Tính các giới hạn sau:

a) Ta có \[I = \mathop {\lim }\limits_{n \to + \infty } \frac{1}{{{n^3}}} + \mathop {\lim }\limits_{n \to + \infty } \frac{1}{{{3^n}}} - \mathop {\lim }\limits_{n \to + \infty } 2023 = 0 + 0 - 2023 = - 2023\].

b) \(\mathop {\lim }\limits_{n \to + \infty } \frac{{\sqrt {4{n^2} + n + 1} + 3n}}{{2{n^{}} - 1}} = \mathop {\lim }\limits_{n \to + \infty } \frac{{\sqrt {4 + \frac{1}{n} + \frac{1}{{{n^2}}}} + 3}}{{2 - \frac{1}{n}}}\)\[ = \frac{{\mathop {\lim }\limits_{n \to + \infty } \sqrt {4 + \frac{1}{n} + \frac{1}{{{n^2}}}} + \mathop {\lim }\limits_{n \to \infty } 3}}{{\mathop {\lim }\limits_{n \to + \infty } \left( {2 - \frac{1}{n}} \right)}} = \frac{{2 + 3}}{2} = \frac{5}{2}\]