Tính giới hạn \(\lim\limits _{x \rightarrow \frac{\pi}{4}} \frac{\tan x-1}{2 \cos x-\sqrt{2}}\)
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Lời giải:
Báo saiTa có
\(\begin{array}{l} \mathop {\lim }\limits_{x \to \frac{\pi }{4}} \frac{{\tan x - 1}}{{2\cos x - \sqrt 2 }} = \mathop {\lim }\limits_{x \to \frac{\pi }{4}} \frac{{\sin x - \cos x}}{{2\cos x\left( {\cos x - \frac{{\sqrt 2 }}{2}} \right)}} = \mathop {\lim }\limits_{x \to \frac{\pi }{4}} \frac{{\sqrt 2 \sin \left( {x - \frac{\pi }{4}} \right)}}{{2\cos x\left( {\cos x - \cos \frac{\pi }{4}} \right)}}\\ = \mathop {\lim }\limits_{x \to \frac{\pi }{4}} \frac{{2\sqrt 2 \sin \left( {\frac{x}{2} - \frac{\pi }{8}} \right)\cos \left( {\frac{x}{2} - \frac{\pi }{8}} \right)}}{{ - 4\cos x\sin \left( {\frac{x}{2} + \frac{\pi }{8}} \right)\sin \left( {\frac{x}{2} - \frac{\pi }{8}} \right)}} = \mathop {\lim }\limits_{x \to \frac{\pi }{4}} \frac{{ - \sqrt 2 \cos \left( {\frac{x}{2} - \frac{\pi }{8}} \right)}}{{2\cos x\sin \left( {\frac{x}{2} + \frac{\pi }{8}} \right)}} = - \sqrt 2 \end{array}\)