Đề kiểm tra Bài tập cuối chương V lớp 11 (có lời giải) - Đề 2

Chọn A

Ta có $\lim \frac{{2{n^3} + {n^2} - 4}}{{a{n^3} + 2}} = \lim \frac{{{n^3}\left( {2 + \frac{1}{n} - \frac{4}{{{n^3}}}} \right)}}{{{n^3}\left( {a + \frac{2}{{{n^3}}}} \right)}} = \frac{2}{a} = \frac{1}{2}$.

Suy ra $a = 4$. Khi đó $a - {a^2} = 4 - {4^2} = - 12$.

Chọn C

Ta có $1 + 2 + 3 + ... + k$ là tổng của cấp số cộng có ${u_1} = 1$, $d = 1$ nên $1 + 2 + 3 + ... + k = \frac{{\left( {1 + k} \right)k}}{2}$

$\Rightarrow \frac{1}{{1 + 2 + ... + k}} = \frac{2}{{k\left( {k + 1} \right)}}$$= \frac{2}{k} - \frac{2}{{k + 1}}$, $\forall k \in {\mathbb{N}^*}$.

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

Chọn B

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

Chọn C

Ta có

$\lim \frac{{3n - 1}}{{3n + 1}} = \lim \frac{{3 - \frac{1}{n}}}{{3 + \frac{1}{n}}} = \frac{3}{3} = 1\,\,$vì $\lim \frac{1}{n} = 0$; $\lim \frac{{2n + 1}}{{2n - 1}} = \lim \frac{{2 + \frac{1}{n}}}{{2 - \frac{1}{n}}} = \frac{2}{2} = 1$ vì $\lim \frac{1}{n} = 0$

$\lim \frac{{4n + 1}}{{3n - 1}} = \lim \frac{{4 + \frac{1}{n}}}{{3 - \frac{1}{n}}} = \frac{4}{3}$ vì $\lim \frac{1}{n} = 0$; $\lim \frac{{n + 1}}{{n - 1}} = \lim \frac{{1 + \frac{1}{n}}}{{1 - \frac{1}{n}}} = 1$ vì $\lim \frac{1}{n} = 0$.

Chọn D

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

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

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

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