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प्रश्न
\[\lim_{x \to \frac{\pi}{4}} \frac{1 - \sin 2x}{1 + \cos 4x}\]
उत्तर
\[\lim_{x \to \frac{\pi}{4}} \frac{1 - \sin 2x}{1 + \cos 4x}\]
\[ = \lim_{h \to 0} \frac{1 - \sin 2\left( \frac{\pi}{4} - h \right)}{1 + \cos 4\left( \frac{\pi}{4} - h \right)}\]
\[ = \lim_{h \to 0} \frac{1 - \cos 2h}{1 - \cos 4h}\]
\[ = \lim_{h \to 0} \frac{2 \sin^2 h}{2 \sin^2 2h}\]
\[ \Rightarrow \lim_{h \to 0} \frac{\frac{\sin^2 h}{h^2}}{\frac{4 \sin^2 2h}{4 h^2}}\]
\[ \Rightarrow \frac{1}{4}\]
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