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Lim X → 1 1 + Cos π X ( 1 − X ) 2 - Mathematics

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प्रश्न

\[\lim_{x \to 1} \frac{1 + \cos \pi x}{\left( 1 - x \right)^2}\] 

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उत्तर

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

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अध्याय 29: Limits - Exercise 29.8 [पृष्ठ ६२]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.8 | Q 20 | पृष्ठ ६२

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