Discussion
\(\lim_{x \to 0} \frac{1 - \cos 4x}{\tan^{2}2x} =\)
Answer: D
The limit is 2.
Use \(1 - \cos 4x = 2\sin^{2}2x\):
\[ \begin{aligned} \frac{1 - \cos 4x}{\tan^{2}2x} \\ &= \frac{2\sin^{2}2x}{\tan^{2}2x} \\ &= 2\cos^{2}2x \to 2 \end{aligned} \]
as \(x \to 0\). Equivalently, \(1 - \cos 4x \approx 8x^{2}\) and \(\tan^{2}2x \approx 4x^{2}\), whose ratio is 2. \(\frac{1}{2}\) inverts the ratio, and 0 or 3 come from mismatched small-angle approximations.
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