lim _{x rightarrow 0} frac{log _{e}(1+x)}{3^{ | vedclass

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(C) Given limit: $\lim _{x \rightarrow 0} \frac{\log _{e}(1+x)}{3^{x}-1}$ This is in the indeterminate form $\frac{0}{0}$. Applying $L$'Hospital's rul

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$\log _{3} e$

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(C) Given limit: $\lim _{x \rightarrow 0} \frac{\log _{e}(1+x)}{3^{x}-1}$ This is in the indeterminate form $\frac{0}{0}$. Applying $L$'Hospital's rule by differentiating the numerator and denominator with respect to $x$: $= \lim _{x \rightarrow 0} \frac{\frac{d}{dx}(\log _{e}(1+x))}{\frac{d}{dx}(3^{x}-1)}$ $= \lim _{x \rightarrow 0} \frac{\frac{1}{1+x}}{3^{x} \cdot \log _{e} 3}$ Substituting $x = 0$: $= \frac{\frac{1}{1+0}}{3^{0} \cdot \log _{e} 3} = \frac{1}{1 \cdot \log _{e} 3} = \frac{1}{\log _{e} 3}$ Using the property $\frac{1}{\log _{a} b} = \log _{b} a$,we get: $= \log _{3} e$

$\lim _{x \rightarrow 0} \frac{\log _{e}(1+x)}{3^{x}-1}$ is equal to

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