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Question
\[\lim_{x \to \frac{\pi}{8}} \frac{\cot 4x - \cos 4x}{\left( \pi - 8x \right)^3}\]
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Solution
\[\lim_{x \to \frac{\pi}{8}} \frac{\cot 4x - \cos 4x}{\left( \pi - 8x \right)^3}\]
\[ = \lim_{h \to 0} \left[ \frac{\cot 4\left( \frac{\pi}{8} - h \right) - \cos 4\left( \frac{\pi}{8} - h \right)}{\left( \pi - 8\left( \frac{\pi}{8} - h \right) \right)^3} \right]\]
\[ = \lim_{h \to 0} \left[ \frac{\tan 4h - \sin 4h}{\left( 8 h \right)^3} \right]\]
\[ = \lim_{h \to 0} \left[ \frac{\sin 4h - \cos 4h \sin 4h}{512 \left( \cos 4h \right) h^3} \right]\]
\[ = \lim_{h \to 0} \left[ \frac{\sin 4h \left( 1 - \cos 4h \right)}{\left( \cos 4h \right) 512 h^3} \right]\]
\[ = \lim_{h \to 0} \left[ \frac{\tan 4h}{4h} \times \frac{2 \sin^2 2h}{32 \times 4 h^2} \right]\]
\[ = \frac{1}{16} \lim_{h \to 0} \left( \frac{\tan 4h}{4h} \right) \times \lim_{h \to 0} \left( \frac{\sin 2h}{2h} \right)^2 \]
\[ = \frac{1}{16} \times 1 \times 1 = \frac{1}{16}\]
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