70 Câu Trắc Nghiệm Về Giới Hạn Của Dãy Số Giải Chi Tiết

Chọn C

Ta có $\lim \frac{{ – 3}}{{4{n^2} – 2n + 1}} = \lim \frac{{\frac{{ – 3}}{{{n^2}}}}}{{4 – \frac{2}{n} + \frac{1}{{{n^2}}}}} = \frac{0}{4} = 0.$

Giải nhanh : Dạng « bậc tử » $ < $ « bậc mẫu » nên kết quả bằng 0.

Chọn B

Ta có $\lim \frac{{3{n^3} – 2n + 1}}{{4{n^4} + 2n + 1}} = \lim \frac{{\frac{3}{n} – \frac{2}{{{n^2}}} + \frac{1}{{{n^4}}}}}{{4 + \frac{2}{{{n^3}}} + \frac{1}{{{n^4}}}}} = \frac{0}{4} = 0.$

Giải nhanh : Dạng « bậc tử » $ < $ « bậc mẫu » nên kết quả bằng 0.

Chọn A

Ta có $\lim \frac{{{v_n}}}{{{u_n}}} = \lim \frac{{n + 1}}{{n + 2}} = \lim \frac{{1 + \frac{1}{n}}}{{1 + \frac{2}{n}}} = \frac{1}{1} = 1.$

Giải nhanh : $\frac{{n + 1}}{{n + 2}} \sim \frac{n}{n} = 1.$

Chọn A

Ta có $\lim {u_n} = \lim \frac{{an + 4}}{{5n + 3}} = \lim \frac{{a + \frac{4}{n}}}{{5 + \frac{3}{n}}} = \frac{a}{5}.$ Khi đó

$\lim {u_n} = 2 \Leftrightarrow \frac{a}{5} = 2 \Leftrightarrow a = 10$

Giải nhanh : $2 \sim \frac{{an + 4}}{{5n + 3}} \sim \frac{{an}}{{5n}} = \frac{a}{5} \Leftrightarrow a = 10.$

Chọn B

Ta có $L = \lim \frac{{{n^2} + n + 5}}{{2{n^2} + 1}} = \lim \frac{{1 + \frac{1}{n} + \frac{5}{{{n^2}}}}}{{2 + \frac{1}{{{n^2}}}}} = \frac{1}{2}$

Giải nhanh: $\frac{{{n^2} + n + 5}}{{2{n^2} + 1}} \sim \frac{{{n^2}}}{{2{n^2}}} = \frac{1}{2}.$

Chọn A

$L = \lim \frac{{{n^2} – 3{n^3}}}{{2{n^3} + 5n – 2}} = \lim \frac{{\frac{1}{n} – 3}}{{2 + \frac{5}{{{n^2}}} – \frac{2}{{{n^3}}}}} = \frac{{ – 3}}{2}$

Giải nhanh: $\frac{{{n^2} – 3{n^3}}}{{2{n^3} + 5n – 2}} \sim \frac{{ – 3{n^3}}}{{2{n^3}}} = – \frac{3}{2}.$

Chọn C

$L = \lim \frac{{5{n^2} – 3a{n^4}}}{{\left( {1 – a} \right){n^4} + 2n + 1}} = \lim \frac{{\frac{5}{{{n^2}}} – 3a}}{{\left( {1 – a} \right) + \frac{2}{{{n^3}}} + \frac{1}{{{n^4}}}}} = \frac{{ – 3a}}{{\left( {1 – a} \right)}} > 0 \Leftrightarrow \left[ \begin{gathered}
a < 0 \hfill \\
a > 1 \hfill \\
\end{gathered} \right..$

Chọn A

Ta có

$L = \lim \frac{{\left( {2n – {n^3}} \right)\left( {3{n^2} + 1} \right)}}{{\left( {2n – 1} \right)\left( {{n^4} – 7} \right)}} = \lim \frac{{{n^3}\left( {\frac{2}{{{n^2}}} – 1} \right).{n^2}\left( {3 + \frac{1}{{{n^2}}}} \right)}}{{n\left( {2 – \frac{1}{n}} \right).{n^4}\left( {1 – \frac{7}{{{n^4}}}} \right)}}$

$ = \lim \frac{{\left( {\frac{2}{{{n^2}}} – 1} \right)\left( {3 + \frac{1}{{{n^2}}}} \right)}}{{\left( {2 – \frac{1}{n}} \right)\left( {1 – \frac{7}{{{n^4}}}} \right)}} = \frac{{ – 1.3}}{{2.1}} = – \frac{3}{2}$

Giải nhanh: $\frac{{\left( {2n – {n^3}} \right)\left( {3{n^2} + 1} \right)}}{{\left( {2n – 1} \right)\left( {{n^4} – 7} \right)}} \sim \frac{{ – {n^3}.3{n^2}}}{{2n.{n^4}}} = – \frac{3}{2}.$

Chọn C

$\lim \frac{{{n^3} – 2n}}{{1 – 3{n^2}}} = \lim \frac{{{n^3}\left( {1 – \frac{2}{{{n^2}}}} \right)}}{{{n^2}\left( {\frac{1}{{{n^2}}} – 3} \right)}} = \lim n.\frac{{1 – \frac{2}{{{n^2}}}}}{{\frac{1}{{{n^2}}} – 3}}.$ Ta có

$\left\{ \begin{gathered}
\lim n = + \infty \hfill \\
\lim \frac{{1 – \frac{2}{{{n^2}}}}}{{\frac{1}{{{n^2}}} – 3}} = – \frac{1}{3} < 0 \hfill \\
\end{gathered} \right.\xrightarrow[{}]{}im\frac{{{n^3} – 2n}}{{1 – 3{n^2}}} = \lim n.\frac{{1 – \frac{2}{{{n^2}}}}}{{\frac{1}{{{n^2}}} – 3}} = – \infty $

Giải nhanh : $\frac{{{n^3} – 2n}}{{1 – 3{n^2}}} \sim \frac{{{n^3}}}{{ – 3{n^2}}} = – \frac{1}{3}n\xrightarrow[{}]{} – \infty .$

Chọn B

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

$ = \lim n.\frac{{\frac{2}{{{n^2}}} + 3}}{{4 + \frac{2}{n} + \frac{1}{{{n^2}}}}}$

Ta có

$\left\{ \begin{gathered}
\lim n = + \infty \hfill \\
\lim \frac{{\frac{2}{{{n^2}}} + 3}}{{4 + \frac{2}{n} + \frac{1}{{{n^2}}}}} = \frac{3}{4} > 0 \hfill \\
\end{gathered} \right.$

$\xrightarrow[{}]{}im\frac{{2n + 3{n^3}}}{{4{n^2} + 2n + 1}} = \lim n.\frac{{\frac{2}{{{n^2}}} + 3}}{{4 + \frac{2}{n} + \frac{1}{{{n^2}}}}} = + \infty $

Giải nhanh : $\frac{{2n + 3{n^3}}}{{4{n^2} + 2n + 1}} \sim \frac{{3{n^3}}}{{4{n^2}}} = \frac{3}{4}.n\xrightarrow[{}]{} + \infty .$

Chọn C

$\lim \frac{{3n – {n^4}}}{{4n – 5}} = \lim \frac{{{n^4}\left( {\frac{3}{{{n^3}}} – 1} \right)}}{{n\left( {4 – \frac{5}{n}} \right)}} = \lim {n^3}.\frac{{\frac{3}{{{n^3}}} – 1}}{{4 – \frac{5}{n}}}.$
Ta có

$\left\{ \begin{gathered}
\lim {n^3} = + \infty \hfill \\
\lim \frac{{\frac{3}{{{n^3}}} – 1}}{{4 – \frac{5}{n}}} = – \frac{1}{4} < 0 \hfill \\
\end{gathered} \right.$

$\xrightarrow[{}]{}\lim \frac{{3n – {n^4}}}{{4n – 5}} = \operatorname{l} \lim {n^3}.\frac{{\frac{3}{{{n^3}}} – 1}}{{4 – \frac{5}{n}}} = – \infty $

Giải nhanh : $\frac{{3n – {n^4}}}{{4n – 5}} \sim \frac{{ – {n^4}}}{{4n}} = – \frac{1}{4}.{n^3}\xrightarrow[{}]{} – \infty .$

Chọn B

. Theo dấu hiệu ở đã nêu ở phần Chú ý trên thì ta chọn giới hạn nào rơi vào trường hợp « bậc tử » $ < $ « bậc mẫu » !

