Giới hạn \(\mathop {\lim }\limits_{x \to  + \infty...

B

- Hướng dẫn giải

\(\mathop {\lim }\limits_{x \to  + \infty } \frac{{{{\left( {x - 1} \right)}^2}\left( {2{x^3} + 3x} \right)}}{{4x - {x^5}}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{{{\left( {1 - \frac{1}{x}} \right)}^2}\left( {2 + \frac{3}{{{x^2}}}} \right)}}{{\frac{4}{{{x^4}}} - 1}} =  - 2\)

Suy ra \(A = {2^2} - {1^2} = 3\).