Cho hàm số \(f\left( x \right) = \frac{{{x^2} + 3x + 2}}{{x - 2}}\). Hàm số liên tục trên khoảng nào sau đây?

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

Trả lời: −0,5.

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

Suy ra \(\frac{a}{2} =  - 2 \Leftrightarrow a =  - 4\).

Trả lời: −4.