Nilai \( \displaystyle \lim_{x \to 0} \ \frac{\sqrt{1-\cos 4x^2}}{1-\cos 2x} = \cdots \)
- 1
- \( \sqrt{2} \)
- 2
- \( 2 \sqrt{2} \)
- 4
(UM UGM 2019)
Pembahasan:
\begin{aligned} \lim_{x \to 0} \ \frac{\sqrt{1-\cos 4x^2}}{1-\cos 2x} &= \lim_{x \to 0} \ \frac{\sqrt{1-\cos (2 \cdot 2x^2)}}{2\sin^2 x} \\[8pt] &= \lim_{x \to 0} \ \frac{\sqrt{2\sin^2 2x^2}}{2\sin^2 x} \\[8pt] &= \lim_{x \to 0} \ \frac{\sqrt{2} \ \sin 2x^2}{2\sin^2 x} \cdot \frac{x^2}{x^2} \\[8pt] &= \frac{\sqrt{2}}{2} \cdot \lim_{x \to 0} \ \frac{\sin 2x^2}{x^2} \cdot \lim_{x \to 0} \ \frac{x^2}{\sin^2 x} \\[8pt] &= \frac{\sqrt{2}}{2} \cdot 2 \cdot (1)^2 = \sqrt{2} \end{aligned}
Jawaban C.