Explanation

(c)
$\lim_{x \rightarrow 0}{\frac{\log{(x + a) - \log a}}{x} + k\lim_{x \rightarrow e}\frac{\log{x - 1}}{x - e} = 1}$
Using L’ Hospital’s rule
$\lim_{x \rightarrow 0}\frac{\frac{1}{x + a}}{1} + k\lim_{x \rightarrow 0}\frac{\frac{1}{x}}{1} = 1$
$\Rightarrow \ \ \ \ \ \ \frac{1}{a} + \frac{k}{e} = 1$
$\Rightarrow \ \ \ \ \ \ \ k = e\left( 1 - \frac{1}{a} \right)$