Với x→+∞ thì x+2>0
⇒f(x) = $\sqrt{\frac{(x+2)^2(x-1)}{x^4+x^2+1}}$
$=\sqrt{\frac{x^4(1+\frac{2}{x})^2(\frac{1}{x}-\frac{1}{x^2})}{x^4(1+\frac{1}{x^2}+\frac{1}{x^4})}}$
$=\sqrt{\frac{(1+\frac{2}{x})^2(\frac{1}{x}-\frac{1}{x^2})}{(1+\frac{1}{x^2}+\frac{1}{x^4})}}$
$\lim_{x \to +\infty} \frac{2}{x}=$$\lim_{x \to +\infty} \frac{1}{x}=$$\lim_{x \to +\infty} \frac{1}{x^2}=$$\lim_{x \to +\infty} \frac{1}{x^4}=0$
Suy ra: $\lim_{x \to +\infty} f(x)=$ $\frac{(1+0)^2.(0-0)}{1+0+0}=0$
