Đáp án đúng: C
Giải chi tiết:\(\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 + 2x}  - \cos x - x}}{{{x^2}}}\)
\(\begin{array}{l} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 + 2x}  - \cos x - x}}{{{x^2}}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 + 2x}  - x - 1 + 1 - \cos x}}{{{x^2}}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 + 2x}  - \left( {x + 1} \right)}}{{{x^2}}} + \mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \cos x}}{{{x^2}}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{1 + 2x - {{\left( {x + 1} \right)}^2}}}{{{x^2}\left[ {\sqrt {1 + 2x}  + \left( {x + 1} \right)} \right]}} + \mathop {\lim }\limits_{x \to 0} \dfrac{{2{{\sin }^2}\dfrac{x}{2}}}{{{x^2}}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{ - 1}}{{\sqrt {1 + 2x}  + \left( {x + 1} \right)}} + \mathop {\lim }\limits_{x \to 0} \dfrac{{2{{\left( {\dfrac{{\sin \dfrac{x}{2}}}{{\dfrac{x}{2}}}} \right)}^2}.\dfrac{{{x^2}}}{4}}}{{{x^2}}}\\ = \dfrac{{ - 1}}{{\sqrt 1  + 1}} + \mathop {\lim }\limits_{x \to 0} \dfrac{1}{2}{\left( {\dfrac{{\sin \dfrac{x}{2}}}{{\dfrac{x}{2}}}} \right)^2}\\ = \dfrac{{ - 1}}{2} + \dfrac{1}{2} = 0\end{array}\)
Chọn C.