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प्रश्न
\[\lim_{x \to 1} \frac{\sqrt{x^2 - 1} + \sqrt{x - 1}}{\sqrt{x^2 - 1}}, x > 1\]
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उत्तर
\[\lim_{x \to 1} \left[ \frac{\sqrt{x^2 - 1} + \sqrt{x - 1}}{\sqrt{x^2 - 1}} \right]\]
\{[\text{ It is of the form } \frac{0}{0} . \]
\[ \lim_{x \to 1} \left[ \frac{\sqrt{x^2 - 1}}{\sqrt{x^2 - 1}} + \frac{\sqrt{x - 1}}{\sqrt{x^2 - 1}} \right]\]
\[ = \lim_{x \to 1} \left[ 1 + \frac{\sqrt{x - 1}}{\left( \sqrt{x - 1} \right)\left( \sqrt{x + 1} \right)} \right]\]
\[ = 1 + \frac{1}{\sqrt{1 + 1}}\]
\[ = 1 + \frac{1}{\sqrt{2}}\]
\[ = \frac{\sqrt{2} + 1}{\sqrt{2}}\]
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