$f(x) = \frac{1 - \cos 4x}{8x^{2}}$ for $x \ne 0$, and $f(0) = k$. For which $k$ is $f$ continuous at $x = 0$?

Explanation

1. 01

   $1 - \cos 4x = 2\sin^{2} 2x$, so the expression is $\frac{2\sin^{2} 2x}{8x^{2}} = \frac{1}{4}\left(\frac{\sin 2x}{x}\right)^{2}$.

2. 02

   $\frac{\sin 2x}{x} \to 2$, so the limit is $\frac{1}{4} \times 4 = 1$.