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