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प्रश्न
\[\lim_{x \to 0} \frac{5^x - 1}{\sqrt{4 + x} - 2}\]
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उत्तर
\[\lim_{x \to 0} \left[ \frac{5^x - 1}{\sqrt{4 + x} - 2} \right]\]
Rationalising the denominator, we get:
\[ = \lim_{x \to 0} \left[ \frac{\left( 5^x - 1 \right) \left( \sqrt{4 + x} + 2 \right)}{4 + x - 4} \right]\]
\[ = \lim_{x \to 0} \left[ \left( \frac{5^x - 1}{x} \right) \left( \sqrt{4 + x} + 2 \right) \right] \left\{ \because \lim_{x \to 0} \left( \frac{a^x - 1}{x} \right) = \log a \right\}\]
\[ = \log 5 \times \left( \sqrt{4 + 0} + 2 \right)\]
\[ = 4 \log 5\]
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