Cho hàm số \(y = f(x) = \sqrt {4{x^2} + 2x + 3} \).

\(\mathop {lim}\limits_{x \to  + \infty } \frac{{f(x)}}{x} = \mathop {lim}\limits_{x \to  + \infty } \frac{{\sqrt {4{x^2} + 2x + 3} }}{x} = \mathop {lim}\limits_{x \to  + \infty } \frac{{\sqrt {4 + \frac{2}{x} + \frac{3}{{{x^2}}}} }}{1} = 2\)

\(\begin{array}{l}
\mathop {lim}\limits_{x \to  + \infty } f(x) - ax = \mathop {lim}\limits_{x \to  + \infty } \sqrt {4{x^2} + 2x + 3}  - 2x = \mathop {lim}\limits_{x \to  + \infty } \frac{{4{x^2} + 2x + 3 - 4{x^2}}}{{\sqrt {4{x^2} + 2x + 3}  - 2x}}\\
 = \mathop {lim}\limits_{x \to  + \infty } \frac{{2x + 3}}{{\sqrt {4{x^2} + 2x + 3}  - 2x}} = \mathop {lim}\limits_{x \to  + \infty } \frac{{2 + \frac{3}{x}}}{{\sqrt {4 + \frac{2}{x} + \frac{3}{{{x^2}}}}  - 2}} =  + \infty 
\end{array}\)

Vậy \(a = 2,b =  + \infty \)