Cho dãy số $({u_n})$xác định bởi $\left\{ \begin{array}{ccccc}u{ _1} = \dfrac{1}{2}\\{u_{n + 1}} = \dfrac{1}{{2 - {u_n}}},\,\,\left( {n \ge 1} \right)\end{array} \right.\,\,$. Khi đó mệnh đề nào sau đây là đúng?
$({u_n}):\,\,\,\,\left\{ \begin{array}{ccccc}u{ _1} = \dfrac{1}{2}\\{u_{n + 1}} = \dfrac{1}{{2 - {u_n}}},\,\,(n \ge 1)\end{array} \right.\,\,$
$\begin{array}{l}{u_2} = \dfrac{1}{{2 - \dfrac{1}{2}}} = \dfrac{1}{{\dfrac{3}{2}}} = \dfrac{2}{3} = \dfrac{2}{{2 + 1}}\\{u_3} = \dfrac{1}{{2 - \dfrac{2}{3}}} = \dfrac{1}{{\dfrac{4}{3}}} = \dfrac{3}{4} = \dfrac{3}{{3 + 1}}\end{array}$
Chứng minh bằng quy nạp: ${u_n} = \dfrac{n}{{n + 1}},\,\,\forall n = 1;2;...\,\,\,\,(*)$
* Với $n = 1,\,n = 2$: (*) đúng
* Giả sử (*) đúng với $n = k$, tức là \({u_k} = \dfrac{k}{{k + 1}}\) , ta chứng minh (*) đúng với $n = k + 1$ , tức là cần chứng minh \({u_{k + 1}} = \dfrac{{k + 1}}{{k + 2}}\)
Ta có: ${u_{k + 1}} = \dfrac{1}{{2 - {u_k}}} = \dfrac{1}{{2 - \dfrac{k}{{k + 1}}}} = \dfrac{1}{{\dfrac{{2k + 2 - k}}{{k + 1}}}} = \dfrac{{k + 1}}{{k + 2}}$
Theo nguyên lý quy nạp, ta chứng minh được (*) đúng với mọi n = 1, 2, …
Như vậy, công thức tổng quát của dãy $({u_n})$là: ${u_n} = \dfrac{n}{{n + 1}},\,\,\forall n = 1;2;...\,\,\,\,(*)$
Từ (*) ta có \({u_{n + 1}} - {u_n} = \dfrac{{n + 1}}{{n + 2}} - \dfrac{n}{{n + 1}} = \dfrac{{{n^2} + 2n + 1 - {n^2} - 2n}}{{\left( {n + 2} \right)\left( {n + 1} \right)}} = \dfrac{1}{{\left( {n + 2} \right)\left( {n + 1} \right)}} > 0\) \( \Rightarrow \left( {{u_n}} \right)\) là dãy tăng và \(\lim {u_n} = \lim \dfrac{n}{{n + 1}} = \lim \dfrac{1}{{1 + \dfrac{1}{n}}} = 1 \Rightarrow \) $({u_n})$ là dãy tăng tới 1 khi $n \to + \infty $
Tính giới hạn của dãy số ${u_n} = q + 2{q^2} + ... + n{q^n}$ với \(\left| q \right| < 1\)
Ta có: ${u_n} - q{u_n} = q + 2{q^2} + ... + n{q^n} - q.\left( {q + 2{q^2} + ... + n{q^n}} \right) $ $= q + {q^2} + {q^3} + ... + {q^n} - n{q^{n + 1}}$
Do \(q,\;{q^2},\;{q^3},.....,\;{q^n}\) là cấp số nhân có công bội \(q\)
\(\begin{array}{l} \Rightarrow {u_n} - q{u_n} = \left( {1 - q} \right){u_n} = q.\dfrac{{1 - {q^n}}}{{1 - q}} - n{q^{n + 1}}\\ \Rightarrow {u_n} = q.\dfrac{{1 - {q^n}}}{{{{\left( {1 - q} \right)}^2}}} - \dfrac{{n.{q^{n + 1}}}}{{1 - q}}.\end{array}\)
Do \(\left| q \right| < 1\) nên \(\mathop {\lim }\limits_{} {q^n} = \mathop {\lim }\limits_{} {q^{n + 1}} = 0\)
Suy ra \(\lim {u_n} = \lim \left[ {\dfrac{{q\left( {1 - {q^n}} \right)}}{{{{\left( {1 - q} \right)}^2}}} - \dfrac{{n.{q^{n + 1}}}}{{1 - q}}} \right] = \dfrac{q}{{{{\left( {1 - q} \right)}^2}}}\).
