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