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प्रश्न
\[\lim_{x \to \sqrt{6}} \frac{\sqrt{5 + 2x} - \left( \sqrt{3} + \sqrt{2} \right)}{x^2 - 6}\]
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उत्तर
\[\lim_{x \to \sqrt{6}} \left[ \frac{\sqrt{5 + 2x} - \left( \sqrt{3} + \sqrt{2} \right)}{x^2 - 6} \right]\]
= \[\lim_{x \to \sqrt{6}} \left[ \frac{\sqrt{5 + 2x} - \sqrt{\left( \sqrt{3} + \sqrt{2} \right)^2}}{x^2 - \left( \sqrt{6} \right)^2} \right]\]
= \[\lim_{x \to \sqrt{6}} \left[ \frac{\sqrt{5 + 2x} - \sqrt{3 + 2 + 2\sqrt{6}}}{\left( x - \sqrt{6} \right)\left( x + \sqrt{6} \right)} \right]\]
= \[\lim_{x \to \sqrt{6}} \left[ \frac{\sqrt{5 + 2x} - \sqrt{5 + 2\sqrt{6}}}{\left( x - \sqrt{6} \right)\left( x + \sqrt{6} \right)} \right]\]
Rationalising the numerator:
\[\lim_{x \to \sqrt{6}} \left[ \frac{\left( \sqrt{5 + 2x} - \sqrt{5 + 2\sqrt{6}} \right)\left( \sqrt{5 + 2x} + \sqrt{5 + 2\sqrt{6}} \right)}{\left( x - \sqrt{6} \right)\left( x + \sqrt{6} \right)\left( \sqrt{5 + 2x} + \sqrt{5 + 2\sqrt{6}} \right)} \right]\]
= \[\lim_{x \to \sqrt{6}} \left[ \frac{\left( 5 + 2x \right) - \left( 5 + 2\sqrt{6} \right)}{\left( x - \sqrt{6} \right)\left( x + \sqrt{6} \right)\left( \sqrt{5 + 2x} + \sqrt{5 + 2\sqrt{6}} \right)} \right]\]
= \[\lim_{x \to \sqrt{6}} \left[ \frac{2\left( x - \sqrt{6} \right)}{\left( x - \sqrt{6} \right)\left( x + \sqrt{6} \right)\left( \sqrt{5 + 2x} + \sqrt{5 + 2\sqrt{6}} \right)} \right]\]
= \[\frac{2}{\left( \sqrt{6} + \sqrt{6} \right)\left( \sqrt{5 + 2\sqrt{6}} + \sqrt{5 + 2\sqrt{6}} \right)}\]
= \[\frac{1}{2\sqrt{6}\left( \sqrt{\left( \sqrt{3} + \sqrt{2} \right)^2} \right)}\]
= \[\frac{1}{2\sqrt{6}\left( \sqrt{3} + \sqrt{2} \right)}\]
= \[\frac{1}{2\sqrt{6}\left( \sqrt{3} + \sqrt{2} \right)} \times \frac{\left( \sqrt{3} - \sqrt{2} \right)}{\left( \sqrt{3} - \sqrt{2} \right)}\]
= \[\frac{\sqrt{3} - \sqrt{2}}{2\sqrt{6}\left( 3 - 2 \right)}\]
= \[\frac{\sqrt{3} - \sqrt{2}}{2\sqrt{6}}\]
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