Tính căn 2x + 9} - 3 / {x} bằng
Giải thích
Chọn B
Ta có:\[\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {2x + 9} - 3}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{{{\left( {\sqrt {2x + 9} } \right)}^2} - {3^2}}}{{x\left( {\sqrt {2x + 9} + 3} \right)}} = \mathop {\lim }\limits_{x \to 0} \frac{2}{{\sqrt {2x + 9} + 3}} = \frac{2}{6} = \frac{1}{3}\].