$\lim \frac{{3 + 2{n^3}}}{{2{n^2} – 1}} = + \infty $ : « bậc tử » $ > $ « bậc mẫu » và ${a_m}{b_k} = 2.2 = 4 > 0.$

$\lim \frac{{2{n^2} – 3}}{{ – 2{n^3} – 4}} = 0$ : « bậc tử » $ < $ « bậc mẫu ».

$\lim \frac{{2n – 3{n^3}}}{{ – 2{n^2} – 1}} = + \infty $ : « bậc tử » $ > $ « bậc mẫu » và ${a_n}{b_k} = \left( { – 3} \right).\left( { – 2} \right) > 0.$

$\lim \frac{{2{n^2} – 3{n^4}}}{{ – 2{n^4} + {n^2}}} = \frac{{ – 3}}{{ – 2}} = \frac{3}{2}$ : « bậc tử » $ = $ « bậc mẫu » và $\frac{{{a_m}}}{{{b_k}}} = \frac{{ – 3}}{{ – 2}} = \frac{3}{2}.$

Chọn C

Ta chọn đáp án dạng « bậc tử » $ = $ « bậc mẫu » và ${a_m}{b_k} < 0.$

${u_n} = \frac{{2{n^2} – 3{n^4}}}{{{n^2} + 2{n^3}}}$ : « bậc tử » $ > $ « bậc mẫu » và ${a_m}{b_k} = – 3.2 = – 6 < 0\xrightarrow[{}]{}\lim {u_n} = – \infty .$

Chú ý : (i) $\lim \left( {{a_m}{n^m} + {a_{n – 1}}{n^{m – 1}} + \cdots + {a_1}n + {a_0}} \right) = \left\{ \begin{gathered}
+ \infty \,\,\,\,khi\,\,{a_n} > 0 \hfill \\
– \infty \,\,\,\,khi\,\,{a_n} < 0 \hfill \\
\end{gathered} \right..$

(ii) Giả sử $\,\left| q \right| > \max \left\{ {\left| {{q_i}} \right|:i = 1;2 \ldots ;m} \right\}$ thì

$\lim \left( {a.{q^n} + {a_m}q_m^n + \cdots + {a_1}q_1^n + {a_0}} \right) = \left\{ \begin{gathered}
{a_0}\,\,\,\,\,\,\,\,\,\,\,khi\,\,\left| q \right| < 1 \hfill \\
+ \infty \,\,\,\,\,khi\,\,\,a > 0,\,\,q > 1 \hfill \\
– \infty \,\,\,\,\,khi\,\,a < 0,\,\,q > 1 \hfill \\
\end{gathered} \right..$

Ta dùng « dấu hiệu nhanh » này để đưa ra kết quả nhanh chóng cho các bài sau.

Chọn D

. $L = \lim \left( {3{n^2} + 5n – 3} \right) = \lim {n^2}\left( {2 + \frac{5}{n} – \frac{3}{{{n^2}}}} \right) = + \infty $ vì $\left\{ \begin{gathered}
\lim {n^2} = + \infty \hfill \\
\lim \left( {2 + \frac{5}{n} – \frac{3}{{{n^2}}}} \right) = 2 > 0 \hfill \\
\end{gathered} \right..$

Giải nhanh :$3{n^2} + 5n – 3 \sim 3{n^2}\xrightarrow[{}]{} + \infty .$

Chọn A

Ta có

$\lim \left( {5n – 3\left( {{a^2} – 2} \right){n^3}} \right) = \lim {n^3}\left( {\frac{5}{{{n^2}}} – 3\left( {{a^2} – 2} \right)} \right) = – \infty $

$ \Leftrightarrow \lim \left( {\frac{5}{{{n^2}}} – 3\left( {{a^2} – 2} \right)} \right) = {a^2} – 2 > 0 \Leftrightarrow \left[ \begin{gathered}
a < – \sqrt 2 \hfill \\
a > \sqrt 2 \hfill \\
\end{gathered} \right.$

Chọn D

Ta có

$\lim \left( {3{n^4} + 4{n^2} – n + 1} \right) = \lim {n^4}\left( {3 + \frac{4}{{{n^2}}} – \frac{1}{{{n^3}}} + \frac{1}{{{n^4}}}} \right) = + \infty $ vì $\left\{ \begin{gathered}
\lim {n^4} = + \infty \hfill \\
\lim \left( {3 + \frac{4}{{{n^2}}} – \frac{1}{{{n^3}}} + \frac{1}{{{n^4}}}} \right) = 3 > 0 \hfill \\
\end{gathered} \right..$

Giải nhanh : $3{n^4} + 4{n^2} – n + 1 \sim 3{n^4}\xrightarrow[{}]{} + \infty .$

Chọn A

$\sqrt {n + 5} – \sqrt {n + 1} \sim \sqrt n – \sqrt n = 0\xrightarrow[{}]{}$nhân lượng liên hợp :

$\lim \left( {\sqrt {n + 5} – \sqrt {n + 1} } \right) = \lim \frac{4}{{\sqrt {n + 5} + \sqrt {n + 1} }} = 0$

Chọn C

$\lim \left( {\sqrt {{n^2} – 1} – \sqrt {3{n^2} + 2} } \right) = \lim n\left( {\sqrt {1 – \frac{1}{{{n^2}}}} – \sqrt {3 + \frac{2}{{{n^2}}}} } \right) = – \infty $ vì

$\lim n = + \infty ,\,\,\lim \left( {\sqrt {1 – \frac{1}{{{n^2}}}} – \sqrt {3 + \frac{2}{{{n^2}}}} } \right) = 1 – \sqrt 3 < 0.$

Giải nhanh : $\sqrt {{n^2} – 1} – \sqrt {3{n^2} + 2} \sim \sqrt {{n^2}} – \sqrt {3{n^2}} = \left( {1 – \sqrt 3 } \right)n\xrightarrow[{}]{} – \infty .$

Chọn B

$\sqrt {{n^2} + 2n} – \sqrt {{n^2} – 2n} \sim \sqrt {{n^2}} – \sqrt {{n^2}} = 0\xrightarrow[{}]{}$nhân lượng liên hợp :

$\lim \left( {\sqrt {{n^2} + 2n} – \sqrt {{n^2} – 2n} } \right) = \lim \frac{{4n}}{{\sqrt {{n^2} + 2n} + \sqrt {{n^2} – 2n} }}$

$ = \lim \frac{4}{{\sqrt {1 + \frac{2}{n}} + \sqrt {1 – \frac{2}{n}} }} = 2$

Giải nhanh : $\sqrt {{n^2} + 2n} – \sqrt {{n^2} – 2n} = \frac{{4n}}{{\sqrt {{n^2} + 2n} + \sqrt {{n^2} – 2n} }} \sim \frac{{4n}}{{\sqrt {{n^2}} + \sqrt {{n^2}} }} = 2.$