Cho dãy số $({u_n})$xác định bởi
\(\left\{ \begin{array}{l}{u_1} = \dfrac{1}{2}\\{u_{n + 1}} = \dfrac{{\sqrt {u_n^2 + 4{u_n}} + {u_n}}}{2},\left( {n \ge 1} \right)\end{array} \right.\)
Đặt ${v_n} = \sum\limits_{i = 1}^n {\dfrac{1}{{u_{_i}^2}}}, $ khẳng định nào sau đây đúng?
Xét \({u_{n + 1}} - {u_n} = \dfrac{{\sqrt {u_n^2 + 4{u_n}} + {u_n}}}{2} - {u_n} \)
\(= \dfrac{{\sqrt {u_n^2 + 4{u_n}} - {u_n}}}{2} > \dfrac{{\sqrt {u_n^2} - {u_n}}}{2} = 0 \)
\(\Rightarrow \left( {{u_n}} \right)\) là dãy tăng.
Giả sử $\lim {u_n} = a$ thì $a > 0$ và $a = \dfrac{{\sqrt {{a^2} + 4a} + a}}{2} \Leftrightarrow a = \sqrt {{a^2} + 4a} $ $ \Leftrightarrow {a^2} = {a^2} + 4a \Rightarrow a = 0$ (vô lý).
Suy ra $\lim {u_n} = + \infty $
\(\begin{array}{l}{u_n} = \dfrac{{\sqrt {u_{n - 1}^2 + 4{u_{n - 1}}} + {u_{n - 1}}}}{2} \\ \Leftrightarrow 2{u_n} - {u_{n - 1}} = \sqrt {u_{n - 1}^2 + 4{u_{n - 1}}} \\ \Leftrightarrow 4u_n^2 - 4{u_n}{u_{n - 1}} + u_{n - 1}^2 = u_{n - 1}^2 + 4{u_{n - 1}} \\ \Leftrightarrow u_n^2 = \left( {{u_n} + 1} \right){u_{n - 1}}\end{array}\)
$\begin{array}{l}
\Rightarrow \frac{1}{{u_n^2}} = \frac{1}{{\left( {{u_n} + 1} \right){u_{n - 1}}}} = \frac{{\left( {{u_n} + 1} \right) - {u_n}}}{{\left( {{u_n} + 1} \right){u_{n - 1}}}}\\
= \frac{{{u_n} + 1}}{{\left( {{u_n} + 1} \right){u_{n - 1}}}} - \frac{{{u_n}}}{{\left( {{u_n} + 1} \right){u_{n - 1}}}} = \frac{1}{{{u_{n - 1}}}} - \frac{{{u_n}}}{{u_n^2}}\\
= \frac{1}{{{u_{n - 1}}}} - \frac{1}{{{u_n}}}\\
\Rightarrow \frac{1}{{u_n^2}} = \frac{1}{{{u_{n - 1}}}} - \frac{1}{{{u_n}}}
\end{array}$
Do đó:
\({v_n} = \sum\limits_{i = 1}^n {\dfrac{1}{{u_i^2}}} = \dfrac{1}{{u_1^2}} + \left( {\dfrac{1}{{{u_1}}} - \dfrac{1}{{{u_2}}}} \right) + \left( {\dfrac{1}{{{u_2}}} - \dfrac{1}{{{u_3}}}} \right) + ... + \left( {\dfrac{1}{{{u_{n - 1}}}} - \dfrac{1}{{{u_n}}}} \right) = \dfrac{1}{{u_1^2}} + \dfrac{1}{{{u_1}}} - \dfrac{1}{{{u_n}}} = 6 - \dfrac{1}{{{u_n}}} \) \(\Rightarrow \lim {v_n} = \lim \left( {6 - \dfrac{1}{{{u_n}}}} \right) = 6 - 0 = 6.\)
Cho dãy \(({x_k})\) được xác định như sau: \({x_k} = \dfrac{1}{{2!}} + \dfrac{2}{{3!}} + ... + \dfrac{k}{{(k + 1)!}}\)
Tìm \(\lim {u_n}\) với \({u_n} = \sqrt[n]{{x_1^n + x_2^n + ... + x_{2011}^n}}\).
