mathop {lim }limits_{x to 0} frac{{log _e}(1 | vedclass

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(D) Given limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{\log _e}(1 + x)}{{3^x - 1}}$This is in the $\frac{0}{0}$ indeterminate form.Applying $L'	ex

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$\log _3 e$

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(D) Given limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{\log _e}(1 + x)}{{3^x - 1}}$This is in the $\frac{0}{0}$ indeterminate form.Applying $L'	ext{Hospital's rule}$ by differentiating the numerator and denominator with respect to $x$:$\mathop {\lim }\limits_{x 	o 0} \frac{\frac{d}{dx}[\log _e(1 + x)]}{\frac{d}{dx}[3^x - 1]} = \mathop {\lim }\limits_{x 	o 0} \frac{\frac{1}{1 + x}}{3^x \log _e 3}$Substituting $x = 0$:$= \frac{\frac{1}{1 + 0}}{3^0 \log _e 3} = \frac{1}{1 \cdot \log _e 3} = \frac{1}{\log _e 3}$Using the property $\frac{1}{\log _a b} = \log _b a$,we get:$= \log _3 e$.

$\mathop {\lim }\limits_{x \to 0} \frac{{\log _e}(1 + x)}{{3^x - 1}} = $

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