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