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