Nilai \( \displaystyle \lim_{x \to \pi/2} \ \frac{\sin x \tan (2x-\pi)}{2\pi - 4x} = \cdots \)
- \( -\frac{1}{2} \)
- \( \frac{1}{2} \)
- \( \frac{1}{3}\sqrt{3} \)
- \( 1 \)
- \( \sqrt{3} \)
(SPMB 2005)
Pembahasan:
\begin{aligned} \lim_{x \to \pi/2} \ \frac{\sin x \tan (2x-\pi)}{2\pi - 4x} &= \lim_{x \to \pi/2} \ \frac{\sin x \ (-\tan (\pi-2x))}{2(\pi - 2x)} \\[8pt] &= -\frac{1}{2} \cdot \lim_{x \to \pi/2} \ \sin x \cdot \lim_{x \to \pi/2} \ \frac{\tan (\pi-2x)}{(\pi - 2x)} \\[8pt] &= -\frac{1}{2} \cdot \sin (\pi/2) \cdot 1 = -\frac{1}{2} \cdot 1 \cdot 1 \\[8pt] &= -\frac{1}{2} \end{aligned}
Jawaban A.