Chọn B

$\sqrt {{n^2} + {a^2}n} – \sqrt {{n^2} + \left( {a + 2} \right)n + 1} \sim \sqrt {{n^2}} – \sqrt {{n^2}} = 0\xrightarrow[{}]{}$nhân lượng liên hợp:

Ta có $\lim \left( {\sqrt {{n^2} + {a^2}n} – \sqrt {{n^2} + \left( {a + 2} \right)n + 1} } \right) = \lim \frac{{\left( {{a^2} – a – 2} \right)n – 1}}{{\sqrt {{n^2} + n} + \sqrt {{n^2} + 1} }}$

$ = \lim \frac{{{a^2} – a – 2 – \frac{1}{n}}}{{\sqrt {1 + \frac{1}{n}} + \sqrt {1 + \frac{1}{{{n^2}}}} }} = \frac{{{a^2} – a – 2}}{2} = 0 \Leftrightarrow \left[ \begin{gathered}
a = – 1 \hfill \\
a = 2 \hfill \\
\end{gathered} \right..$

Chọn B

$\sqrt {2{n^2} – n + 1} – \sqrt {2{n^2} – 3n + 2} \sim \sqrt {2{n^2}} – \sqrt {2{n^2}} = 0\xrightarrow[{}]{}$nhân lượng liên hợp :

$\lim \left( {\sqrt {2{n^2} – n + 1} – \sqrt {2{n^2} – 3n + 2} } \right) = \lim \frac{{2n – 1}}{{\sqrt {2{n^2} – n + 1} + \sqrt {2{n^2} – 3n + 2} }}$

$ = \lim \frac{{2 – \frac{1}{n}}}{{\sqrt {2 – \frac{1}{n} + \frac{1}{{{n^2}}}} + \sqrt {2 – \frac{3}{n} + \frac{2}{{{n^2}}}} }} = \frac{1}{{\sqrt 2 }}$

Giải nhanh :

$\sqrt {2{n^2} – n + 1} – \sqrt {2{n^2} – 3n + 2} $

$ = \frac{{2n – 1}}{{\sqrt {2{n^2} – n + 1} + \sqrt {2{n^2} – 3n + 2} }} \sim \frac{{2n}}{{\sqrt {2{n^2}} + \sqrt {2{n^2}} }} = \frac{1}{{\sqrt 2 }}$

Chọn C

Giải nhanh :$\sqrt {{n^2} + 2n – 1} – \sqrt {2{n^2} + n} \sim \sqrt {{n^2}} – \sqrt {2{n^2}} = \left( {1 – \sqrt 2 } \right)n\xrightarrow[{}]{} – \infty .$

Cụ thể : $\lim \left( {\sqrt {{n^2} + 2n – 1} – \sqrt {2{n^2} + n} } \right) = \lim n.\left( {\sqrt {1 + \frac{2}{n} – \frac{1}{{{n^2}}}} – \sqrt {2 + \frac{1}{n}} } \right) = – \infty $

vì

$\lim n = + \infty ,\,\,\lim \left( {\sqrt {1 + \frac{2}{n} – \frac{1}{{{n^2}}}} – \sqrt {2 + \frac{1}{n}} } \right) = 1 – \sqrt 2 < 0$

Chọn B

Nếu $\sqrt {{n^2} – 8n} – n + {a^2} \sim \sqrt {{n^2}} – n = 0\xrightarrow[{}]{}$nhân lượng liên hợp :

Ta có $\lim \left( {\sqrt {{n^2} – 8n} – n + {a^2}} \right) = \lim \frac{{\left( {2{a^2} – 8} \right)n}}{{\sqrt {{n^2} + n} + n}} = \lim \frac{{2{a^2} – 8}}{{\sqrt {1 + \frac{1}{n}} + 1}}$

$ = {a^2} – 4 = 0 \Leftrightarrow a = \pm 2.$

Chọn A

$\sqrt {{n^2} – 2n + 3} – n \sim \sqrt {{n^2}} – n = 0\xrightarrow[{}]{}$nhân lượng liên hợp :

$\lim \left( {\sqrt {{n^2} – 2n + 3} – n} \right) = \lim \frac{{ – 2n + 3}}{{\sqrt {{n^2} – 2n + 3} + n}}$

$ = \lim \frac{{ – 2 + \frac{3}{n}}}{{\sqrt {1 – \frac{2}{n} + \frac{3}{{{n^2}}}} + 1}} = – 1$

Giải nhanh : $\sqrt {{n^2} – 2n + 3} – n = \frac{{ – 2n + 3}}{{\sqrt {{n^2} – 2n + 3} + n}} \sim \frac{{ – 2n}}{{\sqrt {{n^2}} + n}} = – 1.$

Chọn C

$\sqrt {{n^2} + an + 5} – \sqrt {{n^2} + 1} \sim \sqrt {{n^2}} – \sqrt {{n^2}} = 0\xrightarrow[{}]{}$nhân lượng liên hợp :

$ – 1 = \lim {u_n} = \lim \left( {\sqrt {{n^2} + an + 5} – \sqrt {{n^2} + 1} } \right)$

$ = \lim \frac{{an + 4}}{{\sqrt {{n^2} + an + 5} + \sqrt {{n^2} + 1} }}$

$ = \lim \frac{{a + \frac{4}{n}}}{{\sqrt {1 + \frac{a}{n} + \frac{5}{{{n^2}}}} + \sqrt {1 + \frac{1}{{{n^2}}}} }} = \frac{a}{2} \Leftrightarrow a = – 2.$

Giải nhanh :

$ – 1 \sim \sqrt {{n^2} + an + 5} – \sqrt {{n^2} + 1} = \frac{{an + 4}}{{\sqrt {{n^2} + an + 5} + \sqrt {{n^2} + 1} }} \sim \frac{{an}}{{\sqrt {{n^2}} + \sqrt {{n^2}} }} = \frac{a}{2} \Leftrightarrow a = – 2.$

Chọn C

$\sqrt[3]{{{n^3} + 1}} – \sqrt[3]{{{n^3} + 2}} \sim \sqrt[3]{{{n^3}}} – \sqrt[3]{{{n^3}}} = 0\xrightarrow[{}]{}$nhân lượng liên hợp :

$\lim \left( {\sqrt[3]{{{n^3} + 1}} – \sqrt[3]{{{n^3} + 2}}} \right) = \lim \frac{{ – 1}}{{\sqrt[3]{{{{\left( {{n^3} + 1} \right)}^2}}} + \sqrt[3]{{{n^3} + 1}}.\sqrt[3]{{{n^3} + 2}} + \sqrt[3]{{\left( {{n^3} + 2} \right)}}}} = 0.$

Chọn B

$\sqrt[3]{{{n^3} – 2{n^2}}} – n \sim \sqrt[3]{{{n^3}}} – n = 0\xrightarrow[{}]{}$nhân lượng liên hợp :

$\lim \left( {\sqrt[3]{{{n^3} – 2{n^2}}} – n} \right) = \lim \frac{{ – 2{n^2}}}{{\sqrt[3]{{{{\left( {{n^3} – 2{n^2}} \right)}^2}}} + n.\sqrt[3]{{{n^3} – 2{n^2}}} + {n^2}}}$

$ = \lim \frac{{ – 2}}{{\sqrt[3]{{{{\left( {1 – \frac{2}{n}} \right)}^2}}} + \sqrt[3]{{1 – \frac{2}{n}}} + 1}} = – \frac{2}{3}$

Giải nhanh : $\sqrt[3]{{{n^3} – 2{n^2}}} – n = \frac{{ – 2{n^2}}}{{\sqrt[3]{{{{\left( {{n^3} – 2{n^2}} \right)}^2}}} + n.\sqrt[3]{{{n^3} – 2{n^2}}} + {n^2}}}$

