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प्रश्न
\[\lim_{x \to \infty} \left[ \left\{ \sqrt{x + 1} - \sqrt{x} \right\} \sqrt{x + 2} \right]\]
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उत्तर
\[\lim_{x \to \infty} \left[ \left\{ \sqrt{x + 1} - \sqrt{x} \right\}\sqrt{x + 2} \right]\]
\[\text{ Rationalising the numerator }: \]
\[ \lim_{x \to \infty} \left[ \frac{\left( \sqrt{x + 2} \right) \left\{ \sqrt{x + 1} - \sqrt{x} \right\} \left\{ \sqrt{x + 1} + \sqrt{x} \right\}}{\left( \sqrt{x + 1} + \sqrt{x} \right)} \right]\]
\[ \Rightarrow \lim_{x \to \infty} \left[ \frac{\left( \sqrt{x + 2} \right) \left( x + 1 - x \right)}{\left( \sqrt{x + 1} + \sqrt{x} \right)} \right]\]
\[\text{ Dividing the numerator and the denominator by } \sqrt{x}: \]
\[ \lim_{x \to \infty} \left[ \frac{\frac{\sqrt{x + 2}}{\sqrt{x}}}{\frac{\sqrt{x + 1} + \sqrt{x}}{\sqrt{x}}} \right]\]
\[ = \lim_{x \to \infty} \left[ \frac{\sqrt{1 + \frac{2}{x}}}{\sqrt{1 + \frac{1}{x}} + 1} \right]\]
\[ = \frac{1}{\sqrt{1} + 1}\]
\[ = \frac{1}{2}\]
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