Đáp án đúng: C
Giải chi tiết:\(L = \mathop {\lim }\limits_{x \to \frac{\pi }{4}} \frac{{\sqrt 2 - 2\cos x}}{{\sin \left( {x - \frac{\pi }{4}} \right)}}\)
Đặt \(t = x - \frac{\pi }{4}\), khi \(x \to \frac{\pi }{4}\) thì \(t \to 0\), khi đó ta có:
\(\begin{array}{l}L = \mathop {\lim }\limits_{t \to 0} \frac{{\sqrt 2 - 2\cos \left( {t + \frac{\pi }{4}} \right)}}{{\sin t}}\\\,\,\,\,\, = \mathop {\lim }\limits_{t \to 0} \frac{{\sqrt 2 - 2\left( {\frac{{\sqrt 2 }}{2}\cos t - \frac{{\sqrt 2 }}{2}\sin t} \right)}}{{\sin t}}\\\,\,\,\,\, = \mathop {\lim }\limits_{t \to 0} \frac{{\sqrt 2 - \sqrt 2 \cos t + \sqrt 2 \sin t}}{{\sin t}}\\\,\,\,\,\, = \mathop {\lim }\limits_{t \to 0} \frac{{\sqrt 2 \left( {1 - \cos t} \right)}}{{\sin t}} + \mathop {\lim }\limits_{t \to 0} \frac{{\sqrt 2 \sin t}}{{\sin t}}\\\,\,\,\,\, = \mathop {\lim }\limits_{t \to 0} \frac{{2\sqrt 2 {{\sin }^2}\frac{t}{2}}}{{2\sin \frac{t}{2}\cos \frac{t}{2}}} + \sqrt 2 \\\,\,\,\,\, = \mathop {\lim }\limits_{t \to 0} \frac{{\sqrt 2 \sin \frac{t}{2}}}{{\cos \frac{t}{2}}} + \sqrt 2 = \sqrt 2 \end{array}\)
Chọn C.