$ \sim \frac{{ – 2{n^2}}}{{\sqrt[3]{{{n^6}}} + n.\sqrt[3]{{{n^3}}} + {n^2}}} = – \frac{2}{3}$

Chọn D

$\sqrt n \left( {\sqrt {n + 1} – \sqrt {n – 1} } \right) \sim \sqrt n \left( {\sqrt n – \sqrt n } \right) = 0\xrightarrow[{}]{}$nhân lượng liên hợp :

$\lim \sqrt n \left( {\sqrt {n + 1} – \sqrt {n – 1} } \right) = \lim \frac{{2\sqrt n }}{{\sqrt {n + 1} + \sqrt {n – 1} }} = \lim \frac{2}{{\sqrt {1 + \frac{1}{n}} + \sqrt {1 – \frac{1}{n}} }} = 1$

Giải nhanh : $\sqrt n \left( {\sqrt {n + 1} – \sqrt {n – 1} } \right) = \frac{{2\sqrt n }}{{\sqrt {n + 1} + \sqrt {n – 1} }} \sim \frac{{2\sqrt n }}{{\sqrt n + \sqrt n }} = 1.$

Chọn B

$n\left( {\sqrt {{n^2} + 1} – \sqrt {{n^2} – 3} } \right) \sim n\left( {\sqrt {{n^2}} – \sqrt {{n^2}} } \right) = 0\xrightarrow[{}]{}$nhân lượng liên hợp :

$\lim n\left( {\sqrt {{n^2} + 1} – \sqrt {{n^2} – 3} } \right) = \lim \frac{{4n}}{{\sqrt {{n^2} + 1} + \sqrt {{n^2} – 3} }}$

$ = \lim \frac{4}{{\sqrt {1 + \frac{1}{{{n^2}}}} + \sqrt {1 – \frac{3}{{{n^2}}}} }} = 2$

Giải nhanh : $n\left( {\sqrt {{n^2} + 1} – \sqrt {{n^2} – 3} } \right) = \frac{{4n}}{{\sqrt {{n^2} + 1} + \sqrt {{n^2} – 3} }} \sim \frac{{4n}}{{\sqrt {{n^2}} + \sqrt {{n^2}} }} = 2.$

Chọn C

$n\left( {\sqrt {{n^2} + n + 1} – \sqrt {{n^2} + n – 6} } \right) \sim n\left( {\sqrt {{n^2}} – \sqrt {{n^2}} } \right) = 0\xrightarrow[{}]{}$nhân lượng liên hợp :

$\lim n\left( {\sqrt {{n^2} + n + 1} – \sqrt {{n^2} + n – 6} } \right) = \lim \frac{{7n}}{{\sqrt {{n^2} + n + 1} + \sqrt {{n^2} + n – 6} }}$

$ = \lim \frac{7}{{\sqrt {1 + \frac{1}{n} + \frac{1}{{{n^2}}}} + \sqrt {1 + \frac{1}{n} – \frac{6}{{{n^2}}}} }} = \frac{7}{2}$

Giải nhanh : $n\left( {\sqrt {{n^2} + n + 1} – \sqrt {{n^2} + n – 6} } \right)$

$ = \frac{{7n}}{{\sqrt {{n^2} + n + 1} + \sqrt {{n^2} + n – 6} }} \sim \frac{{7n}}{{\sqrt {{n^2}} + \sqrt {{n^2}} }} = \frac{7}{2}$

Chọn C

$\sqrt {{n^2} + 2} – \sqrt {{n^2} + 4} \sim \sqrt {{n^2}} – \sqrt {{n^2}} = 0\xrightarrow[{}]{}$nhân lượng liên hợp :

$\lim \frac{1}{{\sqrt {{n^2} + 2} – \sqrt {{n^2} + 4} }} = \lim – \frac{1}{2}\left( {\sqrt {{n^2} + 2} + \sqrt {{n^2} + 4} } \right)$

$ = \lim n.\left[ { – \frac{1}{2}\left( {\sqrt {1 + \frac{2}{{{n^2}}}} + \sqrt {1 + \frac{4}{{{n^2}}}} } \right)} \right] = – \infty $

vì $\lim n = + \infty ,\,\,\lim \left[ { – \frac{1}{2}\left( {\sqrt {1 + \frac{2}{{{n^2}}}} + \sqrt {1 + \frac{4}{{{n^2}}}} } \right)} \right] = – 1 < 0$

Giải nhanh :

$\frac{1}{{\sqrt {{n^2} + 2} – \sqrt {{n^2} + 4} }} = – \frac{1}{2}\left( {\sqrt {{n^2} + 2} + \sqrt {{n^2} + 4} } \right)$

$ \sim – \frac{1}{2}\left( {\sqrt {{n^2}} + \sqrt {{n^2}} } \right) = – n\xrightarrow[{}]{} – \infty $

Chọn A

$\sqrt {9{n^2} – n} – \sqrt {n + 2} \sim \sqrt {9{n^2}} = 3n\not = 0\xrightarrow[{}]{}$giải nhanh :

$\frac{{\sqrt {9{n^2} – n} – \sqrt {n + 2} }}{{3n – 2}} \sim \frac{{\sqrt {9{n^2}} }}{{3n}} = 1$

Cụ thể : $\lim \frac{{\sqrt {9{n^2} – n} – \sqrt {n + 2} }}{{3n – 2}} = $

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

Chọn B

$\sqrt[3]{{{n^3} + 1}} – n \sim \sqrt[3]{{{n^3}}} – n = 0\xrightarrow[{}]{}$nhân lượng liên hợp :

$\lim \left( {\sqrt[3]{{{n^3} + 1}} – n} \right) = \lim \frac{1}{{\sqrt[3]{{{{\left( {{n^3} + 1} \right)}^2}}} + n\sqrt[3]{{{n^3} + 1}} + {n^2}}} = 0$

Chọn A

Cụ thể : $\lim \frac{{2 – {5^{n + 2}}}}{{{3^n} + {{2.5}^n}}} = \lim \frac{{2{{\left( {\frac{1}{5}} \right)}^n} – 25}}{{{{\left( {\frac{3}{5}} \right)}^n} + 2}} = – \frac{{25}}{2}.$

Giải nhanh : $\frac{{2 – {5^{n + 2}}}}{{{3^n} + {{2.5}^n}}} \sim \frac{{ – {5^{n + 2}}}}{{{{2.5}^n}}} = – \frac{{25}}{2}$

Chọn B

Giải nhanh : $\frac{{{3^n} – 1}}{{{2^n} – {{2.3}^n} + 1}} \sim \frac{{{3^n}}}{{ – {{2.3}^n}}} = – \frac{1}{2}$

Cụ thể : $\lim \frac{{{3^n} – 1}}{{{2^n} – {{2.3}^n} + 1}} = \lim \frac{{1 – {{\left( {\frac{1}{3}} \right)}^n}}}{{{{\left( {\frac{2}{3}} \right)}^n} – 2 + {{\left( {\frac{1}{3}} \right)}^n}}} = – \frac{1}{2}.$

Chọn B

$\lim \left( {\frac{{{{\left( {\sqrt 5 } \right)}^n} – {2^{n + 1}} + 1}}{{{{5.2}^n} + {{\left( {\sqrt 5 } \right)}^{n + 1}} – 3}} + + \frac{{2{n^2} + 3}}{{{n^2} – 1}}} \right)$

