\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\log (a+x)-\log a}{x}+k\underset{x\to e}{\mathop{\lim }}\,\frac{\log x-1}{x-e}=1,\]then [IIT Screening]

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

\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\log (a+x)-\log a}{x}+k\und

1.	\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\log (a+x)-\log a}{x}+k\underset{x\to e}{\mathop{\lim }}\,\frac{\log x-1}{x-e}=1,\]then [IIT Screening]
A.	\[k=e\left( 1-\frac{1}{a} \right)\]
B.	\[k=e(1+a)\]
C.	\[k=e(2-a)\]
D.	The equality is not possible
Answer» B. \[k=e(1+a)\]

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\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\log (a+x)-\log a}{x}+k\underset{x\to e}{\mathop{\lim }}\,\frac{\log x-1}{x-e}=1,\]then [IIT Screening]

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