Tính $\mathop {\lim }\limits_{x \to 3} \dfrac{{\sqrt {x + 1}  - 2}}{{\sqrt {3x}  - 3}}$ bằng?

$\begin{array}{l}\mathop {\lim }\limits_{x \to 3} \dfrac{{\sqrt {x + 1}  - 2}}{{\sqrt {3x}  - 3}} \\= \mathop {\lim }\limits_{x \to 3} \dfrac{{(\sqrt {x + 1}  - 2)(\sqrt {x + 1}  + 2)(\sqrt {3x}  + 3)}}{{(\sqrt {3x}  - 3)(\sqrt {3x}  + 3)(\sqrt {x + 1}  + 2)}} \\= \mathop {\lim }\limits_{x \to 3} \dfrac{{(x + 1 - 4)(\sqrt {3x}  + 3)}}{{(3x - 9)(\sqrt {x + 1}  + 2)}}\\ = \mathop {\lim }\limits_{x \to 3} \dfrac{{(x - 3)(\sqrt {3x}  + 3)}}{{3(x - 3)(\sqrt {x + 1}  + 2)}} \\= \mathop {\lim }\limits_{x \to 3} \dfrac{{\sqrt {3x}  + 3}}{{3(\sqrt {x + 1}  + 2)}} \\= \dfrac{{\sqrt {3.3}  + 3}}{{3(\sqrt {3 + 1}  + 2)}} = \dfrac{1}{2}\end{array}$