$ = \lim \left( {\frac{{1 – 2.{{\left( {\frac{2}{{\sqrt 5 }}} \right)}^n} + {{\left( {\frac{1}{{\sqrt 5 }}} \right)}^n}}}{{5.{{\left( {\frac{2}{{\sqrt 5 }}} \right)}^n} + \sqrt 5 – .{{\left( {\frac{1}{{\sqrt 5 }}} \right)}^n}}} + \frac{{2 + \frac{3}{{{n^2}}}}}{{1 – \frac{1}{{{n^2}}}}}} \right)$

$ = \frac{1}{{\sqrt 5 }} + 2 = \frac{{\sqrt 5 }}{5} + 2.$

Giải nhanh :

$\frac{{{{\left( {\sqrt 5 } \right)}^n} – {2^{n + 1}} + 1}}{{{{5.2}^n} + {{\left( {\sqrt 5 } \right)}^{n + 1}} – 3}} + \frac{{2{n^2} + 3}}{{{n^2} – 1}} \sim \frac{{{{\left( {\sqrt 5 } \right)}^n}}}{{{{\left( {\sqrt 5 } \right)}^{n + 1}}}} + \frac{{2{n^2}}}{{{n^2}}}$

$ = \frac{1}{{\sqrt 5 }} + 2 = \frac{{\sqrt 5 }}{5} + 2\xrightarrow[{}]{}\left\{ \begin{gathered}
a = 1 \hfill \\
b = 5 \hfill \\
c = 2 \hfill \\
\end{gathered} \right.$

Vậy $S = {1^2} + {5^2} + {2^2} = 30.$

Chọn D

Giải nhanh: $\frac{{{\pi ^n} + {3^n} + {2^{2n}}}}{{3{\pi ^n} – {3^n} + {2^{2n + 2}}}} = \frac{{{\pi ^n} + {3^n} + {4^n}}}{{3{\pi ^n} – {3^n} + {{4.4}^n}}} \sim \frac{{{4^n}}}{{{{4.4}^n}}} = \frac{1}{4}$

Cụ thể : $\lim \frac{{{\pi ^n} + {3^n} + {2^{2n}}}}{{3{\pi ^n} – {3^n} + {2^{2n + 2}}}} = \lim \frac{{{{\left( {\frac{\pi }{4}} \right)}^n} + {{\left( {\frac{3}{4}} \right)}^n} + 1}}{{3.{{\left( {\frac{\pi }{4}} \right)}^n} – 3.{{\left( {\frac{3}{4}} \right)}^n} + 4}} = \frac{1}{4}.$

Chọn D

Giải nhanh : Vì $3 > \sqrt 5 $ nên ${3^n} – {\sqrt 5 ^n} \sim {3^n}\xrightarrow[{}]{} + \infty .$

Cụ thể : $\lim \left[ {{3^n} – {{\sqrt 5 }^n}} \right] = \lim {3^n}\left( {1 – {{\left( {\frac{{\sqrt 5 }}{3}} \right)}^n}} \right) = + \infty $ vì $\left\{ \begin{gathered}
\lim {3^n} = + \infty \hfill \\
\lim 1 – {\left( {\frac{{\sqrt 5 }}{3}} \right)^n} = 1 > 0 \hfill \\
\end{gathered} \right..$

Chọn C

Giải nhanh : ${3^4}{.2^{n + 1}} – {5.3^n} \sim – {5.3^n} = – \infty \,\,\left( { – 5 < 0} \right).$

Cụ thể : $\lim \left( {{3^4}{{.2}^{n + 1}} – {{5.3}^n}} \right) = \lim {3^n}\left( {162.{{\left( {\frac{2}{3}} \right)}^n} – 5} \right) = – \infty $ vì $\left\{ \begin{gathered}
\lim {3^n} = + \infty \hfill \\
\lim \left( {162.{{\left( {\frac{2}{3}} \right)}^n} – 5} \right) = – 5 < 0 \hfill \\
\end{gathered} \right..$

Chọn A

Giải nhanh : $\frac{{{3^n} – {{4.2}^{n + 1}} – 3}}{{3.2n + {4^n}}} \sim \frac{{{3^n}}}{{{4^n}}} = {\left( {\frac{3}{4}} \right)^n}\xrightarrow[{}]{}0.$

Cụ thể : $0 \leqslant \left| {\frac{{{3^n} – {{4.2}^{n + 1}} – 3}}{{3.2n + {4^n}}}} \right| \leqslant \frac{{{{8.3}^{n + 1}}}}{{{4^n}}} = 24.{\left( {\frac{3}{4}} \right)^n}$

$ \to 0\xrightarrow[{}]{}\lim \frac{{{3^n} – {{4.2}^{n + 1}} – 3}}{{3.2n + {4^n}}} = 0$

Chọn A

Ta có ${2^n} = \sum\limits_{k = 0}^n {C_n^k} \Rightarrow {2^n} \geqslant C_n^3 = \frac{{n\left( {n – 1} \right)\left( {n – 2} \right)}}{6} \sim \frac{{{n^3}}}{6}$

$ \Rightarrow \left\{ \begin{gathered}
\frac{n}{{{2^n}}} \to 0 \hfill \\
\frac{{{2^n}}}{{{n^2}}} \to + \infty \hfill \\
\end{gathered} \right.$

Khi đó:

$\lim \frac{{{2^{n + 1}} + 3n + 10}}{{3{n^2} – n + 2}} = \lim \frac{{{2^n}}}{{{n^2}}}.\frac{{2 + 3.\frac{n}{{{2^n}}} + 10.{{\left( {\frac{1}{2}} \right)}^n}}}{{3 – \frac{1}{n} + \frac{2}{{{n^2}}}}} = + \infty $ vì $\left\{ \begin{gathered}
\lim \frac{{{2^n}}}{{{n^2}}} = + \infty \hfill \\
\lim \frac{{2 + 3.\frac{n}{{{2^n}}} + 10.{{\left( {\frac{1}{2}} \right)}^n}}}{{3 – \frac{1}{n} + \frac{2}{{{n^2}}}}} = \frac{2}{3} > 0 \hfill \\
\end{gathered} \right..$

Chọn B

$\lim \sqrt[4]{{\frac{{{4^n} + {2^{n + 1}}}}{{{3^n} + {4^{n + a}}}}}} = \lim \sqrt[4]{{\frac{{1 + 2.{{\left( {\frac{1}{2}} \right)}^n}}}{{{{\left( {\frac{3}{4}} \right)}^n} + {4^a}}}}} = \sqrt {\frac{1}{{{4^a}}}} = \sqrt {\frac{1}{{{{\left( {{2^a}} \right)}^2}}}} = \frac{1}{{{2^a}}}.$

Giải nhanh: $\sqrt[4]{{\frac{{{4^n} + {2^{n + 1}}}}{{{3^n} + {4^{n + 2}}}}}} \sim \sqrt[4]{{\frac{{{4^n}}}{{{4^{n + a}}}}}} = \frac{1}{{{2^a}}} \leqslant \frac{1}{{1024}}$

$ \Leftrightarrow {2^a} \geqslant 1024 = {2^{10}} \Leftrightarrow a \geqslant 10$

Mà $a \in \left( {0;2018} \right)$ và $a \in \mathbb{Z}$ nên $a \in \left\{ {10;2017} \right\}\xrightarrow[{}]{}$có 2008 giá trị $a.$

Chọn C

Ta có $\lim \left( {\frac{{\sqrt {{n^2} + 2n} }}{{3n – 1}} + \frac{{{{\left( { – 1} \right)}^n}}}{{{3^n}}}} \right) = \lim \frac{{\sqrt {{n^2} + 2n} }}{{3n – 1}} + \lim \frac{{{{\left( { – 1} \right)}^n}}}{{{3^n}}}.$
Ta có

