Tìm giới hạn \[C = \mathop {\lim }\limits_{x \to 3} \frac{{\sqrt {2x + 3} - x}}{{{x^2} - 4x + 3}}\].

Chọn B

\[\begin{array}{l}C = \mathop {\lim }\limits_{x \to 3} \frac{{\sqrt {2x + 3} - x}}{{{x^2} - 4x + 3}} = \mathop {\lim }\limits_{x \to 3} \frac{{2x + 3 - {x^2}}}{{\left( {x - 1} \right)\left( {x - 3} \right)\left( {\sqrt {2x + 3} + x} \right)}}\\ = \mathop {\lim }\limits_{x \to 3} \frac{{ - \left( {x + 1} \right)\left( {x - 3} \right)}}{{\left( {x - 1} \right)\left( {x - 3} \right)\left( {\sqrt {2x + 3} + x} \right)}} = \mathop {\lim }\limits_{x \to 3} \frac{{ - \left( {x + 1} \right)}}{{\left( {x - 1} \right)\left( {\sqrt {2x + 3} + x} \right)}}\\ = \frac{{ - \left( {3 + 1} \right)}}{{\left( {3 - 1} \right)\left( {\sqrt {2.3 + 3} + 3} \right)}} = \frac{{ - 1}}{3}\end{array}\]