Tìm các giới hạn sau
a) \(\mathop {\lim }\limits_{x \to {2^ - }} {{{x^2} - 3x + 2} \over {\sqrt {2 - x} }}\) b) \(\mathop {\lim }\limits_{x \to {0^ + }} {{3\sqrt x - x} \over {\sqrt {2x} + x}}\)
a) \({{{x^2} - 3x + 2} \over {\sqrt {2 - x} }} = {{\left( {x - 1} \right)\left( {x - 2} \right)} \over {\sqrt {2 - x} }} = \left( {1 - x} \right)\sqrt {2 - x} \) với mọi \(x < 2.\)
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Do đó
\(\mathop {\lim }\limits_{x \to {2^ - }} {{{x^2} - 3x + 2} \over {\sqrt {2 - x} }} = \mathop {\lim }\limits_{x \to {2^ - }} \left( {1 - x} \right)\sqrt {2 - x} = 0;\)
b) Với mọi x > 0 ta có:
\(\eqalign{
& {{3\sqrt x - x} \over {\sqrt {2x} + x}} = {{\sqrt x \left( {3 - \sqrt x } \right)} \over {\sqrt x \left( {\sqrt 2 + \sqrt x } \right)}} = {{3 - \sqrt x } \over {\sqrt 2 + \sqrt x }} \cr
& \mathop {\lim }\limits_{x \to {0^ + }} {{3\sqrt x - x} \over {\sqrt {2x} + x}} = \mathop {\lim }\limits_{x \to {0^ + }} {{3 - \sqrt x } \over {\sqrt 2 + \sqrt x }} = {{3\sqrt 2 } \over 2} \cr} \)