We are given, $\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} – 4}}{{\sqrt {3x – 2} – \sqrt {x + 2} }}$
Rationalize the denominator to get,
$ = \mathop {\lim }\limits_{x \to 2} \frac{{{x^2} – 4}}{{\sqrt {3x – 2} – \sqrt {x + 2} }} \times \left( {\frac{{\sqrt {3x – 2} + \sqrt {x + 2} }}{{\sqrt {3x – 2} + \sqrt {x + 2} }}} \right)$
$ = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {{x^2} – 4} \right)(\sqrt {3x – 2} + \sqrt {x + 2} )}}{{(3x – 2) – (x + 2)}}$
$ = \mathop {\lim }\limits_{x \to 2} \frac{{(x – 2)(x + 2)(\sqrt {3x – 2} + \sqrt {x + 2} )}}{{2x – 4}}$
Grouping the like terms and simplifying to get,
$ = \mathop {\lim }\limits_{x \to 2} \frac{{(x – 2)(x + 2)(\sqrt {3x – 2} + \sqrt {x + 2} )}}{{2(x – 2)}}$
Substituting the limits to get,
$ = \mathop {\lim }\limits_{x \to 2} \frac{{(x + 2)(\sqrt {3x – 2} + \sqrt {x + 2} )}}{2}$
Therefore, $\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} – 4}}{{\sqrt {3x – 2} – \sqrt {x + 2} }} = 8$.