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

JEE MATHEMATICS MCQ: LIMITS

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Analyzing +4 | -1

\lim\limits_{x \to 0} \frac{{\log \left( {x + a} \right) - \log a}}{x} + k\lim\limits_{x \to e} \frac{{\log x - 1}}{{x - e}} = 1

, then the value of

k

View answer and explanation

Correct option: C

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

Option C (

e(1 - \frac{1}{a})

) is correct because it is obtained by applying L'Hopital's rule to both limits and solving for k.

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