Phần 3. Trắc nghiệm trả lời ngắn

Tìm \(a\) để \(\lim \frac{{a{n^3} + {n^2} - 4}}{{2{n^3} + 1}} =  - 2\).

\(\mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} \frac{{1 - \sqrt {x + 3} }}{{x + 2}}\)\[ = \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} \frac{{\left( {1 - \sqrt {x + 3} } \right)\left( {1 + \sqrt {x + 3} } \right)}}{{\left( {x + 2} \right)\left( {1 + \sqrt {x + 3} } \right)}}\]\[ = \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} \frac{{ - \left( {x + 2} \right)}}{{\left( {x + 2} \right)\left( {1 + \sqrt {x + 3} } \right)}}\]\[ = \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} \frac{{ - 1}}{{\left( {1 + \sqrt {x + 3} } \right)}} =  - 0,5\].

Trả lời: −0,5.

\(\lim \left( {{u_n}{v_n}} \right) = 2 \cdot 3 = 6\). Chọn B.