Giải thích các bước giải:
$\begin{array}{l} \mathop {\lim }\limits_{x \to 2} \frac{{\sqrt {x + 2} - \sqrt {2x} }}{{\sqrt {x - 1} - \sqrt {3 - x} }}\\ = \mathop {\lim }\limits_{x \to 2} \frac{{(\sqrt {x + 2} - \sqrt {2x} )(\sqrt {x - 1} + \sqrt {3 - x} )(\sqrt {x + 2} + \sqrt {2x} )}}{{(\sqrt {x - 1} - \sqrt {3 - x} )(\sqrt {x - 1} + \sqrt {3 - x} )(\sqrt {x + 2} + \sqrt {2x} )}}\\ = \mathop {\lim }\limits_{x \to 2} \frac{{(x + 2 - 2x)(\sqrt {x - 1} + \sqrt {3 - x} )}}{{(x - 1 - 3 + x)(\sqrt {x + 2} + \sqrt {2x} )}}\\ = \mathop {\lim }\limits_{x \to 2} \frac{{(2 - x)(\sqrt {x - 1} + \sqrt {3 - x} )}}{{2(x - 2)(\sqrt {x + 2} + \sqrt {2x} )}}\\ = \mathop {\lim }\limits_{x \to 2} \frac{{ - (\sqrt {x - 1} + \sqrt {3 - x} )}}{{2(\sqrt {x + 2} + \sqrt {2x} )}}\\ = \frac{{ - (\sqrt {2 - 1} + \sqrt {3 - 2} )}}{{2(\sqrt {2 + 2} + \sqrt {2.2} )}} = \frac{{ - 1}}{4} \end{array}$
