Tìm giới hạn sau \[A = \mathop {\lim }\limits_{n \to  + \infty } \frac{{3n + 2023}}{{4n - 2024}}\]       

a,

\(A = \mathop {\lim }\limits_{n \to + \infty } \frac{{3n + 2023}}{{4n - 2024}} = \mathop {\lim }\limits_{n \to + \infty } \frac{{3 + \frac{{2023}}{n}}}{{4 - \frac{{2024}}{n}}} = \frac{3}{4}\)

 

b, Ta có \[\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {4{x^2} - 7x + 12} }}{{a\left| x \right| - 17}} = \]\[\mathop {\lim }\limits_{x \to - \infty } \frac{{ - x\sqrt {4 - \frac{7}{x} + \frac{{12}}{{{x^2}}}} }}{{ - x\left( {a + \frac{{17}}{x}} \right)}} = \]

\[\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {4 - \frac{7}{x} + \frac{{12}}{{{x^2}}}} }}{{a + \frac{{17}}{x}}} = \frac{2}{a}\]\[ = \frac{2}{3} \Rightarrow a = 3\]