Nilai \( \displaystyle \lim_{x\to 0} \ \frac{x^2 \tan 2x}{x-x \cos 4x} = \cdots \)
- -3/4
- -2/4
- -1/4
- 1/4
- 2/4
Pembahasan:
\begin{aligned} \lim_{x\to 0} \ \frac{x^2 \tan 2x}{x-x \cos 4x} &= \lim_{x\to 0} \ \frac{x \tan 2x}{1- \cos 2(2x)} \\[8pt] &= \lim_{x\to 0} \ \frac{x \tan 2x}{2\sin^2 2x} \\[8pt] &= \frac{1}{2} \cdot \frac{x}{\sin 2x} \cdot \frac{\tan 2x}{\sin 2x} \\[8pt] &= \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{2}{2} =\frac{1}{4} \end{aligned}
Jawaban D.