Cho hai dãy số (un) và (vn) có un = 4n2 – n + 3; vn = 3n2 + 7.

17/22

a) \(\mathop {\lim }\limits_{n \to + \infty } \frac{{{u_n}}}{{{v_n}}} = \frac{4}{3}\).

b) \(\mathop {\lim }\limits_{n \to + \infty } \frac{{{u_n}}}{{{{\left( {{v_n}} \right)}^2}}} = \frac{4}{9}\).

c) \(\mathop {\lim }\limits_{n \to + \infty } \frac{{{{\left( {{u_n}} \right)}^2}}}{{{v_n}}} = \frac{{16}}{3}\).

d) \(\mathop {\lim }\limits_{n \to + \infty } \frac{{{u_n} + a{n^2} + 7}}{{{v_n}}} = 8\) khi đó a = 20.

Nhập câu trả lời ở đây

0/3000 ký tự

Giải thích

a) \(\mathop {\lim }\limits_{n \to + \infty } \frac{{{u_n}}}{{{v_n}}} = \mathop {\lim }\limits_{n \to + \infty } \frac{{4{n^2} - n + 3}}{{3{n^2} + 7}} = \mathop {\lim }\limits_{n \to + \infty } \frac{{{n^2}\left( {4 - \frac{1}{n} + \frac{3}{{{n^2}}}} \right)}}{{{n^2}\left( {3 + \frac{7}{{{n^2}}}} \right)}} = \frac{4}{3}\).

b) \(\mathop {\lim }\limits_{n \to + \infty } \frac{{{u_n}}}{{{{\left( {{v_n}} \right)}^2}}} = \mathop {\lim }\limits_{n \to + \infty } \frac{{4{n^2} - n + 3}}{{{{\left( {3{n^2} + 7} \right)}^2}}} = \mathop {\lim }\limits_{n \to + \infty } \frac{{{n^2}\left( {4 - \frac{1}{n} + \frac{3}{{{n^2}}}} \right)}}{{{n^4}{{\left( {3 + \frac{7}{{{n^2}}}} \right)}^2}}} = \mathop {\lim }\limits_{n \to + \infty } \left[ {\frac{1}{{{n^2}}}.\frac{{\left( {4 - \frac{1}{n} + \frac{3}{{{n^2}}}} \right)}}{{{{\left( {3 + \frac{7}{{{n^2}}}} \right)}^2}}}} \right] = 0\).

c) \(\mathop {\lim }\limits_{n \to + \infty } \frac{{{{\left( {{u_n}} \right)}^2}}}{{{v_n}}} = \mathop {\lim }\limits_{n \to + \infty } \frac{{{{\left( {4{n^2} - n + 3} \right)}^2}}}{{3{n^2} + 7}} = \mathop {\lim }\limits_{n \to + \infty } \frac{{{n^4}{{\left( {4 - \frac{1}{n} + \frac{3}{{{n^2}}}} \right)}^2}}}{{{n^2}\left( {3 + \frac{7}{{{n^2}}}} \right)}} = \mathop {\lim }\limits_{n \to + \infty } \left[ {{n^2}.\frac{{{{\left( {4 - \frac{1}{n} + \frac{3}{{{n^2}}}} \right)}^2}}}{{\left( {3 + \frac{7}{{{n^2}}}} \right)}}} \right] = + \infty \).

d) \(\mathop {\lim }\limits_{n \to + \infty } \frac{{{u_n} + a{n^2} + 7}}{{{v_n}}} = \mathop {\lim }\limits_{n \to + \infty } \frac{{4{n^2} - n + 3 + a{n^2} + 7}}{{3{n^2} + 7}}\)\( = \mathop {\lim }\limits_{n \to + \infty } \frac{{\left( {a + 4} \right){n^2} - n + 10}}{{3{n^2} + 7}}\)\( = \mathop {\lim }\limits_{n \to + \infty } \frac{{{n^2}\left[ {\left( {a + 4} \right) - \frac{1}{n} + \frac{{10}}{{{n^2}}}} \right]}}{{{n^2}\left( {3 + \frac{7}{{{n^2}}}} \right)}}\)\( = \mathop {\lim }\limits_{n \to + \infty } \frac{{\left( {a + 4} \right) - \frac{1}{n} + \frac{{10}}{{{n^2}}}}}{{3 + \frac{7}{{{n^2}}}}} = \frac{{a + 4}}{3}\).

Khi đó \(\frac{{a + 4}}{3} = 8 \Leftrightarrow a = 20\).

Đáp án: a) Đúng; b) Sai; c) Sai; d) Đúng.