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प्रश्न
\[\lim_{x \to 0} \frac{\log \left( 1 + x \right)}{3^x - 1}\]
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उत्तर
\[\lim_{x \to 0} \left[ \frac{\log \left( 1 + x \right)}{3^x - 1} \right]\]
Dividing the numerator and the denominator by x:
\[\lim_{x \to 0} \left[ \frac{\log \left( 1 + x \right)}{x \cdot \left( \frac{3^x - 1}{x} \right)} \right] \left[ \because \lim_{x \to 0} \frac{\log \left( 1 + x \right)}{x} = 1 \lim_{x \to 0} \left( \frac{a^x - 1}{x} \right) = \log a \right]\]
\[ = \frac{1}{\log 3}\]
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