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Lim X → 0 √ 1 + Sin X − √ 1 − Sin X X

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प्रश्न

\[\lim_{x \to 0} \frac{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}{x}\] 

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उत्तर

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

 

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अध्याय 29: Limits - Exercise 29.7 [पृष्ठ ५१]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.7 | Q 44 | पृष्ठ ५१

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