Biết lim x → + ∞ (a x + √ x 2 − 3 x + 5)/( 2 x − 7) = 2 . Khi đó

11/22

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

\[ - 1 \le a \le 2\].

\[a < - 1\].

\[a \ge 5\].

\[2 < a < 5\].

Giải thích

Chọn D

Ta có\(\mathop {\lim }\limits_{x \to + \infty } \frac{{ax + \sqrt {{x^2} - 3x + 5} }}{{2x - 7}} = 2\)\( \Leftrightarrow \mathop {\lim }\limits_{x \to + \infty } \frac{{a + \sqrt {1 - \frac{3}{x} + \frac{5}{{{x^2}}}} }}{{2 - \frac{7}{x}}} = 2\)\( \Leftrightarrow \frac{{a + 1}}{2} = 2\)\( \Leftrightarrow \frac{{a + 1}}{2} = 3\).

\( \Leftrightarrow a + 1 = 6 \Leftrightarrow a = 5\)