Tính \(\lim\limits_{x\rightarrow3}\frac{\sqrt{x^2-2x+6}-\sqrt{x^2+2x-6}}{x^2-4x+3}\).
\(\dfrac{1}{3}\).\(\dfrac{2}{3}\).\(-\dfrac{1}{3}\).\(-\dfrac{2}{3}\).Hướng dẫn giải:Có \(\frac{\sqrt{x^2-2x+6}-\sqrt{x^2+2x-6}}{x^2-4x+3}=\frac{1}{\left(x-1\right)\left(x-3\right)}.\frac{\left(x^2-2x+6\right)-\left(x^2+2x-6\right)}{\sqrt{x^2-2x+6}+\sqrt{x^2+2x-6}}\)\(=\frac{-4\left(x-3\right)}{\left(x-1\right)\left(x-3\right)\left(\sqrt{x^2-2x+6}+\sqrt{x^2+2x-6}\right)}=-\frac{4}{\left(x-1\right)\left(\sqrt{x^2-2x+6}+\sqrt{x^2+2x-6}\right)}\).
Khi \(x\rightarrow3,\) hàm số có giới hạn bằng \(-\frac{4}{\left(3-1\right)\left(\sqrt{9-6+6}+\sqrt{9+6-6}\right)}=-\frac{1}{3}.\)