Continuity Condition

A function $f(x)$ is continuous at $x = 0$ if the limit of the function as $x$ approaches 0 equals the function's value at $x = 0$. Therefore, we must have:

$ \lim_{x \to 0} f(x) = f(0) $

Function Value at x = 0

According to the function definition:

$ f(0) = k $

Limit Calculation

Now, we calculate the limit of $f(x)$ as $x$ approaches 0:

$ \lim_{x \to 0} \frac{\log_e \left(1 + \frac{x}{a}\right) - \log_e \left(1 - \frac{x}{b}\right)}{x} $

This limit is of the indeterminate form $\frac{0}{0}$. We use the standard limit result: $\lim_{y \to 0} \frac{\log_e(1+y)}{y} = 1$.

Separate the terms and adjust them to fit the standard limit form:

$ \lim_{x \to 0} \left( \frac{\log_e \left(1 + \frac{x}{a}\right)}{x} - \frac{\log_e \left(1 - \frac{x}{b}\right)}{x} \right) $

Rewrite the expression:

$ \lim_{x \to 0} \left( \frac{1}{a} \cdot \frac{\log_e \left(1 + \frac{x}{a}\right)}{\frac{x}{a}} - \left(-\frac{1}{b}\right) \cdot \frac{\log_e \left(1 - \frac{x}{b}\right)}{-\frac{x}{b}} \right) $

Let $y = \frac{x}{a}$ and $z = -\frac{x}{b}$. As $x \to 0$, both $y \to 0$ and $z \to 0$.

Substitute these into the limit expression:

$ \frac{1}{a} \lim_{y \to 0} \frac{\log_e(1+y)}{y} + \frac{1}{b} \lim_{z \to 0} \frac{\log_e(1+z)}{z} $

Apply the standard limit value (1):

$ \left(\frac{1}{a} \times 1\right) + \left(\frac{1}{b} \times 1\right) = \frac{1}{a} + \frac{1}{b} $

Determining the Value of k

For continuity at $x = 0$, we set the limit equal to $f(0)$:

$ k = \frac{1}{a} + \frac{1}{b} $