$f(x) = \frac{e^{3x} - 1}{x}$ for $x \ne 0$, and $f(0) = k$. For which $k$ is $f$ continuous at $x = 0$?

Explanation

1. 01

   $\frac{e^{3x} - 1}{x} = 3 \cdot \frac{e^{3x} - 1}{3x}$.

2. 02

   $\frac{e^{t} - 1}{t} \to 1$, so the limit is 3.