If mathop {lim }limits_{x to 0} frac{{log (a | vedclass

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(A) Let $f(x) = \log x$. Then $f'(x) = \frac{1}{x}$.The first limit is $\mathop {\lim }\limits_{x 	o 0} \frac{{\log (a + x) - \log a}}{x} = f'(a) = \

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$k = e\left( {1 - \frac{1}{a}} \right)$

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(A) Let $f(x) = \log x$. Then $f'(x) = \frac{1}{x}$.The first limit is $\mathop {\lim }\limits_{x 	o 0} \frac{{\log (a + x) - \log a}}{x} = f'(a) = \frac{1}{a}$.The second limit is $\mathop {\lim }\limits_{x 	o e} \frac{{\log x - 1}}{{x - e}}$. Since $\log e = 1$,this is $\mathop {\lim }\limits_{x 	o e} \frac{{\log x - \log e}}{{x - e}} = f'(e) = \frac{1}{e}$.Substituting these into the given equation: $\frac{1}{a} + k \cdot \frac{1}{e} = 1$.$\frac{k}{e} = 1 - \frac{1}{a} = \frac{a - 1}{a}$.Therefore,$k = e \left( \frac{a - 1}{a} ight) = e \left( 1 - \frac{1}{a} ight)$.

If $\mathop {\lim }\limits_{x \to 0} \frac{{\log (a + x) - \log a}}{x} + k\mathop {\lim }\limits_{x \to e} \frac{{\log x - 1}}{{x - e}} = 1$,then

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