Ta có: \(\dfrac{k}{{(k + 1)!}} = \dfrac{1}{{k!}} - \dfrac{1}{{(k + 1)!}}\) nên \({x_k} = 1 - \dfrac{1}{{(k + 1)!}}\)
$\begin{array}{l} \Rightarrow {x_k} - {x_{k + 1}} = 1 - \dfrac{1}{{\left( {k + 1} \right)!}} - 1 + \dfrac{1}{{(k + 2)!}} = \dfrac{1}{{(k + 2)!}} - \dfrac{1}{{(k + 1)!}} < 0\\ \Rightarrow {x_k} < {x_{k + 1}} \Rightarrow {x_1} < {x_2} < ... < {x_{2011}}\\ \Rightarrow x_1^n < x_2^n < ... < x_{2011}^n \Rightarrow \sqrt[n]{{x_1^n + x_2^n + ... + x_{2011}^n}} < \sqrt[n]{{x_{2011}^n + x_{2011}^n + ... + x_{2011}^n}} = \sqrt[n]{{2011.x_{2011}^n}} = \sqrt[n]{{2011}}.{x_{2011}}\end{array}$
Lại có: \({x_{2011}} = \sqrt[n]{{x_{2011}^n}} < \sqrt[n]{{x_1^n + x_2^n + ... + x_{2011}^n}}\)
Vậy: \({x_{2011}} < \sqrt[n]{{x_1^n + x_2^n + ... + x_{2011}^n}} < \sqrt[n]{{2011}}{x_{2011}}\)
Ta có: \({x_{2011}} = 1 - \dfrac{1}{{\left( {2011 + 1} \right)!}} = 1 - \dfrac{1}{{2012!}}\)
\(\begin{array}{l} \Rightarrow \lim {x_{2011}} = {x_{2011}} = 1 - \dfrac{1}{{2012!}}\\ \Rightarrow \lim \sqrt[n]{{2011}}{x_{2011}} = \lim \,\,{2011^{\dfrac{1}{n}}}{x_{2011}} = {2011^0}.{x_{2011}} = {x_{2011}} = 1 - \dfrac{1}{{2012!}}\end{array}\)
Vậy \(\lim {u_n} = 1 - \dfrac{1}{{2012!}}\).
Cho ${u_n} = \dfrac{{1 - 4n}}{{5n}}$. Khi đó $\lim {u_n}$bằng?
$\lim {u_n} = \lim \dfrac{{1 - 4n}}{{5n}} = \lim \dfrac{{\dfrac{1}{n} - 4}}{5} = \dfrac{{ - 4}}{5} = - \dfrac{4}{5}.$
Cho ${u_n} = \dfrac{{{n^2} - 3n}}{{1 - 4{n^2}}}$. Khi đó $\lim {u_n}$bằng?
$\lim {u_n} = \lim \dfrac{{{n^2} - 3n}}{{1 - 4{n^2}}} = \lim \dfrac{{1 - \dfrac{3}{n}}}{{\dfrac{1}{{{n^2}}} - 4}} = \dfrac{1}{{ - 4}} = - \dfrac{1}{4}.$
Cho ${u_n} = \dfrac{{{n^2} - 3n}}{{1 - 4{n^3}}}$. Khi đó $\lim {u_n}$bằng?
$\lim {u_n} = \lim \dfrac{{{n^2} - 3n}}{{1 - 4{n^3}}} = \lim \dfrac{{\dfrac{1}{n} - \dfrac{3}{{{n^2}}}}}{{\dfrac{1}{{{n^3}}} - 4}} = \dfrac{0}{{ - 4}} = 0.$
Cho ${u_n} = \dfrac{{{3^n} + {5^n}}}{{{5^n}}}$. Khi đó $\lim {u_n}$bằng?
$\lim {u_n} = \lim \dfrac{{{3^n} + {5^n}}}{{{5^n}}} = \lim \dfrac{{{{\left( {\dfrac{3}{5}} \right)}^n} + 1}}{1} = \dfrac{1}{1} = 1.$
Trong các giới hạn sau giới hạn nào bằng $-1$?
