मराठी

P Lim X → 0 Sin 3 X − Sin X Sin X

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प्रश्न

\[\lim_{x \to 0} \frac{\sin 3x - \sin x}{\sin x}\] 

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उत्तर

\[\lim_{x \to 0} \left[ \frac{\sin 3x - \sin x}{\sin x} \right]\]

\[= \lim_{x \to 0} \left[ \frac{2 \cos\left( \frac{3x + x}{2} \right) \sin\left( \frac{3x - x}{2} \right)}{\sin x} \right] \left[ \because \sin C - \sin D = 2\cos\left( \frac{C + D}{2} \right)\sin\left( \frac{C - D}{2} \right) \right]\]
\[ = \lim_{x \to 0} \left[ \frac{2 \cos2x . \sin x}{\sin x} \right]\]
\[ = 2\cos0\]
\[ = 2\]

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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 22 | पृष्ठ ५०

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

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