Biết \(\mathop {\lim }\limits_{x \to - \infty } \left( {ax + \sqrt {{x^2} + bx + 1} } \right) = \frac{1}{2}\). Tính \(A = 2a + b\)

* Hướng dẫn giải

Dễ thấy nếu \(a \le 0\) thì \(\mathop {\lim }\limits_{x \to  - \infty } \left( {ax + \sqrt {{x^2} + bx + 1} } \right) =  + \infty \) nên không thỏa mãn.

Ta xét \(a > 0\).

Đặt

\(\begin{array}{l}L = \mathop {\lim }\limits_{x \to  - \infty } \left( {ax + \sqrt {{x^2} + bx + 1} } \right)\\ = \mathop {\lim }\limits_{x \to  - \infty } \frac{{{a^2}{x^2} - \left( {{x^2} + bx + 1} \right)}}{{ax - \sqrt {{x^2} + bx + 1} }}\\ = \mathop {\lim }\limits_{x \to  - \infty } \frac{{\left( {{a^2} - 1} \right){x^2} - bx - 1}}{{ax - \sqrt {{x^2} + bx + 1} }}\\ = \mathop {\lim }\limits_{x \to  - \infty } \frac{{\left( {{a^2} - 1} \right){x^2} - bx - 1}}{{ax - \left| x \right|\sqrt {1 + \frac{b}{x} + \frac{1}{{{x^2}}}} }}\\ = \mathop {\lim }\limits_{x \to  - \infty } \frac{{\left( {{a^2} - 1} \right){x^2} - bx - 1}}{{ax + x\sqrt {1 + \frac{b}{x} + \frac{1}{{{x^2}}}} }}\\ = \mathop {\lim }\limits_{x \to  - \infty } \frac{{\left( {{a^2} - 1} \right){x^2} - bx - 1}}{{x\left( {a + \sqrt {1 + \frac{b}{x} + \frac{1}{{{x^2}}}} } \right)}}\end{array}\)

Nếu \({a^2} - 1 \ne 0\) và \(a > 0\) thì \(L = \infty \) nên loại.

Do đó \({a^2} = 1 \Leftrightarrow a = 1\) (vì \(a > 0\)). Khi đó,

\(L = \mathop {\lim }\limits_{x \to  - \infty } \frac{{ - bx - 1}}{{x\left( {1 + \sqrt {1 + \frac{b}{x} + \frac{1}{{{x^2}}}} } \right)}}\) \( =  - \frac{b}{2}\)

\( \Rightarrow L = \frac{1}{2} \Leftrightarrow  - \frac{b}{2} = \frac{1}{2}\) \( \Leftrightarrow b =  - 1\).

\( \Rightarrow A = 2a + b = 2.1 + \left( { - 1} \right) = 1\)