$\left\{ \begin{gathered}
\lim \frac{{\sqrt {{n^2} + 2n} }}{{3n – 1}} = \lim \frac{{\sqrt {1 + \frac{2}{n}} }}{{3 – \frac{1}{n}}} = \frac{1}{3} \hfill \\
0 \leqslant \left| {\frac{{{{\left( { – 1} \right)}^n}}}{{{3^n}}}} \right| \leqslant {\left( {\frac{1}{3}} \right)^n} \to 0 \Rightarrow \lim \frac{{{{\left( { – 1} \right)}^n}}}{{{3^n}}} = 0 \hfill \\
\end{gathered} \right.$

$ \Rightarrow \lim \left( {\frac{{\sqrt {{n^2} + 2n} }}{{3n – 1}} + \frac{{{{\left( { – 1} \right)}^n}}}{{{3^n}}}} \right) = \frac{1}{3}$

Chọn B

$\lim \left( {\frac{{\sqrt {3n} + {{\left( { – 1} \right)}^n}\cos 3n}}{{\sqrt n – 1}}} \right) = \lim \left( {\frac{{\sqrt {3n} }}{{\sqrt n – 1}} + \frac{{{{\left( { – 1} \right)}^n}\cos 3n}}{{\sqrt n }}} \right).$
Ta có :

$\left\{ \begin{gathered}
\lim \frac{{\sqrt {3n} }}{{\sqrt n – 1}} = \frac{{\sqrt 3 }}{1} = \sqrt 3 \hfill \\
0 \leqslant \left| {\frac{{{{\left( { – 1} \right)}^n}\cos 3n}}{{\sqrt n – 1}}} \right| \leqslant \frac{1}{{\sqrt n – 1}} \to 0 \Rightarrow \lim \frac{{{{\left( { – 1} \right)}^n}\cos 3n}}{{\sqrt n – 1}} = 0 \hfill \\
\end{gathered} \right.$

$ \Rightarrow \lim \left( {\frac{{\sqrt {3n} + {{\left( { – 1} \right)}^n}\cos 3n}}{{\sqrt n – 1}}} \right) = \sqrt 3 $

Chọn D

Ta có $\lim \sqrt {{{2.3}^n} – n + 2} = \lim \sqrt {{3^n}} .\sqrt {2 – \frac{n}{{{3^n}}} + 2.{{\left( {\frac{1}{3}} \right)}^n}} .$ Vì

$\left. \begin{gathered}
\lim \sqrt {{3^n}} = + \infty \hfill \\
0 \leqslant \frac{n}{{{3^n}}} \leqslant \frac{n}{{C_n^2}} = \frac{n}{{\frac{{n\left( {n – 1} \right)}}{2}}} = \frac{2}{{n – 1}} \to 0 \Rightarrow \lim \frac{n}{{{3^n}}} = 0 \hfill \\
\lim {\left( {\frac{1}{3}} \right)^n} = 0 \hfill \\
\end{gathered} \right\}$

$\xrightarrow[{}]{}\left\{ \begin{gathered}
\lim \sqrt {{3^n}} = + \infty \hfill \\
\lim \sqrt {2 – \frac{n}{{{3^n}}} + 2.{{\left( {\frac{1}{3}} \right)}^n}} = \sqrt 2 > 0 \hfill \\
\end{gathered} \right.$

Do đó $\lim \sqrt {{{2.3}^n} – n + 2} = + \infty $

Chọn A

Gọi $q$ là công bội của cấp số nhân, ta có :

$\left\{ \begin{gathered}
\frac{{{u_1}}}{{1 – q}} = 2 \hfill \\
{S_3} = {u_1}.\frac{{1 – {q^3}}}{{1 – q}} = \frac{9}{4} \hfill \\
\end{gathered} \right. \Leftrightarrow \left\{ \begin{gathered}
{u_1} = 2\left( {1 – q} \right) \hfill \\
2\left( {1 – {q^3}} \right) = \frac{9}{4} \hfill \\
\end{gathered} \right.$

$ \Leftrightarrow \left\{ \begin{gathered}
q = – \frac{1}{2} \hfill \\
{u_1} = 2\left( {1 + \frac{1}{2}} \right) = 3 \hfill \\
\end{gathered} \right.$

Chọn A

Ta có

$S = 9 + 3 + 1 + \frac{1}{3} + \frac{1}{9} + \cdots + \frac{1}{{{3^{n – 3}}}} + \cdots $

$ = 9\left( {\underbrace {1 + \frac{1}{3} + \frac{1}{{{3^2}}} + \frac{1}{{{3^4}}} + \cdots + \frac{1}{{{3^{n – 1}}}} + \cdots }_{CSN\,lvh:\,\,{u_1} = 1,\,q = \frac{1}{3}}} \right)$

$ = 9\left( {\frac{1}{{1 – \frac{1}{3}}}} \right) = \frac{{27}}{2}$

Chọn C

Ta có

$S = \sqrt 2 \left( {\underbrace {1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots + \frac{1}{{{2^n}}} + \cdots }_{CSN\,lvh:\,\,{u_1} = 1,\,\,q = \frac{1}{2}}} \right) = \sqrt 2 \left( {\frac{1}{{1 – \frac{1}{2}}}} \right) = 2\sqrt 2 .$

Chọn A

Ta có

$S = 1 + \frac{2}{3} + \frac{4}{9} + \cdots + \frac{{{2^n}}}{{{3^n}}} + \cdots $

$ = \underbrace {1 + \frac{2}{3} + {{\left( {\frac{2}{3}} \right)}^2} + \cdots + {{\left( {\frac{2}{3}} \right)}^n} + \cdots }_{CSN\,\,lvh:\,\,\,{u_1} = 1,\,\,q = \frac{2}{3}} = \frac{1}{{1 – \frac{2}{3}}} = 3$

Chon D

. Ta có :

$S = \frac{1}{2} – \frac{1}{6} + \frac{1}{{18}} + \cdots + \frac{{{{\left( { – 1} \right)}^{n + 1}}}}{{{{2.3}^{n – 1}}}} + \cdots $

$ = \frac{1}{2}\left( {\underbrace {1 – \frac{1}{3} + \frac{1}{{{3^2}}} + \cdots + \frac{{{{\left( { – 1} \right)}^{n + 1}}}}{{{3^{n – 1}}}}}_{CSN\,\,lvh:\,{u_1} = 1,\,\,q = – \frac{1}{3}}} \right) = \frac{1}{2}\left( {\frac{1}{{1 + \frac{1}{3}}}} \right) = \frac{3}{8}$

Chọn D

Ta có

$S = \left( {\frac{1}{2} – \frac{1}{3}} \right) + \left( {\frac{1}{4} – \frac{1}{9}} \right) + … + \left( {\frac{1}{{{2^n}}} – \frac{1}{{{3^n}}}} \right) + …$

$ = \left( {\underbrace {\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{{{2^n}}} + \cdots }_{CSN\,lvh:\,\,{u_1} = q = \frac{1}{2}}} \right) – \left( {\underbrace {\frac{1}{3} + \frac{1}{9} + \cdots + \frac{1}{{{3^n}}} + \cdots }_{CSN\,\,lvh:\,{u_1} = q = \frac{1}{3}}} \right)$

$ = \frac{{\frac{1}{2}}}{{1 – \frac{1}{2}}} – \frac{{\frac{1}{3}}}{{1 – \frac{1}{3}}} = 1 – \frac{1}{2} = \frac{1}{2}$

