Cho $\underset{x \rightarrow - \infty}{lim} \dfrac{a \sqrt{x^{2} + 1} + 2023}{x + 2024} = - \dfrac{1}{2}$ và $\underset{x \rightarrow + \infty}{lim} \left( \sqrt{x^{2} + b x + 1} - x \right) = 1$. Tính $P = 4 a + b$.

A.  

$P = 2$.

B.  

$P = 0$.

C.  

$P = 4$.

D.  

$P = 3$.

Đáp án đúng là: C

Cho $\underset{x \rightarrow - \infty}{lim} \dfrac{a \sqrt{x^{2} + 1} + 2023}{x + 2024} = - \dfrac{1}{2}$ và $\underset{x \rightarrow + \infty}{lim} \left( \sqrt{x^{2} + b x + 1} - x \right) = 1$. Tính $P = 4 a + b$.
A. $P = 2$. B. $P = 0$. C. $P = 4$. D. $P = 3$.
Lời giải
Ta có
$\underset{x \rightarrow - \infty}{lim} \dfrac{a \sqrt{x^{2} + 1} + 2023}{x + 2024} = \underset{x \rightarrow - \infty}{lim} \dfrac{- a x \sqrt{1 + \dfrac{1}{x^{2}}} + 2023}{x \left( 1 + \dfrac{2024}{x} \right)} = \underset{x \rightarrow - \infty}{lim} \dfrac{x \left( - a \sqrt{1 + \dfrac{1}{x^{2}}} + \dfrac{2023}{x} \right)}{x \left( 1 + \dfrac{2024}{x} \right)} \\ = \underset{x \rightarrow - \infty}{lim} \dfrac{- a \sqrt{1 + \dfrac{1}{x^{2}}} + \dfrac{2023}{x}}{1 + \dfrac{2024}{x}} = \dfrac{- a \sqrt{1 + 0} + 0}{1 + 0} = - a = - \dfrac{1}{2} \Leftrightarrow a = \dfrac{1}{2}$
Ta có
$\underset{x \rightarrow + \infty}{lim} \left( \sqrt{x^{2} + b x + 1} - x \right) = \underset{x \rightarrow + \infty}{lim} \dfrac{b x + 1}{\sqrt{x^{2} + b x + 1} + x} = \underset{x \rightarrow + \infty}{lim} \dfrac{x \left( b + \dfrac{1}{x} \right)}{x \sqrt{1 + \dfrac{b}{x} + \dfrac{1}{x^{2}}} + x} \\ = \underset{x \rightarrow + \infty}{lim} \dfrac{x \left( b + \dfrac{1}{x} \right)}{x \left( \sqrt{1 + \dfrac{b}{x} + \dfrac{1}{x^{2}}} + 1 \right)} = \underset{x \rightarrow + \infty}{lim} \dfrac{b + \dfrac{1}{x}}{\sqrt{1 + \dfrac{b}{x} + \dfrac{1}{x^{2}}} + 1} = \dfrac{b + 0}{\sqrt{1 + 0 + 0} + 1} = \dfrac{b}{2} = 1 \Leftrightarrow b = 2$
Vậy $P = 4 a + b = 4$.