Tìm giới hạn lim x→0 2sinx−sin2x/ 2tanx−tan2x.

Question

Tìm giới hạn \({\mathop{\lim }\limits_{x\to 0}} \frac{2\sin x-\sin 2x}{2\tan x-\tan 2x} .\)

\(A. -2. \)

\(B. \frac{-1}{2} .\)

\(C. 1. \)

\(D. 2.\)

Answer 1

Chọn B

Ta có \({\mathop{\lim }\limits_{x\to 0}} \frac{2\sin x-\sin 2x}{2\tan x-\tan 2x} ={\mathop{\lim }\limits_{x\to 0}} \frac{2\sin x-2\sin x\cos x}{\frac{2\sin x}{\cos x} -\frac{\sin 2x}{\cos 2x} } ={\mathop{\lim }\limits_{x\to 0}} \frac{2\sin x\left(1-\cos x\right)}{\frac{2\sin x}{\cos x} -\frac{2\sin x\cos x}{\cos 2x} }  \)

\({\mathop{\lim }\limits_{x\to 0}} \frac{2\sin x\left(1-\cos x\right)}{2\sin x\left(\frac{1}{\cos x} -\frac{\cos x}{\cos 2x} \right)} ={\mathop{\lim }\limits_{x\to 0}} \frac{1-\cos x}{\frac{1}{\cos x} -\frac{\cos x}{\cos 2x} } ={\mathop{\lim }\limits_{x\to 0}} \frac{1-\cos x}{\frac{\cos ^{2} x-1}{\cos x\cos 2x} } ={\mathop{\lim }\limits_{x\to 0}} \frac{-\cos x\cos 2x}{\cos x+1} =\frac{-1}{2} .   \)