Nilai \( \displaystyle \lim_{x \to 0} \ \frac{\sin x \cos x}{\sqrt{\pi+2\sin x} - \sqrt{\pi}} = \cdots \)
- \( -2 \sqrt{\pi} \)
- \( -\sqrt{\pi} \)
- 0
- \( \sqrt{\pi} \)
- \( 2 \sqrt{\pi} \)
(SBMPTN 2018)
Pembahasan:
\begin{aligned} \lim_{x \to 0} \ \frac{\sin x \cos x}{\sqrt{\pi+2\sin x} - \sqrt{\pi}} &= \lim_{x \to 0} \ \frac{\sin x \cos x}{\sqrt{\pi+2\sin x} - \sqrt{\pi}} \times \frac{\sqrt{\pi+2\sin x} + \sqrt{\pi}}{\sqrt{\pi+2\sin x} + \sqrt{\pi}} \\[8pt] &= \lim_{x \to 0} \ \frac{(\sqrt{\pi+2\sin x} + \sqrt{\pi}) \sin x \cos x}{(\pi+2\sin x) - \pi} \\[8pt] &= \lim_{x \to 0} \ \frac{(\sqrt{\pi+2\sin x} + \sqrt{\pi}) \sin x \cos x}{2\sin x} \\[8pt] &= \frac{1}{2} \cdot \lim_{x \to 0} \ (\sqrt{\pi+2\sin x} + \sqrt{\pi}) \cos x \\[8pt] &= \frac{1}{2} \cdot (\sqrt{\pi+2\sin 0} + \sqrt{\pi}) \cos 0 \\[8pt] &= \frac{1}{2} \cdot (\sqrt{\pi} + \sqrt{\pi}) \cdot 1 \\[8pt] &= \sqrt{\pi} \end{aligned}
Jawaban D.