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प्रश्न
\[\lim_{x \to \frac{\pi}{2}} \frac{\left( \frac{\pi}{2} - x \right) \sin x - 2 \cos x}{\left( \frac{\pi}{2} - x \right) + \cot x}\]
उत्तर
\[\lim_{x \to \frac{\pi}{2}} \frac{\left( \frac{\pi}{2} - x \right) \sin x - 2 \cos x}{\left( \frac{\pi}{2} - x \right) + \cot x}\]
\[ = \lim_{h \to 0} \frac{\left[ \frac{\pi}{2} - \left( \frac{\pi}{2} - h \right) \right] \sin \left( \frac{\pi}{2} - h \right) - 2 \cos \left( \frac{\pi}{2} - h \right)}{\left[ \frac{\pi}{2} - \left( \frac{\pi}{2} - h \right) \right] + \cot \left( \frac{\pi}{2} - h \right)} \left( Plugging x = \frac{\pi}{2} - h \right)\]
\[ = \lim_{h \to 0} \frac{h \cos h - 2 \sin h}{h + \tan h}\]
\[\text{ Dividing the numerator and the denominator by } h:\]
\[ \lim_{h \to 0} \frac{\cos h - 2 \frac{\sin h}{h}}{1 + \frac{\tan h}{h}}\]
\[ \Rightarrow \frac{1 - 2\left( 1 \right)}{1 + 1} = \frac{- 1}{2}\]
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