Nilai \( \displaystyle \lim_{x \to 0} \ \sqrt{\frac{x \tan x}{\sin^2 x - \cos 2x + 1}} = \cdots \)
- \( 3 \)
- \( \sqrt{3} \)
- \( \frac{\sqrt{3}}{3} \)
- \( \frac{1}{3} \)
- \( \frac{\sqrt{3}}{2} \)
(SBMPTN 2013)
Pembahasan:
\begin{aligned} \lim_{x \to 0} \ \sqrt{\frac{x \tan x}{\sin^2 x - \cos 2x + 1}} &= \lim_{x \to 0} \ \sqrt{\frac{x \tan x}{\sin^2 x + 1 - \cos 2x}} \\[8pt] &= \lim_{x \to 0} \ \sqrt{\frac{x \tan x}{\sin^2 x + 2 \sin^2 x}} \\[8pt] &= \lim_{x \to 0} \ \sqrt{\frac{x \tan x}{3 \sin^2 x}} = \sqrt{\lim_{x \to 0} \ \frac{x \tan x}{3 \sin^2 x}} \\[8pt] &= \sqrt{\frac{1}{3} \cdot \lim_{x \to 0} \ \frac{x}{\sin x} \cdot \lim_{x \to 0} \ \frac{\tan x}{\sin x}} \\[8pt] &= \sqrt{\frac{1}{3} \cdot 1 \cdot 1} = \sqrt{\frac{1}{3}} = \frac{1}{3} \sqrt{3} \end{aligned}
Jawaban C.