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प्रश्न
\[\lim_{x \to \frac{\pi}{2}} \frac{\sqrt{2 - \sin x} - 1}{\left( \frac{\pi}{2} - x \right)^2}\]
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उत्तर
\[\lim_{x \to \frac{\pi}{2}} \frac{\sqrt{2 - \sin x} - 1}{\left( \frac{\pi}{2} - x \right)^2}\]
\[ = \lim_{h \to 0} \frac{\sqrt{2 - \sin \left( \frac{\pi}{2} - h \right)} - 1}{\left( \frac{\pi}{2} - \left( \frac{\pi}{2} - h \right) \right)^2}\]
\[ = \lim_{h \to 0} \frac{\sqrt{2 - \cos h} - 1}{h^2}\]
\[\text{ Dividing the numerator and thedenominator by } \sqrt{2 - \cos h} + 1: \]
\[ \lim_{h \to 0} \frac{\left( \sqrt{2 - \cos h} - 1 \right) \left( \sqrt{2 - \cos h} + 1 \right)}{\left( \sqrt{2 - \cos h} + 1 \right) h^2}\]
\[ = \lim_{h \to 0} \frac{2 - \cos h - 1}{h^2 \left( \sqrt{2 - \cos h} + 1 \right)}\]
\[ = \lim_{h \to 0} \frac{1 - \cos h}{h^2 \left[ \sqrt{2 - \cos h} + 1 \right]}\]
\[ = \lim_{h \to 0} \frac{2 \sin^2 \frac{h}{2}}{\frac{4 h^2}{4} \left( \sqrt{2 - \cos h} + 1 \right)}\]
\[ = \frac{1}{2} \lim_{h \to 0} \left( \frac{\sin \frac{h}{2}}{\frac{h}{2}} \right)^2 \times \lim_{h \to 0} \left( \frac{1}{\sqrt{2 - \cos h} + 1} \right)\]
\[ = \frac{1}{2 \left( \sqrt{2 - 1} + 1 \right)}\]
\[ = \frac{1}{4}\]
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