PHẦN II. TRẮC NGHIỆM ĐÚNG – SAI

Cho hai dãy số (un), (vn) với un = 4.3n – 7n + 1 ; vn = 7n.

a) \(\mathop {\lim }\limits_{n \to  + \infty } \frac{1}{{{v_n}}} = 0\).

b) \(\mathop {\lim }\limits_{n \to  + \infty } {v_n} =  + \infty \).

c) \(\mathop {\lim }\limits_{n \to  + \infty } \frac{{{u_n} - {v_n}}}{{3{u_n} + 2{v_n}}} = \frac{8}{{19}}\).

d) \(\mathop {\lim }\limits_{n \to  + \infty } {u_n} =  + \infty \).

a) \(\mathop {\lim }\limits_{n \to + \infty } \frac{1}{{{v_n}}} = \mathop {\lim }\limits_{n \to + \infty } {\left( {\frac{1}{7}} \right)^n} = 0\).

b) \(\mathop {\lim }\limits_{n \to + \infty } {7^n} = + \infty \).

c) \(\mathop {\lim }\limits_{n \to + \infty } \frac{{{u_n} - {v_n}}}{{3{u_n} + 2{v_n}}} = \mathop {\lim }\limits_{n \to + \infty } \frac{{{{4.3}^n} - {{8.7}^n}}}{{{{12.3}^n} - {{19.7}^n}}} = \mathop {\lim }\limits_{n \to + \infty } \frac{{4.{{\left( {\frac{3}{7}} \right)}^n} - 8}}{{12.{{\left( {\frac{3}{7}} \right)}^n} - 19}} = \frac{8}{{19}}\).

d) Ta có \(\mathop {\lim }\limits_{n \to + \infty } {u_n} = \mathop {\lim }\limits_{n \to + \infty } \left( {{{4.3}^n} - {7^{n + 1}}} \right) = \mathop {\lim }\limits_{n \to + \infty } {7^n}\left[ {4.{{\left( {\frac{3}{7}} \right)}^n} - 7} \right]\).

Vì \(\mathop {\lim }\limits_{n \to + \infty } {7^n} = + \infty \); \(\mathop {\lim }\limits_{n \to + \infty } \left[ {4.{{\left( {\frac{3}{7}} \right)}^n} - 7} \right] = - 7 < 0\) nên \(\mathop {\lim }\limits_{n \to + \infty } {u_n} = - \infty \).

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