[man111]

$\lim_{x\rightarrow \infty}\left(\sqrt{(x-a)(x-b)}-x\right)$

Using $\bf{A.M\geq G.M\geq H.M}$

$\displaystyle \frac{x-a+x-b}{2}\geq \sqrt{(x-a)(x-b)}\geq \frac{2}{\frac{1}{x-a}+\frac{1}{x-b}} = 2\frac{x^2-(a+b)x+ab}{2x-a-b}$

$\displaystyle \frac{2x^2-2(a+b)x+2ab}{2x-a-b}-x\leq \left[\sqrt{(x-a)(x-b)}-x\right]\leq \frac{2x-a-b}{2}-x$

$\displaystyle \lim_{x\rightarrow \infty}\frac{2x^2-2(a+b)x+2ab-2x^2+(a+b)x}{2x-a-b}\leq \left[\sqrt{(x-a)(x-b)}-x\right] \leq \lim_{x\rightarrow \infty}\frac{2x-a-b}{2}-x$

So $\displaystyle \lim_{x\rightarrow \infty}\left(\sqrt{(x-a)(x-b)}-x\right) = -\left(\frac{a+b}{2}\right)$
[RIGHT][I][B]Nguồn: MathScope.ORG[/B][/I][/RIGHT]