Cho hàm số \(f(x)\) xác định trên \(\mathbb{R}\) thỏa mãn \[\emptyset \]. Tính giới hạn \(\mathop {\lim }\limits_{x \to 2} \frac{{\sqrt[3]{{5f(x) - 16}} - 4}}{{{x^2} + 2x - 8}}\)

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

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

Gọi \(E\) là trung điểm của \(AD\).

Trong

Xét tam giác \(ESF\) ta có: \(\frac{{EG}}{{ES}} = \frac{1}{3}\;\)

Do \(AE//BF \Rightarrow \frac{{NE}}{{NF}} = \frac{{AN}}{{BN}} = \frac{1}{2} \Rightarrow \frac{{NE}}{{EF}} = \frac{1}{3}.\)

Suy ra \(\frac{{EG}}{{ES}} = \frac{{NE}}{{EF}} \Rightarrow NG//SF.\)

Mà \(SF \subset \left( {SBC} \right)\)\( \Rightarrow NG//\left( {SBC} \right).\)