Đáp án đúng: D
Giải chi tiết:\(\mathop {\lim }\limits_{x \to \dfrac{\pi }{4}} \tan 2x.\tan \left( {\dfrac{\pi }{4} - x} \right)\)
\(\begin{array}{l} = \mathop {\lim }\limits_{x \to \dfrac{\pi }{4}} \dfrac{{\sin 2x}}{{\cos 2x}}.\dfrac{{\sin \left( {\dfrac{\pi }{4} - x} \right)}}{{\cos \left( {\dfrac{\pi }{4} - x} \right)}}\\ = \mathop {\lim }\limits_{x \to \dfrac{\pi }{4}} \dfrac{{\sin 2x}}{{\cos \left( {\dfrac{\pi }{4} - x} \right)}}.\dfrac{{\dfrac{{\sqrt 2 }}{2}\cos x - \dfrac{{\sqrt 2 }}{2}\sin x}}{{\left( {\cos x - \sin x} \right)\left( {\cos x + \sin x} \right)}}\\ = \mathop {\lim }\limits_{x \to \dfrac{\pi }{4}} \dfrac{{\sin 2x}}{{\cos \left( {\dfrac{\pi }{4} - x} \right)}}.\dfrac{{\dfrac{{\sqrt 2 }}{2}}}{{\cos x + \sin x}}\\ = \dfrac{{\sin \dfrac{\pi }{2}}}{{\cos 0}}.\dfrac{{\dfrac{{\sqrt 2 }}{2}}}{{\cos \dfrac{\pi }{4} + \sin \dfrac{\pi }{4}}} = \dfrac{1}{2}\end{array}\)
Đặt \(t = x - \dfrac{\pi }{4} \Rightarrow t \to 0\), ta có:
\(\begin{array}{l}\mathop {\lim }\limits_{x \to \dfrac{\pi }{4}} \tan 2x.\tan \left( {\dfrac{\pi }{4} - x} \right)\\ = \mathop {\lim }\limits_{x \to \dfrac{\pi }{4}} \tan 2\left( {t + \dfrac{\pi }{4}} \right).\tan \left( { - t} \right)\\ = \mathop {\lim }\limits_{x \to \dfrac{\pi }{4}} \left[ { - \tan \left( {2t + \dfrac{\pi }{2}} \right).\tan t} \right]\\ = \mathop {\lim }\limits_{x \to \dfrac{\pi }{4}} \left[ { - \tan \left( {2t + \dfrac{\pi }{2}} \right).\tan t} \right] = 0\end{array}\)
Chọn D.