$\lim \limits_{x\to \infty}x \bigg [ (\sqrt{x^2+2x}-2\sqrt{x^2+x}+x) \bigg ]$
$= \lim \limits_{x\to \infty}x.\dfrac{(\sqrt{x^2+2x}+x)^2 -4(x^2+x)}{ \sqrt{x^2+2x}+x +2\sqrt{x^2+x}} =\lim \limits_{x\to \infty} 2x^2.\dfrac{\sqrt{x^2+2x}-(x+1)}{ \sqrt{x^2+2x}+x +2\sqrt{x^2+x}}$
$=\lim \limits_{x\to \infty}\dfrac{-2x^2}{\bigg [\sqrt{x^2+2x}+x +2\sqrt{x^2+x} \bigg]. \bigg [\sqrt{x^2+2x}+(x+1)\bigg ]}$
Chia 2 trường hợp $\lim \limits_{x\to +\infty}=-\dfrac{1}{4};\ \lim \limits_{x\to -\infty}=-\infty$