\(\mathop {\lim }\limits_{x \to \dfrac{\pi }{4}} \dfrac{{\sqrt 2 - 2\cos x}}{{\sin \left( {x - \dfrac{\pi }{4}} \right)}}\)
Tìm các giới hạn sau:
Câu 470531: \(\mathop {\lim }\limits_{x \to \dfrac{\pi }{4}} \dfrac{{\sqrt 2 - 2\cos x}}{{\sin \left( {x - \dfrac{\pi }{4}} \right)}}\)
A. \(0\)
B. \(1\)
C. \(\dfrac{1}{2}\)
D. \(\sqrt 2 \)
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Đáp án : D(0) bình luận (0) lời giải
Giải chi tiết:
\(\mathop {\lim }\limits_{x \to \dfrac{\pi }{4}} \dfrac{{\sqrt 2 - 2\cos x}}{{\sin \left( {x - \dfrac{\pi }{4}} \right)}}\)
Đặt \(t = x - \dfrac{\pi }{4} \Rightarrow t \to 0\), ta có:
\(\begin{array}{l}\mathop {\lim }\limits_{x \to \dfrac{\pi }{4}} \dfrac{{\sqrt 2 - 2\cos x}}{{\sin \left( {x - \dfrac{\pi }{4}} \right)}} = \mathop {\lim }\limits_{t \to 0} \dfrac{{\sqrt 2 - 2\cos \left( {t + \dfrac{\pi }{4}} \right)}}{{\sin t}}\\ = \mathop {\lim }\limits_{t \to 0} \dfrac{{\sqrt 2 - 2\left( {\cos t.\dfrac{{\sqrt 2 }}{2} - \sin t.\dfrac{{\sqrt 2 }}{2}} \right)}}{{\sin t}}\\ = \mathop {\lim }\limits_{t \to 0} \dfrac{{\sqrt 2 - \sqrt 2 \left( {\cos t - \sin t} \right)}}{{\sin t}}\\ = \mathop {\lim }\limits_{t \to 0} \dfrac{{\sqrt 2 - \sqrt 2 \cos t + \sqrt 2 \sin t}}{{\sin t}}\\ = \mathop {\lim }\limits_{t \to 0} \left[ {\dfrac{{\sqrt 2 \left( {1 - \cos t} \right)}}{{\sin t}} + \sqrt 2 } \right]\\ = \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{{\sqrt 2 .2{{\sin }^2}\dfrac{t}{2}}}{{2\sin \dfrac{t}{2}\cos \dfrac{t}{2}}} + \sqrt 2 } \right)\\ = \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{{\sqrt 2 \sin \dfrac{t}{2}}}{{\cos \dfrac{t}{2}}} + \sqrt 2 } \right)\\ = \dfrac{{\sqrt 2 .0}}{1} + \sqrt 2 = \sqrt 2 \end{array}\)
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