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