Đáp án:
$\lim\limits_{x\to 3}\dfrac{\sqrt{x^2 - 2x + 6} - \sqrt{x^2 + 2x - 6}}{x^2 - 4x + 3}=-\dfrac13$
Giải thích các bước giải:
\(\begin{array}{l}
\quad \lim\limits_{x\to 3}\dfrac{\sqrt{x^2 - 2x + 6} - \sqrt{x^2 + 2x - 6}}{x^2 - 4x + 3}\\
= \lim\limits_{x\to 3}\dfrac{(\sqrt{x^2 - 2x + 6} - \sqrt{x^2 + 2x - 6})(\sqrt{x^2 - 2x + 6} + \sqrt{x^2 + 2x - 6})}{(x-1)(x-3)(\sqrt{x^2 - 2x + 6} + \sqrt{x^2 + 2x - 6})}\\
= \lim\limits_{x\to 3}\dfrac{x^2 - 2x + 6 - (x^2 + 2x - 6)}{(x-1)(x-3)(\sqrt{x^2 - 2x + 6} + \sqrt{x^2 + 2x - 6})}\\
= \lim\limits_{x\to 3}\dfrac{-4x+12}{(x-1)(x-3)(\sqrt{x^2 - 2x + 6} + \sqrt{x^2 + 2x - 6})}\\
= \lim\limits_{x\to 3}\dfrac{-4}{(x-1)(\sqrt{x^2 - 2x + 6} + \sqrt{x^2 + 2x - 6})}\\
= \dfrac{-4}{(3 -1)(\sqrt{3^2 - 2.3 + 6} + \sqrt{3^2 + 2.3 - 6})}\\
= -\dfrac13
\end{array}\)
