Ta có
$\underset{x \to 3^-}{\lim} \dfrac{3-x}{\sqrt{27 - x^3}} = \underset{x \to 3^-}{\lim} \dfrac{3-x}{\sqrt{(3-x)(x^2 + 3x + 9)}}$
$= \underset{x \to 3^-}{\lim} \dfrac{\sqrt{3-x}}{\sqrt{x^2 + 3x + 9}} = -\dfrac{0}{\sqrt{9 + 9 + 9}} = 0$
Vậy
$\underset{x \to 3^-}{\lim} \dfrac{3-x}{\sqrt{27 - x^3}} = 0$