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प्रश्न
\[\lim_{x \to \infty} \left[ \sqrt{x}\left\{ \sqrt{x + 1} - \sqrt{x} \right\} \right]\]
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उत्तर
\[\lim_{x \to \infty} \left[ \sqrt{x}\left\{ \sqrt{x + 1} - \sqrt{x} \right\} \right]\]
\[ = \lim_{x \to \infty} \left[ \sqrt{x}\left\{ \left( \sqrt{x + 1} - \sqrt{x} \right)\frac{\sqrt{x + 1} + \sqrt{x}}{\left( \sqrt{x + 1} + \sqrt{x} \right)} \right\} \right]\]
\[ = \lim_{x \to \infty} \left[ \frac{\sqrt{x}\left\{ \left( x + 1 \right) - x \right\}}{\left( \sqrt{x + 1} + \sqrt{x} \right)} \right]\]
Dividing the numerator and the denominator by\[\sqrt{x}\]
\[\lim_{x \to \infty} \left[ \frac{1}{\frac{\sqrt{x + 1} + \sqrt{x}}{\sqrt{x}}} \right]\]
\[ = \lim_{x \to \infty} \left[ \frac{1}{\sqrt{1 + \frac{1}{x}} + 1} \right]\]
\[As x \to \infty , \frac{1}{x} \to 0\]
\[ = \frac{1}{\sqrt{1 + 0} + 1}\]
\[ = \frac{1}{2}\]
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