\(\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 + 2x} - 1}}{x}\)
Giải chi tiết:
\(\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 + 2x} - 1}}{x} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {\sqrt {1 + 2x} - 1} \right)\left( {\sqrt {1 + 2x} + 1} \right)}}{{x\left( {\sqrt {1 + 2x} + 1} \right)}}\)
\(\begin{array}{l} = \mathop {\lim }\limits_{x \to 0} \dfrac{{1 + 2x - 1}}{{x\left( {\sqrt {1 + 2x} + 1} \right)}} = \mathop {\lim }\limits_{x \to 0} \dfrac{2}{{\sqrt {1 + 2x} + 1}}\\ = \dfrac{2}{{\sqrt {1 + 0} + 1}} = \dfrac{2}{2} = 1\end{array}\).
Chọn C.