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lim x → π 2 ( π 2 − x ) sin x − 2 cos x ( π 2 − x ) + cot x - Mathematics

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प्रश्न

\[\lim_{x \to \frac{\pi}{2}} \frac{\left( \frac{\pi}{2} - x \right) \sin x - 2 \cos x}{\left( \frac{\pi}{2} - x \right) + \cot x}\]

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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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अध्याय 29: Limits - Exercise 29.8 [पृष्ठ ६३]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.8 | Q 36 | पृष्ठ ६३

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