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Lim X → 0 Sin X 2 ( 1 − Cos X 2 ) X 6 - Mathematics

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प्रश्न

\[\lim_{x \to 0} \frac{\sin x^2 \left( 1 - \cos x^2 \right)}{x^6}\] 

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उत्तर

\[\lim_{x \to 0} \left[ \frac{\sin x^2 \left( 1 - \cos x^2 \right)}{x^6} \right]\] 

\[= \lim_{x \to 0} \left[ \frac{\sin x^2 \times 2 \sin^2 \left( \frac{x^2}{2} \right)}{x^6} \right] \left[ \because 1 - \cos A = 2 \sin^2 \left( \frac{A}{2} \right) \right]\]
\[ = 2 \lim_{x \to 0} \left[ \frac{\sin x^2}{x^2} \times \frac{\sin \left( \frac{x^2}{2} \right)}{2 \times \frac{x^2}{2}} \times \frac{\sin \left( \frac{x^2}{2} \right)}{2 \times \frac{x^2}{2}} \right]\]
\[ = \frac{2}{2 \times 2}\]
\[ = \frac{1}{2}\]

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अध्याय 29: Limits - Exercise 29.7 [पृष्ठ ५०]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.7 | Q 17 | पृष्ठ ५०

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