$f(x) = \frac{\log(1 + 3x) - \log(1 - 2x)}{x}$ for $x \ne 0$, and the value of $f$ at $x = 0$ is $k$. For which $k$ is $f$ continuous at $x = 0$?

Explanation

1. 01

   $\frac{\log(1 + 3x)}{x} = 3 \cdot \frac{\log(1 + 3x)}{3x} \to 3$ and $\frac{\log(1 - 2x)}{x} = -2 \cdot \frac{\log(1 - 2x)}{-2x} \to -2$.

2. 02

   The limit is $3 - (-2) = 5$, so $k = 5$.