Nilai \( \displaystyle \lim_{x\to \frac{\pi}{2}} \ \frac{1-\sin^2 x}{(\sin^2 \frac{1}{2} x - \cos^2 \frac{1}{2} x)} = \cdots \)
- -2
- -1
- 0
- 1
- 2
Pembahasan:
\begin{aligned} \lim_{x\to \frac{\pi}{2}} \ \frac{1-\sin^2 x}{(\sin \frac{1}{2} x - \cos \frac{1}{2} x)^2} &= \lim_{x\to \frac{\pi}{2}} \ \frac{(1-\sin x)(1+\sin x)}{\sin^2 \frac{1}{2} x + \cos^2 \frac{1}{2} x - 2 \sin \frac{1}{2} x \cos \frac{1}{2} x} \\[8pt] &= \lim_{x\to \frac{\pi}{2}} \ \frac{(1-\sin x)(1+\sin x)}{1 - \sin x} \\[8pt] &= \lim_{x\to \frac{\pi}{2}} \ (1+\sin x) \\[8pt] &= 1 + \sin \frac{\pi}{2} \\[8pt] &= 1 + 1 = 2 \end{aligned}
Jawaban E.