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