$f(x) = \frac{\sin 5x}{2x}$ for $x \ne 0$, and $f(0) = k$. For which $k$ is $f$ continuous at $x = 0$?

Explanation

1. 01

   $\frac{\sin 5x}{2x} = \frac{5}{2} \cdot \frac{\sin 5x}{5x}$.

2. 02

   As $x \to 0$, $\frac{\sin 5x}{5x} \to 1$, so the limit is $\frac{5}{2}$.