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Lim X → π 2 Cos 2 X 1 − Sin X

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प्रश्न

\[\lim_{x \to \frac{\pi}{2}} \frac{\cos^2 x}{1 - \sin x}\]

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उत्तर

\[\lim_{x \to \frac{\pi}{2}} \frac{\cos^2 x}{1 - \sin x}\]
\[ = \lim_{x \to \frac{\pi}{2}} \frac{1 - \sin^2 x}{1 - \sin x} \binom{ \because \cos^2 x = 1 - \sin^2 x }{ a^2 - b^2 = \left( a - b \right) \left( a + b \right)}\]
\[ = \lim_{x \to \frac{\pi}{2}} \frac{\left( 1 - \sin x \right) \left( 1 + \sin x \right)}{\left( 1 - \sin x \right)}\]
\[ = \lim_{x \to \frac{\pi}{2}} \left( 1 + \sin x \right)\]
\[ \Rightarrow 1 + 1 = 2\]

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अध्याय 29: Limits - Exercise 29.8 [पृष्ठ ६२]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.8 | Q 3 | पृष्ठ ६२

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