Tính giới hạn tổng hai dãy số và so sánh

Tính \(\mathop{\lim}\limits_{n \to +\infty}(u_n + v_n)\): \(u_n + v_n = 2 + \dfrac{1}{n} + 3 - \dfrac{2}{n} = 5 - \dfrac{1}{n}\) \(\mathop{\lim}\limits_{n \to +\infty}(u_n + v_n) = \mathop{\lim}\limits_{n \to +\infty}\left(5 - \dfrac{1}{n}\right) = 5 - 0 = 5\) Tính riêng từng giới hạn: \(\mathop{\lim}\limits_{n \to +\infty} u_n = \mathop{\lim}\limits_{n \to +\infty}\left(2 + \dfrac{1}{n}\right) = 2 + 0 = 2\) \(\mathop{\lim}\limits_{n \to +\infty} v_n = \mathop{\lim}\limits_{n \to +\infty}\left(3 - \dfrac{2}{n}\right) = 3 - 0 = 3\) \(\mathop{\lim}\limits_{n \to +\infty} u_n + \mathop{\lim}\limits_{n \to +\infty} v_n = 2 + 3 = 5\) Vậy \(\mathop{\lim}\limits_{n \to +\infty}(u_n + v_n) = \mathop{\lim}\limits_{n \to +\infty} u_n + \mathop{\lim}\limits_{n \to +\infty} v_n = 5\).