$\begin{array}{l}\lim \dfrac{{2{n^2} - 3}}{{ - 2{n^3} - 4}} = \lim \dfrac{{\dfrac{2}{n} - \dfrac{3}{{{n^3}}}}}{{ - 2 - \dfrac{4}{{{n^3}}}}} = \dfrac{0}{{ - 2}} = 0.\\\lim \dfrac{{2{n^2} - 3}}{{ - 2{n^2} - 1}} = \lim \dfrac{{2 - \dfrac{3}{{{n^2}}}}}{{ - 2 - \dfrac{1}{{{n^2}}}}} = \dfrac{2}{{ - 2}} = - 1.\\\lim \dfrac{{2{n^2} - 3}}{{2{n^2} + 1}} = \lim \dfrac{{2 - \dfrac{3}{{{n^2}}}}}{{2 + \dfrac{1}{{{n^2}}}}} = \dfrac{2}{2} = 1.\\\lim \dfrac{{2{n^3} - 3}}{{2{n^2} - 1}} = \lim \dfrac{{2 - \dfrac{3}{{{n^3}}}}}{{\dfrac{2}{n} - \dfrac{1}{{{n^3}}}}} = + \infty .\end{array}$
Giá trị \(\lim \left( {{n^3} - 2n + 1} \right)\) bằng
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 \(\lim \left( {{n^3} - 2n + 1} \right) = + \infty \)
Giới hạn $\lim \dfrac{{{2^{n + 1}} - {{3.5}^n} + 5}}{{{{3.2}^n} + {{9.5}^n}}}$bằng?
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}.$
Giới hạn $\lim \dfrac{{{{\left( {2 - 5n} \right)}^3}{{\left( {n + 1} \right)}^2}}}{{2 - 25{n^5}}}$bằng?
$\lim \dfrac{{{{(2 - 5n)}^3}{{(n + 1)}^2}}}{{2 - 25{n^5}}}$ $ = \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}}$ $= \dfrac{{{{( - 5)}^3}{{.1}^2}}}{{ - 25}} = 5$.
Giới hạn $\lim \dfrac{{\sqrt {{n^2} - 3n - 5} - \sqrt {9{n^2} + 3} }}{{2n - 1}}$bằng?
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 $n.$
$\lim \frac{{\sqrt {{n^2} - 3n - 5} - \sqrt {9{n^2} + 3} }}{{2n - 1}} = \lim \frac{{\sqrt {1 - \frac{3}{n} - \frac{5}{{{n^2}}}} - \sqrt {9 + \frac{3}{{{n^2}}}} }}{{2 - \frac{1}{n}}} = \lim \frac{{1 - 3}}{2} = - 1$
Giới hạn $\lim \dfrac{{2{n^2} - n + 4}}{{\sqrt {2{n^4} - {n^2} + 1} }}$bằng?
$\lim \dfrac{{2{n^2} - n + 4}}{{\sqrt {2{n^4} - {n^2} + 1} }} = \lim \dfrac{{2 - \dfrac{1}{n} + \dfrac{4}{{{n^2}}}}}{{\sqrt {2 - \dfrac{1}{{{n^2}}} + \dfrac{1}{{{n^4}}}} }} = \dfrac{2}{{\sqrt 2 }} = \sqrt 2 .$
Giới hạn $\lim \left( {\sqrt {{n^2} - n} - n} \right)$ bằng?
\(\lim \left( {\sqrt {{n^2} - n} - n} \right)\) \(= \lim \dfrac{{\left( {\sqrt {{n^2} - n} - n} \right).\left( {\sqrt {{n^2} - n} + n} \right)}}{{\sqrt {{n^2} - n} + n}} \) \(= \lim \dfrac{{{n^2} - n - {n^2}}}{{\sqrt {{n^2} - n} + n}} \) \(= \lim \dfrac{{ - n}}{{\sqrt {{n^2} - n} + n}}\) \(= \lim \dfrac{{ - 1}}{{\sqrt {1 - \dfrac{1}{n}} + 1}}\) \(= \dfrac{{ - 1}}{2} = - \dfrac{1}{2}.\)
Giới hạn $\lim \left( {\sqrt {{n^2} - n + 1} - \sqrt {{n^2} + 1} } \right)$ bằng?
