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प्रश्न
Evaluate the following limit:
\[\lim_{x \to \pi} \frac{1 - \sin\frac{x}{2}}{\cos\frac{x}{2}\left( \cos\frac{x}{4} - \sin\frac{x}{4} \right)}\]
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उत्तर
\[\lim_{x \to \pi} \frac{1 - \sin\frac{x}{2}}{\cos\frac{x}{2}\left( \cos\frac{x}{4} - \sin\frac{x}{4} \right)}\]
Put
\[x = \pi + h\] When \[x \to \pi, h \to 0\]
\[\therefore \lim_{x \to \pi} \frac{1 - \sin\frac{x}{2}}{\cos\frac{x}{2}\left( \cos\frac{x}{4} - \sin\frac{x}{4} \right)}\]
\[ = \lim_{h \to 0} \frac{1 - \sin\left( \frac{\pi + h}{2} \right)}{\cos\left( \frac{\pi + h}{2} \right)\left[ \cos\left( \frac{\pi + h}{4} \right) - \sin\left( \frac{\pi + h}{4} \right) \right]}\]
\[ = \lim_{h \to 0} \frac{1 - \sin\left( \frac{\pi}{2} + \frac{h}{2} \right)}{\cos\left( \frac{\pi}{2} + \frac{h}{2} \right)\left[ \cos\left( \frac{\pi}{4} + \frac{h}{4} \right) - \sin\left( \frac{\pi}{4} + \frac{h}{4} \right) \right]}\]
\[ = \lim_{h \to 0} \frac{1 - \cos\left( \frac{h}{2} \right)}{- \sin\left( \frac{h}{2} \right)\left[ \left( \cos\frac{\pi}{4}\cos\frac{h}{4} - \sin\frac{\pi}{4}\sin\frac{h}{4} \right) - \left( \sin\frac{\pi}{4}\cos\frac{h}{4} + \cos\frac{\pi}{4}\sin\frac{h}{4} \right) \right]}\]
\[= \lim_{h \to 0} \frac{2 \sin^2 \frac{h}{4}}{- 2\sin\frac{h}{4}\cos\frac{h}{4}\left( \frac{1}{\sqrt{2}}\cos\frac{h}{4} - \frac{1}{\sqrt{2}}\sin\frac{h}{4} - \frac{1}{\sqrt{2}}\cos\frac{h}{4} - \frac{1}{\sqrt{2}}\sin\frac{h}{4} \right)}\]
\[ = \lim_{h \to 0} \frac{\sin\frac{h}{4}}{- \cos\frac{h}{4} \times \left( - \sqrt{2}\sin\frac{h}{4} \right)}\]
\[ = \frac{1}{\sqrt{2}} \times \frac{1}{\lim_{h \to 0} \cos\frac{h}{4}}\]
\[ = \frac{1}{\sqrt{2}} \times 1 \left( \lim_\theta \to 0 \cos\theta = 1 \right)\]
\[ = \frac{1}{\sqrt{2}}\]
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