Ta có
$\underset{x \to 1}{\lim} \dfrac{\sqrt{x^2 + 3} - 3x + 1}{x^2-1} = \underset{x \to 1}{\lim} \dfrac{(\sqrt{x^2 + 3} - 2) - 3x + 3}{x^2-1}$
$= \underset{x \to 1}{\lim} \dfrac{\frac{x^2 + 3 - 4}{\sqrt{x^2 + 3} + 2} - 3(x-1)}{(x-1)(x+1)}$
$= \underset{x \to 1}{\lim} \dfrac{\frac{(x-1)(x+1)}{\sqrt{x^2 + 3} + 2} - 3(x-1)}{(x-1)(x+1)}$
$= \underset{x \to 1}{\lim} \dfrac{\frac{x+1}{\sqrt{x^2 +3} + 2} - 3}{x+1}$
$= \dfrac{\frac{2}{4} - 3}{2} = -\dfrac{5}{4}$
Vậy
$\underset{x \to 1}{\lim} \dfrac{\sqrt{x^2 + 3} - 3x + 1}{x^2-1} = -\dfrac{5}{4}$.
