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lim x → π 4 1 − sin 2 x 1 + cos 4 x - Mathematics

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प्रश्न

\[\lim_{x \to \frac{\pi}{4}} \frac{1 - \sin 2x}{1 + \cos 4x}\] 

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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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पाठ 29: Limits - Exercise 29.8 [पृष्ठ ६२]

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आरडी शर्मा Mathematics [English] Class 11
पाठ 29 Limits
Exercise 29.8 | Q 22 | पृष्ठ ६२

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