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Lim X → 0 Log ( 1 + X ) 3 X − 1

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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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अध्याय 29: Limits - Exercise 29.1 [पृष्ठ ७१]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.1 | Q 2 | पृष्ठ ७१

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