Đáp án:
$\begin{array}{l}
a)\mathop {\lim }\limits_{x \to 3} \dfrac{{\sqrt {x + 1} - 2}}{{{x^2} - 9}}\\
= \mathop {\lim }\limits_{x \to 3} \dfrac{{x + 1 - 4}}{{\left( {x - 3} \right)\left( {x + 3} \right).\left( {\sqrt {x + 1} + 2} \right)}}\\
= \mathop {\lim }\limits_{x \to 3} \dfrac{1}{{\left( {x + 3} \right)\left( {\sqrt {x + 1} + 2} \right)}}\\
= \dfrac{1}{{\left( {3 + 3} \right).\left( {\sqrt {3 + 1} + 2} \right)}}\\
= \dfrac{1}{{24}}\\
b)\mathop {\lim }\limits_{x \to 2} \dfrac{{\sqrt {3x - 2} - 2}}{{{x^2} - 4}}\\
= \mathop {\lim }\limits_{x \to 2} \dfrac{{3x - 2 - 4}}{{\left( {x - 2} \right)\left( {x + 2} \right)\left( {\sqrt {3x - 2} + 2} \right)}}\\
= \mathop {\lim }\limits_{x \to 2} \dfrac{3}{{\left( {x + 2} \right)\left( {\sqrt {3x - 2} + 2} \right)}}\\
= \dfrac{3}{{\left( {2 + 2} \right).\left( {\sqrt {3.2 - 2} + 2} \right)}}\\
= \dfrac{3}{{16}}
\end{array}$
