Đáp án đúng: A
Giải chi tiết:\(\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 + \tan x}  - \sqrt {1 + \sin x} }}{{{x^3}}}\)
\(\begin{array}{l} = \mathop {\lim }\limits_{x \to 0} \dfrac{{1 + \tan x - 1 - \sin x}}{{{x^3}\left( {\sqrt {1 + \tan x}  + \sqrt {1 + \sin x} } \right)}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{{\sin x}}{{\cos x}} - \sin x}}{{{x^3}\left( {\sqrt {1 + \tan x}  + \sqrt {1 + \sin x} } \right)}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x\left( {1 - \cos x} \right)}}{{{x^3}\left( {\sqrt {1 + \tan x}  + \sqrt {1 + \sin x} } \right)\cos x}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x.2{{\sin }^2}\dfrac{x}{2}}}{{{x^3}\left( {\sqrt {1 + \tan x}  + \sqrt {1 + \sin x} } \right)\cos x}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{2.\dfrac{{\sin x}}{x}.x.{{\left( {\dfrac{{\sin \dfrac{x}{2}}}{{\dfrac{x}{2}}}} \right)}^2}.\dfrac{{{x^2}}}{4}}}{{{x^3}\left( {\sqrt {1 + \tan x}  + \sqrt {1 + \sin x} } \right)\cos x}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{1}{2}.\dfrac{{\sin x}}{x}.{{\left( {\dfrac{{\sin \dfrac{x}{2}}}{{\dfrac{x}{2}}}} \right)}^2}}}{{\left( {\sqrt {1 + \tan x}  + \sqrt {1 + \sin x} } \right)\cos x}}\\ = \dfrac{{\dfrac{1}{2}.1.1}}{{\left( {1 + 1} \right).1}} = \dfrac{1}{4}\end{array}\)
Chọn A.