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प्रश्न
\[\lim_{x \to 0} \frac{\sqrt{1 + x} - \sqrt{1 - x}}{2x}\]
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उत्तर
\[\lim_{x \to 0} \left[ \frac{\sqrt{1 + x} - \sqrt{1 - x}}{2x} \right]\] It is of the form\[\frac{0}{0} .\]
Rationalising the numerator:
\[\lim_{x \to 0} \left[ \frac{\left( \sqrt{1 + x} - \sqrt{1 - x} \right)\left( \sqrt{1 + x} + \sqrt{1 - x} \right)}{2x\left( \sqrt{1 + x} + \sqrt{1 - x} \right)} \right]\]
\[\lim_{x \to 0} \left[ \frac{\left( 1 + x \right) - \left( 1 - x \right)}{2x\left( \sqrt{1 + x} + \sqrt{1 - x} \right)} \right]\]
\[\lim_{x \to 0} \left[ \frac{2x}{2x\left( \sqrt{1 + x} + \sqrt{1 - x} \right)} \right]\]
\[\frac{1}{\sqrt{1 + 0} + \sqrt{1 - 0}}\]
\[\frac{1}{2}\]
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