Câu hỏi 3

Ta có \(\mathop {\lim }\limits_{x \to  + \infty } y = \mathop {\lim }\limits_{x \to  + \infty } \dfrac{{2x + 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to  + \infty } \dfrac{{2 + \dfrac{1}{x}}}{{1 - \dfrac{1}{x}}} = \dfrac{{\mathop {\lim }\limits_{x \to  + \infty } \left( {2 + \dfrac{1}{x}} \right)}}{{\mathop {\lim }\limits_{x \to  + \infty } \left( {1 - \dfrac{1}{x}} \right)}} = 2.\)

Chọn A.