Chọn B

Ta có $1 + a + {a^2} + … + {a^n}$ là tổng $n + 1$ số hạng đầu tiên của cấp số nhân với số hạng đầu là $1$ và công bội là $a$, nên $1 + a + {a^2} + … + {a^n} = \frac{{1.\left( {1 – {a^{n + 1}}} \right)}}{{1 – a}} = \frac{{1 – {a^{n + 1}}}}{{1 – a}}.$

Tương tự: $1 + b + {b^2} + … + {b^n} = \frac{{1\left( {1 – {b^{n + 1}}} \right)}}{{1 – b}} = \frac{{1 – {b^{n + 1}}}}{{1 – b}}.$

Do đó $\lim \frac{{1 + a + {a^2} + … + {a^n}}}{{1 + b + {b^2} + … + {b^n}}}\, = \lim \frac{{\frac{{1 – {a^{n + 1}}}}{{1 – a}}}}{{\frac{{1 – {b^{n + 1}}}}{{1 – b}}}}$

$ = \lim \frac{{1 – b}}{{1 – a}}.\frac{{1 – {a^{n + 1}}}}{{1 – {b^{n + 1}}}} = \frac{{1 – b}}{{1 – a}}\,\,\,\left( {\left| a \right| < 1,\left| b \right| < 1} \right)$

Chọn C

Ta có

$S = \underbrace {1 + {{\cos }^2}x + {{\cos }^4}x + {{\cos }^6}x + \cdots + {{\cos }^{2n}}x + \cdots }_{CSN\,\,lvh:\,\,{u_1} = 1,\,\,q = {{\cos }^2}x}$

$ = \frac{1}{{1 – {{\cos }^2}x}} = \frac{1}{{{{\sin }^2}x}}.$

Chọn C

Ta có

$S = \underbrace {1 – {{\sin }^2}x + {{\sin }^4}x – {{\sin }^6}x + \cdots + {{\left( { – 1} \right)}^n}.{{\sin }^{2n}}x + \cdots }_{CSN\,\,lvh:\,\,{u_1} = 1,\,\,q = – {{\sin }^2}x}$

$ = \frac{1}{{1 + {{\sin }^2}x}}$

Chọn B

Ta có $\tan \alpha \in \left( {0;1} \right)$ với mọi $\alpha \in \left( {0;\frac{\pi }{4}} \right),$ do đó

$S = \underbrace {1 – \tan \alpha + {{\tan }^2}\alpha – {{\tan }^3}\alpha + \ldots }_{CSN\,\,lvh:\,\,{u_1} = 1,\,q = – \tan \alpha }$

$ = \frac{1}{{1 + \tan \alpha }} = \frac{{\cos \alpha }}{{\sin \alpha + \cos \alpha }} = \frac{{\cos \alpha }}{{\sqrt 2 \sin \left( {\alpha + \frac{\pi }{4}} \right)}}$

Chọn A

Ta có $\left\{ \begin{gathered}
M = \frac{1}{{1 – m}} \hfill \\
N = \frac{1}{{1 – n}} \hfill \\
\end{gathered} \right. \Rightarrow \left\{ \begin{gathered}
m = 1 – \frac{1}{M} \hfill \\
n = 1 – \frac{1}{N} \hfill \\
\end{gathered} \right.,$ khi đó

$A = \frac{1}{{1 – mn}} = \frac{1}{{1 – \left( {1 – \frac{1}{M}} \right)\left( {1 – \frac{1}{N}} \right)}} = \frac{{MN}}{{M + N – 1}}.$

Chọn B

Ta có $0,5111 \cdots = 0,5 + {10^{ – 2}} + {10^{ – 3}} + \cdots + {10^{ – n}} + \cdots $

Dãy số ${10^{ – 2}};{10^{ – 3}};…;{10^{ – n}};…$là một cấp số nhân lùi vô hạn có số hạng đầu bằng ${u_1} = {10^{ – 2}},$ công bội bằng $q = {10^{ – 1}}$ nên $S = \frac{{{u_1}}}{{1 – q}} = \frac{{{{10}^{ – 2}}}}{{1 – {{10}^{ – 1}}}} = \frac{1}{{90}}.$

Vậy $0,5111… = 0,5 + S = \frac{{46}}{{90}} = \frac{{23}}{{45}}\xrightarrow{{}}\left\{ \begin{gathered}
a = 23 \hfill \\
b = 45 \hfill \\
\end{gathered} \right.\xrightarrow{{}}T = a + b = 68.$

Chọn B

Ta có

$A = 0,353535… = 0,35 + 0,0035 + … = \frac{{35}}{{{{10}^2}}} + \frac{{35}}{{{{10}^4}}} + …$

$ = \frac{{\frac{{35}}{{{{10}^2}}}}}{{1 – \frac{1}{{{{10}^2}}}}} = \frac{{35}}{{99}} \Rightarrow \left\{ \begin{gathered}
a = 35 \hfill \\
b = 99 \hfill \\
\end{gathered} \right. \Rightarrow T = 3465$

Chọn A

Ta có

$B = 5,231231… = 5 + 0,231 + 0,000231 + …$

$ = 5 + \frac{{231}}{{{{10}^3}}} + \frac{{231}}{{{{10}^6}}} + … = 5 + \frac{{\frac{{231}}{{{{10}^3}}}}}{{1 – \frac{1}{{{{10}^3}}}}}$

$ = 5 + \frac{{231}}{{999}} = \frac{{1742}}{{333}}\xrightarrow[{}]{}\left\{ \begin{gathered}
a = 1742 \hfill \\
b = 333 \hfill \\
\end{gathered} \right. \Rightarrow T = 1409$

Chọn D

Ta có $0,17232323 \ldots = 0,17 + 23\left( {\frac{1}{{{{10}^4}}} + \frac{1}{{{{10}^6}}} + \frac{1}{{{{10}^8}}} \cdots } \right)$

$ = \frac{{17}}{{100}} + 23.\frac{{\frac{1}{{10000}}}}{{1 – \frac{1}{{100}}}} = \frac{{17}}{{100}} + \frac{{23}}{{100.99}} = \frac{{1706}}{{9900}} = \frac{{853}}{{4950}}$

$\xrightarrow[{}]{}\left\{ \begin{gathered}
a = 853 \hfill \\
b = 4950 \hfill \\
\end{gathered} \right. \Rightarrow {2^{12}} < T = 4097 < {2^{13}}.$

Chọn B

Ta có: $1 + 3 + 5 + … + \left( {2n + 1} \right) = {\left( {n + 1} \right)^2}.$

Vậy: $\lim \frac{{1 + 3 + 5 + … + \left( {2n + 1} \right)}}{{3{n^2} + 4}} = \lim \frac{{{{\left( {n + 1} \right)}^2}}}{{3{n^2} + 4}}$

$ = \lim \frac{{{n^2} + 2n + 1}}{{3{n^2} + 4}} = \lim \frac{{1 + \frac{2}{n} + \frac{1}{{{n^2}}}}}{{3 + \frac{4}{{{n^2}}}}} = \frac{1}{3}.$

Chọn B

Ta có: $\lim \left[ {\frac{1}{{1.2}} + \frac{1}{{2.3}} + … + \frac{1}{{n\left( {n + 1} \right)}}} \right]$

$ = \lim \left( {1 – \frac{1}{2} + \frac{1}{2} – \frac{1}{3} + … + \frac{1}{n} – \frac{1}{{n + 1}}} \right)$

$ = \lim \left( {1 – \frac{1}{{n + 1}}} \right) = 1.$

Chọn c

Ta có: $\lim \left[ {\frac{1}{{1.3}} + \frac{1}{{3.5}} + … + \frac{1}{{n\left( {2n – 1} \right)\left( {2n + 1} \right)}}} \right]$

