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Question
\[\lim_\theta \to 0 \frac{1 - \cos 4\theta}{1 - \cos 6\theta}\]
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Solution
\[\lim_\theta \to 0 \left[ \frac{1 - \cos \left( 4\theta \right)}{1 - \cos 6\theta} \right]\]
\[ = \lim_\theta \to 0 \left[ \frac{2 \sin^2 2\theta}{2 \sin^2 3\theta} \right] \left\{ \because 1 - \cos A = 2 \sin^2 \left( \frac{A}{2} \right) \right\}\]
\[ = \lim_\theta \to 0 \left[ \frac{\sin^2 2\theta}{\left( 2\theta \right)^2} \times \frac{\left( 2\theta \right)^2}{\frac{\sin^2 3\theta}{\left( 3\theta \right)^2} \times \left( 3\theta \right)^2} \right]\]
\[ = \lim_\theta \to 0 \left[ \left( \frac{\sin 2\theta}{2\theta} \right)^2 \times \left( \frac{3\theta}{\sin 3\theta} \right)^2 \times \frac{4}{9} \right]\]
\[ = \frac{4}{9} \left[ \because \lim_{X \to 0} \frac{\sin x}{x} = 1 \right]\]
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