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