Câu 6
Ta có
$\underset{x \to 5}{\lim} \dfrac{\sqrt{3x+1} - 4}{3 - \sqrt{x+4}} = \underset{x \to 5}{\lim} \dfrac{(3x + 1 - 16)(3 + \sqrt{x+4})}{[9 - (x + 4)](\sqrt{3x + 1} + 4)}$
$= \underset{x \to 5}{\lim} \dfrac{3(x-5)(3 + \sqrt{x+4})}{(5-x)(\sqrt{3x+1} + 4)}$
$= \underset{x \to 5}{\lim} \dfrac{-3(3 + \sqrt{x+4})}{\sqrt{3x+1} + 4}$
$= \dfrac{-3(3 + 3)}{4 + 4} = -\dfrac{9}{4}$
