Nilai \( \displaystyle \lim_{x \to \frac{\pi}{8}} \ \frac{\sin^2 2x - \cos^2 2x}{\sin 2x - \cos 2x} = \cdots \)
- \( -\sqrt{2} \)
- \( -\frac{1}{2}\sqrt{2} \)
- \( 0 \)
- \( \frac{1}{2}\sqrt{2} \)
- \( \sqrt{2} \)
Pembahasan:
\begin{aligned} \lim_{x \to \frac{\pi}{8}} \ \frac{\sin^2 2x - \cos^2 2x}{\sin 2x - \cos 2x} &= \lim_{x \to \frac{\pi}{8}} \ \frac{(\sin 2x - \cos 2x)(\sin 2x + \cos 2x)}{\sin 2x - \cos 2x} \\[8pt] &= \lim_{x \to \frac{\pi}{8}} \ (\sin 2x + \cos 2x) \\[8pt] &= \sin \left(2 \cdot \frac{\pi}{8}\right) + \cos \left(2 \cdot \frac{\pi}{8}\right) \\[8pt] &= \sin \frac{\pi}{4} + \cos \frac{\pi}{4} \\[8pt] &= \frac{1}{2} \sqrt{2} + \frac{1}{2} \sqrt{2} = \sqrt{2} \end{aligned}
Jawaban E.