Ta có
$\underset{x \to +\infty}{\lim} [\sqrt{x^2 + 4x} - (x+3)] = \underset{x \to +\infty}{\lim} \dfrac{x^2 + 4x - (x^2 + 6x + 9)}{\sqrt{x^2 + 4x} + x + 3}$
$= \underset{x \to +\infty}{\lim} \dfrac{-2x - 9}{\sqrt{x^2 + 4x} + x +3}$
$= \underset{x \to +\infty}{\lim} \dfrac{-2 - \frac{9}{x}}{\sqrt{1 + \frac{4}{x}} + 1 + \frac{3}{x}}$
$= \dfrac{-2}{1 + 1} = -1$
Vậy
$\underset{x \to +\infty}{\lim} [\sqrt{x^2 + 4x} - (x+3)] = -1$
