Đáp án đúng: A
Giải chi tiết:\(\mathop {\lim }\limits_{x \to \dfrac{\pi }{4}} \dfrac{{\sin \left( {\dfrac{\pi }{4} - x} \right)}}{{1 - \sqrt 2 \sin x}} = \mathop {\lim }\limits_{x \to \dfrac{\pi }{4}} \dfrac{{\sin \left( {\dfrac{\pi }{4} - x} \right)}}{{ - \sqrt 2 \left( {\sin x - \sin \dfrac{\pi }{4}} \right)}}\)
\(\begin{array}{l} = \mathop {\lim }\limits_{x \to \dfrac{\pi }{4}} \dfrac{{\sin \left( {x - \dfrac{\pi }{4}} \right)}}{{\sqrt 2 .2\cos \left( {\dfrac{x}{2} + \dfrac{\pi }{8}} \right)\sin \left( {\dfrac{x}{2} - \dfrac{\pi }{8}} \right)}}\\ = \mathop {\lim }\limits_{x \to \dfrac{\pi }{4}} \dfrac{{2\sin \left( {\dfrac{x}{2} - \dfrac{\pi }{8}} \right)\cos \left( {\dfrac{x}{2} - \dfrac{\pi }{8}} \right)}}{{\sqrt 2 .2\cos \left( {\dfrac{x}{2} + \dfrac{\pi }{8}} \right)\sin \left( {\dfrac{x}{2} - \dfrac{\pi }{8}} \right)}}\\ = \mathop {\lim }\limits_{x \to \dfrac{\pi }{4}} \dfrac{{\cos \left( {\dfrac{x}{2} - \dfrac{\pi }{8}} \right)}}{{\sqrt 2 \cos \left( {\dfrac{x}{2} + \dfrac{\pi }{8}} \right)}} = \dfrac{{\cos \left( {\dfrac{\pi }{8} - \dfrac{\pi }{8}} \right)}}{{\sqrt 2 \cos \left( {\dfrac{\pi }{8} + \dfrac{\pi }{8}} \right)}} = \dfrac{1}{{\sqrt 2 .\dfrac{{\sqrt 2 }}{2}}} = 1\end{array}\)
Chọn A.