$\begin{array}{l}\lim ( {\sqrt {{n^2} - n + 1} - \sqrt {{n^2} + 1} } ) \\ =\lim \frac{{( {\sqrt {{n^2} - n + 1} - \sqrt {{n^2} + 1} }) ( {\sqrt {{n^2} - n + 1} + \sqrt {{n^2} + 1} } )}}{{\sqrt {{n^2} - n + 1} + \sqrt {{n^2} + 1} }}\\ = \lim \frac{{{n^2} - n + 1 - {n^2} - 1}}{{\sqrt {{n^2} - n + 1} + \sqrt {{n^2} + 1} }} \\= \lim \frac{{ - n}}{{\sqrt {{n^2} - n + 1} + \sqrt {{n^2} + 1} }} \\= \lim \frac{{ - 1}}{{\sqrt {1 - \frac{1}{n} + \frac{1}{{{n^2}}}} + \sqrt {1 + \frac{1}{{{n^2}}}} }} \\= -\frac{{1}}{2}\end{array}$
Cho dãy số $({u_n})$với ${u_n} = \dfrac{1}{{1.2}} + \dfrac{1}{{2.3}} + \dfrac{1}{{3.4}} + ... + \dfrac{1}{{n.\left( {n + 1} \right)}}$. Khi đó $\lim {u_n}$ bằng?
$\begin{array}{l}{u_n} = \dfrac{1}{{1.2}} + \dfrac{1}{{2.3}} + \dfrac{1}{{3.4}} + ... + \dfrac{1}{{n.\left( {n + 1} \right)}} \\= \dfrac{{2 - 1}}{{1.2}} + \dfrac{{3 - 2}}{{2.3}} + ... + \dfrac{{n + 1 - n}}{{n.\left( {n + 1} \right)}}\\ = 1 - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + .... + \dfrac{1}{n} - \dfrac{1}{{n + 1}} \\= 1 - \dfrac{1}{{n + 1}}\\ \Rightarrow \lim {u_n} = \lim \left( {1 - \dfrac{1}{{n + 1}}} \right) = 1.\end{array}$
Cho dãy số $({u_n})$ với ${u_n} = \dfrac{1}{{1.3}} + \dfrac{1}{{3.5}} + ... + \dfrac{1}{{\left( {2n - 1} \right).\left( {2n + 1} \right)}}$
Khi đó $\lim {u_n}$ bằng?
$\begin{array}{l}{u_n} = \frac{1}{{1.3}} + \frac{1}{{3.5}}+ ... + \frac{1}{{\left( {2n - 1} \right).\left( {2n + 1} \right)}}\\ = \frac{1}{2}. \left( {1 - \frac{1}{3} + \frac{1}{3} - \frac{1}{5}+ ... + \frac{1}{{2n - 1}} - \frac{1}{{2n + 1}}} \right) \\ = \frac{1}{2}.\left( {1 - \frac{1}{{2n + 1}}} \right)\\ \Rightarrow \lim {u_n} = \lim \frac{1}{2}\left( {1 - \frac{1}{{2n + 1}}} \right) = \frac{1}{2}.\end{array}$
Giá trị \(\lim \dfrac{{\sin \left( {n!} \right)}}{{{n^2} + 1}}\) bằng
Ta có \(\left| {\dfrac{{\sin \left( {n!} \right)}}{{{n^2} + 1}}} \right| \le \dfrac{1}{{{n^2} + 1}}\) mà \(\lim \dfrac{1}{{{n^2} + 1}} = 0\) nên chọn đáp án A.
Cho dãy số $({u_n})$ với ${u_n} = \dfrac{{\left( {2n + 1} \right)\left( {1 - 3n} \right)}}{{\sqrt[3]{{{n^3} + 5n - 1}}}}$. Khi đó $\lim {u_n}$ bằng?
$\lim {u_n} = \lim \dfrac{{\left( {2n + 1} \right)\left( {1 - 3n} \right)}}{{\sqrt[3]{{{n^3} + 5n - 1}}}}$ $ = \lim \dfrac{{ - 6{n^2} - n + 1}}{{\sqrt[3]{{{n^3} + 5n - 1}}}} $ $= \lim \dfrac{{\dfrac{{ - 6{n^2} - n + 1}}{{{n^2}}}}}{{\sqrt[3]{{\dfrac{{{n^3} + 5n - 1}}{{{n^6}}}}}}}$ $=\lim \dfrac{{ - 6 - \dfrac{1}{n} + \dfrac{1}{{{n^2}}}}}{{\sqrt[3]{{\dfrac{1}{{{n^3}}} + \dfrac{5}{{{n^5}}} - \dfrac{1}{{{n^6}}}}}}} = - \infty .$
