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प्रश्न
\[\lim_{x \to 2} \frac{\sqrt{x^2 + 1} - \sqrt{5}}{x - 2}\]
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उत्तर
\[\lim_{x \to 2} \left[ \frac{\sqrt{x^2 + 1} - \sqrt{5}}{x - 2} \right]\] It is of the form \[\frac{0}{0}\]
Rationalising the numerator:
\[\lim_{x \to 2} \left[ \frac{\left( \sqrt{x^2 + 1} - \sqrt{5} \right)\left( \sqrt{x^2 + 1} + \sqrt{5} \right)}{\left( x - 2 \right)\left( \sqrt{x^2 + 1} + \sqrt{5} \right)} \right]\]
= \[\lim_{x \to 2} \left[ \frac{x^2 + 1 - 5}{\left( x - 2 \right)\left( \sqrt{x^2 + 1} + \sqrt{5} \right)} \right]\]
= \[\lim_{x \to 2} \left[ \frac{x^2 - 4}{\left( x - 2 \right)\left( \sqrt{x^2 + 1} + \sqrt{5} \right)} \right]\]
= \[\lim_{x \to 2} \left[ \frac{\left( x - 2 \right)\left( x + 2 \right)}{\left( x - 2 \right)\left( \sqrt{x^2 + 1} + \sqrt{5} \right)} \right]\]
= \[\frac{4}{2\sqrt{5}}\]
= \[\frac{2}{\sqrt{5}}\]
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