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प्रश्न
\[\lim_{x \to 0} \frac{\sin x}{\sqrt{1 + x} - 1} .\]
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उत्तर
\[\lim_{x \to 0} \left[ \frac{\sin x}{\sqrt{1 + x} - 1} \right]\]
\[\text{ Rationalising the denominator, we get }: \]
\[ = \lim_{x \to 0} \left[ \frac{\sin x}{\left( \sqrt{1 + x} - 1 \right)} \times \left( \frac{\sqrt{1 + x} + 1}{\sqrt{1 + x} + 1} \right) \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\sin x \times \left( \sqrt{1 + x} + 1 \right)}{1 + x - 1} \right]\]
\[ = \lim_{x \to 0} \left[ \left( \frac{\sin x}{x} \right) \left( \sqrt{1 + x} + 1 \right) \right]\]
\[ = 1 \times \left( \sqrt{1} + 1 \right)\]
\[ = 2\]
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