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प्रश्न
\[\lim_{x \to 2} \frac{\sqrt{1 + 4x} - \sqrt{5 + 2x}}{x - 2}\]
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उत्तर
\[\lim_{x \to 2} \left[ \frac{\sqrt{1 + 4x} - \sqrt{5 + 2x}}{x - 2} \right]\]It is of the from \[\frac{0}{0}\]
Rationalising the numerator:
\[\lim_{x \to 2} \left[ \frac{\left( \sqrt{1 + 4x} - \sqrt{5 + 2x} \right) \left( \sqrt{1 + 4x} + \sqrt{5 + 2x} \right)}{\left( x - 2 \right) \left( \sqrt{1 + 4x} + \sqrt{5 + 2x} \right)} \right]\]
=\[\lim_{x \to 2} \left[ \frac{\left( 1 + 4x \right) - \left( 5 + 2x \right)}{\left( x - 2 \right)\left( \sqrt{1 + 4x} + \sqrt{5 + 2x} \right)} \right]\]
= \[\lim_{x \to 2} \left[ \frac{2\left( x - 2 \right)}{\left( x - 2 \right)\left( \sqrt{1 + 4x} + \sqrt{5 + 2x} \right)} \right]\]
= \[\frac{2}{\left( \sqrt{1 + 4 \times 2} + \sqrt{5 + 2 \times 2} \right)}\]
= \[\frac{2}{3 + 3}\]
= \[\frac{1}{3}\]
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