$ = \frac{1}{2}\lim \left( {1 – \frac{1}{3} + \frac{1}{3} – \frac{1}{5} + … + \frac{1}{{2n – 1}} – \frac{1}{{2n + 1}}} \right)$

$ = \frac{1}{2}\lim \left( {1 – \frac{1}{{2n + 1}}} \right) = \frac{1}{2}.$

Chọn A

Ta có: $\frac{1}{{1.3}} + \frac{1}{{2.4}} + … + \frac{1}{{n\left( {n + 2} \right)}}$

$ = \frac{1}{2}\left( {1 – \frac{1}{3} + \frac{1}{2} – \frac{1}{4} + \frac{1}{3} – \frac{1}{5} + … + \frac{1}{{n – 1}} – \frac{1}{{n + 1}} + \frac{1}{n} – \frac{1}{{n + 2}}} \right)$

$ = \frac{1}{2}\left( {1 + \frac{1}{2} – \frac{1}{{n + 1}} – \frac{1}{{n + 2}}} \right)$

Vậy $\lim \left[ {\frac{1}{{1.3}} + \frac{1}{{2.4}} + … + \frac{1}{{n\left( {n + 2} \right)}}} \right] = \frac{3}{4}.$

Chọn A

Ta có: $\frac{1}{{1.4}} + \frac{1}{{2.5}} + … + \frac{1}{{n\left( {n + 3} \right)}}$

$ = \frac{1}{3}\left( {\frac{1}{1} – \frac{1}{4} + \frac{1}{2} – \frac{1}{5} + \frac{1}{3} – \frac{1}{6} + \frac{1}{4} – \frac{1}{7} + … + \frac{1}{{n – 3}} – \frac{1}{n} + \frac{1}{{n – 2}} – \frac{1}{{n + 1}} + \frac{1}{{n – 1}} – \frac{1}{{n + 2}} + \frac{1}{n} – \frac{1}{{n + 3}}} \right)$

$ = \frac{1}{3}\left( {1 + \frac{1}{2} + \frac{1}{3} – \frac{1}{{n + 1}} – \frac{1}{{n + 2}} – \frac{1}{{n + 3}}} \right)$

Vậy: $\lim \left[ {\frac{1}{{1.4}} + \frac{1}{{2.5}} + … + \frac{1}{{n\left( {n + 3} \right)}}} \right] = \frac{{11}}{{18}}.$

Chọn B

Dãy số 1, 2, 3, …, n là cấp số cộng có số hạng đầu là ${u_1} = 1$ số hạng cuối cùng ${u_n} = n$, công sai $d = 1$.

Khi đó ${S_n} = 1 + 2 + 3 + … + n = \frac{{n\left( {{u_1} + n} \right)}}{2} = \frac{{n\left( {n + 1} \right)}}{2}.$

Viết lại: ${u_n} = \frac{{n\left( {n + 1} \right)}}{{2\left( {{n^2} + 1} \right)}}$

$\lim {u_n} = \lim \frac{{n\left( {n + 1} \right)}}{{2\left( {{n^2} + 1} \right)}} = \lim \frac{{{n^2}\left( {1 + \frac{1}{n}} \right)}}{{{n^2}\left( {2 + \frac{2}{{{n^2}}}} \right)}} = \lim \frac{1}{2}.$

Chọn B

Ta có: ${U_1} = \frac{1}{2};\,\,{U_2} = \frac{5}{8};\,\,{U_3} = \frac{{57}}{{64}};…$

Ta chứng minh: ${U_n} < 1\,\,\forall n \in {\mathbb{N}^*}$ (bằng phương pháp quy nạp). Vậy dãy bị chặn trên.

Ta chứng minh $\left( {{U_n}} \right)$ là dãy tăng. Thật vậy:

Ta có: ${U_{n + 1}} > {U_n} \Leftrightarrow \frac{1}{2} + \frac{{U_n^2}}{2} > {U_n}$

$ \Leftrightarrow U_n^2 – 2{U_n} + 1 > 0 \Leftrightarrow {\left( {{U_n} – 1} \right)^2} > 0$ luôn đúng $\forall n \in {\mathbb{N}^*}$, vì ${U_n} < 1$.

Vậy dãy có giới hạn. Đặt $a = \lim {U_n} = \lim {U_{n + 1}}$.

Ta có: $\lim {U_{n + 1}} = \lim \left( {\frac{1}{2} + \frac{{U_n^2}}{2}} \right) \Rightarrow a = \frac{1}{2} + \frac{{{a^2}}}{2} \Rightarrow 2a = 1 + {a^2}$

$ \Rightarrow {a^2} – 2a + 1 = 0 \Rightarrow a = 1$.

Chọn B

Ta có: ${U_{n + 1}} = \frac{1}{{{U_n}}} + \frac{1}{2}{U_n} \geqslant \sqrt 2 $ (theo bất đẳng thức Cô-si với ${U_n} > 0$). Vậy $\left( {{U_n}} \right)$ là dãy bị chặn dưới.

Dấu “=” không xảy ra, nên ${U_n} > \sqrt 2 ,\,\,\forall n \in {\mathbb{N}^*}.$

Lại có: $\frac{{{U_{n + 1}}}}{{{U_n}}} = \frac{{2 + U_n^2}}{{2U_n^2}} = \frac{1}{{U_n^2}} + \frac{1}{2}$. Vì ${U_n} > \sqrt 2 \Rightarrow U_n^2 > 2$

$ \Rightarrow \frac{1}{{U_n^2}} < \frac{1}{2} \Rightarrow \frac{1}{{U_n^2}} + \frac{1}{2} < \frac{1}{2} + \frac{1}{2} = 1$

$ \Rightarrow {U_{n + 1}} < {U_n},\,\,\forall n \in {\mathbb{N}^*}.$

Vậy dãy giảm, khi đó ${U_n}$ có giới hạn. Đặt $\lim {U_{n + 1}} = \lim {U_n} = a$ $\left( {a > 0} \right)$.

Ta có: $\lim {U_{n + 1}} = \lim \frac{{2 + U_n^2}}{{2{U_n}}} \Rightarrow a = \frac{{2 + {a^2}}}{{2a}} \Rightarrow 2{a^2} = 2 + {a^2}$

$ \Rightarrow {a^2} = 2 \Rightarrow a = \sqrt 2 $ (vì $a > 0$).

Chọn A

Ta có: ${U_1} = \sqrt 2 ;\,\,{U_2} = \sqrt {2\sqrt 2 } $;…

Ta sẽ chứng minh ${U_n} < 2$; $\forall n \in {\mathbb{N}^*}$ (bằng phương pháp quy nạp).

$n = 1,\,\,{U_1} = \sqrt 2 < 2$. Giả sử ${U_k} < 2,\,\,\forall k \geqslant 1$.

Ta có: ${U_{k + 1}} = \sqrt {2{U_k}} < \sqrt {2.2} = \sqrt 4 = 2.$

Vậy ${U_n} < 2,\,\,\forall n \in \mathbb{N}$. Lại có: ${U_n} > 0,\,\,\forall n \in {\mathbb{N}^*}.$

Lại có: $\frac{{{U_{n + 1}}}}{{{U_n}}} = \frac{{\sqrt {2{U_n}} }}{{{U_n}}} = \sqrt {\frac{2}{{{U_n}}}} > \sqrt {\frac{2}{2}} = 1 \Rightarrow $ dãy tăng.

Vậy dãy đã cho có giới hạn. Đặt $\lim {U_{n + 1}} = \lim {U_n} = a\,\,\left( {a > 0} \right)$

Ta có: $\lim {U_{n + 1}} = \lim \sqrt {2{U_n}} \Rightarrow a = \sqrt {2a} \Rightarrow {a^2} = 2a \Rightarrow a = 2.$