Đề kiểm tra Bài tập cuối chương V (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 \[\lim n\left( {\sqrt {4{n^2} + 3}  - \sqrt[3]{{8{n^3} + n}}} \right) = \frac{3}{4} - \frac{1}{{12}}\]\[ = \frac{2}{3}\].

Chọn C

\(\mathop {\lim }\limits_{x \to 2} \frac{{x - 2}}{{{x^2} - 4}} = \mathop {\lim }\limits_{x \to 2} \frac{{x - 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \mathop {\lim }\limits_{x \to 2} \frac{1}{{x + 2}} = \frac{1}{4}\).

Chọn B

Ta có \(L = \mathop {\lim }\limits_{x \to 3} \frac{{x - 3}}{{x + 3}}\)\( = \frac{{3 - 3}}{{3 + 3}} = 0\).

Chọn D

\(\mathop {\lim }\limits_{x \to  - \infty } \frac{{4x + 1}}{{ - x + 1}}\)\( = \mathop {\lim }\limits_{x \to  - \infty } \frac{{4 + \frac{1}{x}}}{{ - 1 + \frac{1}{x}}}\)\( =  - 4\).