Biết \(\mathop {\lim }\limits_{x \to + \infty } \dfrac{{ax + \sqrt {{x^2} - 3x + 5} }}{{2x - 7}} = 2\). Khi đó:

* Hướng dẫn giải

\(\begin{array}{l}\mathop {\lim }\limits_{x \to  + \infty } \dfrac{{ax + \sqrt {{x^2} - 3x + 5} }}{{2x - 7}} = \mathop {\lim }\limits_{x \to  + \infty } \dfrac{{a + \sqrt {1 - \dfrac{3}{x} + \dfrac{5}{{{x^2}}}} }}{{2 - \dfrac{7}{x}}} = \dfrac{{a + 1}}{2}\\ \Rightarrow \dfrac{{a + 1}}{2} = 2 \Leftrightarrow a + 1 = 4 \Leftrightarrow a = 3\end{array}\)

Vậy \(2 < a < 5\).

Chọn D.