mathop lim limitsx to pi 6 2sin x - 14cos ^2x - 3
Câu hỏi
Nhận biết\(\mathop {\lim }\limits_{x \to \frac{\pi }{6}} \frac{{2\sin x - 1}}{{4{{\cos }^2}x - 3}}\)
Đáp án đúng: D
Lời giải của Tự Học 365
Giải chi tiết:
\(\mathop {\lim }\limits_{x \to \frac{\pi }{6}} \frac{{2\sin x - 1}}{{4{{\cos }^2}x - 3}}\)
\(\begin{array}{l} = \mathop {\lim }\limits_{x \to \frac{\pi }{6}} \frac{{2\left( {\sin x - \sin \frac{\pi }{6}} \right)}}{{4\left( {{{\cos }^2}x - {{\cos }^2}\frac{\pi }{6}} \right)}}\\ = \frac{1}{2}\mathop {\lim }\limits_{x \to \frac{\pi }{6}} \frac{{\sin x - \sin \frac{\pi }{6}}}{{\left( {\cos x - \cos \frac{\pi }{6}} \right)\left( {\cos x + \cos \frac{\pi }{6}} \right)}}\\ = \frac{1}{2}\mathop {\lim }\limits_{x \to \frac{\pi }{6}} \frac{{\sin x - \sin \frac{\pi }{6}}}{{x - \frac{\pi }{6}}}.\frac{{x - \frac{\pi }{6}}}{{\cos x - \cos \frac{\pi }{6}}}.\frac{1}{{\cos x + \cos \frac{\pi }{6}}}\\ = \frac{1}{2}\cos \frac{\pi }{6}.\frac{1}{{ - \sin \frac{\pi }{6}}}.\frac{1}{{2\cos \frac{\pi }{6}}} = - \frac{1}{2}\end{array}\)
Chọn D.
