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$\lim_{x \to 0} \frac{x\cot(4x)}{\sin^2 x \cot^2(2x)}$ is equal to:
(1) $0$
(2) $2$
(3) $4$
(4) $1$
$= \lim_{x\to 0}\frac{x \cdot \tan^2 2x}{\sin^2 x \cdot \tan 4x} = \lim_{x\to 0}\frac{x}{\sin x}\left(\frac{\tan 2x}{2x}\right)^2 \left(\frac{4x}{\tan 4x}\right) \cdot \frac{4}{4^2} \cdot 4 = 1$.
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