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प्रश्न
\[\lim_{x \to 1} \frac{\sqrt{3 + x} - \sqrt{5 - x}}{x^2 - 1}\]
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उत्तर
\[\lim_{x \to 1} \left[ \frac{\sqrt{3 + x} - \sqrt{5 - x}}{x^2 - 1} \right]\] It is of the form \[\frac{0}{0}\]
Rationalising the numerator:
\[\lim_{x \to 1} \left[ \frac{\left( \sqrt{3 + x} - \sqrt{5 - x} \right)\left( \sqrt{3 + x} + \sqrt{5 - x} \right)}{\left( x - 1 \right)\left( x + 1 \right)\left( \sqrt{3 + x} + \sqrt{5 - x} \right)} \right]\]
= \[\lim_{x \to 1} \left[ \frac{\left( 3 + x \right) - \left( 5 - x \right)}{\left( x - 1 \right)\left( x + 1 \right)\left\{ \sqrt{3 + x} + \sqrt{5 - x} \right\}} \right]\]
= \[\lim_{x \to 1} \left[ \frac{2x - 2}{\left( x - 1 \right)\left( x + 1 \right) \left\{ \sqrt{3 + x} + \sqrt{5 - x} \right\}} \right]\]
= \[\lim_{x \to 1} \left[ \frac{2\left( x - 1 \right)}{\left( x - 1 \right)\left( x + 1 \right)\left\{ \sqrt{3 + x} + \sqrt{5 - x} \right\}} \right]\]
= \[\frac{2}{2 \left( \sqrt{4} + \sqrt{4} \right)}\]
= \[\frac{1}{4}\]
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