Nilai \( \displaystyle \lim_{x\to 45^0} \ \frac{\cos 2x}{1-\tan x} = \cdots \)
- 1
- 2
- 3
- 4
- 5
Pembahasan:
\begin{aligned} \lim_{x\to 45^0} \ \frac{\cos 2x}{1-\tan x} &= \lim_{x\to 45^0} \ \frac{\cos^2 x - \sin^2 x}{\frac{\cos x - \sin x}{\cos x}} \\[8pt] &= \lim_{x\to 45^0} \ \frac{(\cos^2 x - \sin^2 x) \cos x}{\cos x - \sin x} \\[8pt] &= \lim_{x\to 45^0} \ \frac{(\cos x - \sin x) (\cos x + \sin x) \cos x}{\cos x - \sin x} \\[8pt] &= \lim_{x\to 45^0} \ (\cos x + \sin x) \cos x \\[8pt] &= (\cos 45^0 + \sin 45^0) \cos 45^0 \\[8pt] &= \left( \frac{1}{2} \sqrt{2} + \frac{1}{2} \sqrt{2} \right) \frac{1}{2} \sqrt{2} \\[8pt] &= 1 \end{aligned}
Jawaban A.