मराठी

Lim X → 0 Sin ( 2 + X ) − Sin ( 2 − X ) X

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प्रश्न

\[\lim_{x \to 0} \frac{\sin \left( 2 + x \right) - \sin \left( 2 - x \right)}{x}\]

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उत्तर

\[\lim_{x \to 0} \left[ \frac{\sin \left( 2 + x \right) - \sin \left( 2 - x \right)}{x} \right]\] 

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

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पाठ 29: Limits - Exercise 29.7 [पृष्ठ ५०]

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आर.डी. शर्मा Mathematics [English] Class 11
पाठ 29 Limits
Exercise 29.7 | Q 26 | पृष्ठ ५०

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