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