Cho hàm số \(f\left( x \right) = \sqrt {4{x^2} + ax + 1}  + bx;a,b \in \mathbb{R}\).

a) \(\mathop {\lim }\limits_{x \to 0} f\left( x \right) = 1\).

b) \(\mathop {\lim }\limits_{x \to  - \infty } \left( {\sqrt {4{x^2} + ax + 1}  + bx} \right) = \mathop {\lim }\limits_{x \to  - \infty } \left[ {x\left( { - \sqrt {4 + \frac{a}{x} + \frac{1}{{{x^2}}}}  + b} \right)} \right]\).

c) Khi b = 2 thì \(\mathop {\lim }\limits_{x \to  - \infty } f\left( x \right) = \frac{a}{4}\).

d) Biết rằng \(\mathop {\lim }\limits_{x \to  - \infty } \left( {\sqrt {4{x^2} + ax + 1}  + bx} \right) =  - 1\). Khi đó biểu thức P = a2 – 2b3 có giá trị bằng 0.

a) \(\mathop {\lim }\limits_{x \to 0} f\left( x \right) = 1\).

b) \(\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {4{x^2} + ax + 1} + bx} \right) = \mathop {\lim }\limits_{x \to - \infty } \left( { - x\sqrt {4 + \frac{a}{x} + \frac{1}{{{x^2}}}} + bx} \right) = \mathop {\lim }\limits_{x \to - \infty } \left[ {x\left( { - \sqrt {4 + \frac{a}{x} + \frac{1}{{{x^2}}}} + b} \right)} \right]\).

c) Khi b = 2 thì \(\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {4{x^2} + ax + 1} + 2x} \right)\)\( = \mathop {\lim }\limits_{x \to - \infty } \frac{{ax + 1}}{{\sqrt {4{x^2} + ax + 1} - 2x}}\)\( = \mathop {\lim }\limits_{x \to - \infty } \frac{{a + \frac{1}{x}}}{{ - \sqrt {4 + \frac{a}{x} + \frac{1}{{{x^2}}}} - 2}}\)\( = - \frac{a}{4}\).

d) \(\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {4{x^2} + ax + 1} + bx} \right) = \mathop {\lim }\limits_{x \to - \infty } \left[ {x\left( { - \sqrt {4 + \frac{a}{x} + \frac{1}{{{x^2}}}} + b} \right)} \right]\).

Nếu b ≠ 2: \(\mathop {\lim }\limits_{x \to - \infty } \left[ {x\left( { - \sqrt {4 + \frac{a}{x} + \frac{1}{{{x^2}}}} + b} \right)} \right] = \left\{ \begin{array}{l} - \infty \;khi\;b > 2\\ + \infty \;khi\;b < 2\end{array} \right.\) mâu thuẫn với giải thiết.

Vậy b = 2.

Khi đó \(\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {4{x^2} + ax + 1} + bx} \right) = - \frac{a}{4}\).

Mà \(\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {4{x^2} + ax + 1} + bx} \right) = - 1\) nên \( - \frac{a}{4} = - 1 \Rightarrow a = 4\).

Vậy a = 4; b = 2. Do đó P = a² – 2b³ = 0.

Đáp án: a) Đúng; b) Đúng;   c) Sai;   d) Đúng.