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