

\(\begin{array}{l}{u_2} = {u_1} + 4.1 + 3\\{u_3} = {u_2} + 4.2 + 3\\...\\{u_n} = {u_{n - 1}} + 4.\left( {n - 1} \right) + 3\end{array}\)

Cộng vế theo vế và rút gọn ta được

\({u_n} = {u_1} + 4.\left( {1 + 2 + ... + n - 1} \right) + 3\left( {n - 1} \right)\)\( = 4\dfrac{{n\left( {n - 1} \right)}}{2} + 3\left( {n - 1} \right)\)\( = 2{n^2} + n - 3\), với mọi $n \ge 1$.

Suy ra                                                           

\(\begin{array}{l}{u_{2n}} = 2{\left( {2n} \right)^2} + 2n - 3\\{u_{{2^2}n}} = 2{\left( {{2^2}n} \right)^2} + {2^2}n - 3\\...\\{u_{{2^{2018}}n}} = 2{\left( {{2^{2018}}n} \right)^2} + {2^{2018}}n - 3\end{array}\)

Và

\(\begin{array}{l}{u_{4n}} = 2{\left( {4n} \right)^2} + 4n - 3\\{u_{{4^2}n}} = 2{\left( {{4^2}n} \right)^2} + {4^2}n - 3\\...\\{u_{{4^{2018}}n}} = 2{\left( {{4^{2018}}n} \right)^2} + {4^{2018}}n - 3\end{array}\)

Do đó \(\lim \dfrac{{\sqrt {{u_n}}  + \sqrt {{u_{4n}}}  + \sqrt {{u_{{4^2}n}}}  + ... + \sqrt {{u_{{4^{2018}}n}}} }}{{\sqrt {{u_n}}  + \sqrt {{u_{2n}}}  + \sqrt {{u_{{2^2}n}}}  + ... + \sqrt {{u_{{2^{2018}}n}}} }}\)

$ = \lim \dfrac{{\sqrt {2 + \dfrac{1}{n} - \dfrac{3}{{{n^2}}}}  + \sqrt {{{2.4}^2} + \dfrac{4}{n} - \dfrac{3}{{{n^2}}}}  + ... + \sqrt {2{{\left( {{4^{2018}}} \right)}^2} + \dfrac{{{4^{2018}}}}{n} - \dfrac{3}{{{n^2}}}} }}{{\sqrt {2 + \dfrac{1}{n} - \dfrac{3}{{{n^2}}}}  + \sqrt {{{2.2}^2} + \dfrac{2}{n} - \dfrac{3}{{{n^2}}}}  + ... + \sqrt {2{{\left( {{2^{2018}}} \right)}^2} + \dfrac{{{2^{2018}}}}{n} - \dfrac{3}{{{n^2}}}} }}$

\( = \dfrac{{\sqrt 2 \left( {1 + 4 + {4^2} + ... + {4^{2018}}} \right)}}{{\sqrt 2 \left( {1 + 2 + {2^2} + ... + {2^{2018}}} \right)}}\)$ = \dfrac{{1\dfrac{{1 - {4^{2019}}}}{{1 - 4}}}}{{\dfrac{{1 - {2^{2019}}}}{{1 - 2}}}}$$ = \dfrac{1}{3}\dfrac{{{4^{2019}} - 1}}{{{2^{2019}} - 1}}$\( = \dfrac{{{2^{2019}} + 1}}{3}\).

Vì \({2^{2019}} > 2019\) cho nên sự xác định ở trên là duy nhất nên \(\left\{ \begin{array}{l}a = 2\\b = 1\\c = 3\end{array} \right.\)

Vậy \(S = a + b - c = 0\).

Cách 1:

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

Cách 2: Chia cả tử và mẫu cho $n.$

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

Ta có \(\dfrac{1}{{n\sqrt {n + 1}  + \left( {n + 1} \right)\sqrt n }}\)\( = \dfrac{1}{{\sqrt n \sqrt {n + 1} \left( {\sqrt n  + \sqrt {n + 1} } \right)}}\) \( = \dfrac{{\sqrt {n + 1}  - \sqrt n }}{{\sqrt n \sqrt {n + 1} }} = \dfrac{1}{{\sqrt n }} - \dfrac{1}{{\sqrt {n + 1} }}\)

Suy ra

\({S_n} = \dfrac{1}{{1\sqrt 2  + 2\sqrt 1 }} + \dfrac{1}{{2\sqrt 3  + 3\sqrt 2 }} + ... + \dfrac{1}{{n\sqrt {n + 1}  + \left( {n + 1} \right)\sqrt n }}\)

\( = \dfrac{1}{1} - \dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 2 }} - \dfrac{1}{{\sqrt 3 }} + ... + \dfrac{1}{{\sqrt n }} - \dfrac{1}{{\sqrt {n + 1} }} = 1 - \dfrac{1}{{\sqrt {n + 1} }}\)

Suy ra \(\lim {S_n} = 1\)

$\lim \dfrac{{2{n^2} - n + 4}}{{\sqrt {2{n^4} - {n^2} + 1} }} = \lim \dfrac{{2 - \dfrac{1}{n} + \dfrac{4}{{{n^2}}}}}{{\sqrt {2 - \dfrac{1}{{{n^2}}} + \dfrac{1}{{{n^4}}}} }} = \dfrac{2}{{\sqrt 2 }} = \sqrt 2 .$

Cách 1:

Xét dãy số \(\left( {{u_n}} \right)\), với \({u_n} = \left( {1 - \dfrac{1}{{{2^2}}}} \right)\left( {1 - \dfrac{1}{{{3^2}}}} \right)...\left( {1 - \dfrac{1}{{{n^2}}}} \right)\), \(n \ge 2,\,n \in \mathbb{N}\).

Ta có:

\({u_2} = 1 - \dfrac{1}{{{2^2}}} = \dfrac{3}{4} = \dfrac{{2 + 1}}{{2.2}}\);

\({u_3} = \left( {1 - \dfrac{1}{{{2^2}}}} \right).\left( {1 - \dfrac{1}{{{3^2}}}} \right) = \dfrac{3}{4}.\dfrac{8}{9} = \dfrac{4}{6} = \dfrac{{3 + 1}}{{2.3}}\);

\({u_4} = \left( {1 - \dfrac{1}{{{2^2}}}} \right).\left( {1 - \dfrac{1}{{{3^2}}}} \right)\left( {1 - \dfrac{1}{{{4^2}}}} \right) = \dfrac{3}{4}.\dfrac{8}{9}.\dfrac{{15}}{{16}} = \dfrac{5}{8} = \dfrac{{4 + 1}}{{2.4}}\)

\( \cdots  \cdots \)

\({u_n} = \dfrac{{n + 1}}{{2n}}\).

Dễ dàng chứng minh bằng phương pháp qui nạp để khẳng định \({u_n} = \dfrac{{n + 1}}{{2n}},\,\forall n \ge 2\)

Khi đó \(\lim \left[ {\left( {1 - \dfrac{1}{{{2^2}}}} \right)\left( {1 - \dfrac{1}{{{3^2}}}} \right)...\left( {1 - \dfrac{1}{{{n^2}}}} \right)} \right] = \lim \dfrac{{n + 1}}{{2n}} = \dfrac{1}{2}\).

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