Cho hàm số \(y = f\left( x \right)\) xác định trên \(\mathbb{R}\) thỏa mãn \(\mathop {\lim }\limits_{x \to 2} \frac{{f\left( x \right) - 2}}{{x - 2}} = 2023.\)

Ta có: \(\mathop {\lim }\limits_{n \to + \infty } \frac{{2022{n^2} - 2023n}}{{2024 + 2025n - 2026{n^2}}} = \mathop {\lim }\limits_{n \to + \infty } \frac{{{n^2}\left( {2022 - \frac{{2023}}{n}} \right)}}{{{n^2}\left( {\frac{{2024}}{{{n^2}}} + \frac{{2025}}{n} - 2026} \right)}}\)

              \( = \mathop {\lim }\limits_{n \to + \infty } \frac{{2022 - \frac{{2023}}{n}}}{{\frac{{2024}}{{{n^2}}} + \frac{{2025}}{n} - 2026}}\)\( = \frac{{2022}}{{ - 2026}} = - \frac{{1011}}{{1013}}.\)

Chọn B

\(\lim \left( {{u_n} + 3} \right) = 0 \Leftrightarrow \lim {u_n} + \lim 3 = 0 \Leftrightarrow \lim {u_n} + 3 = 0 \Leftrightarrow \lim {u_n} = - 3.\)