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