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