How does $f(x) = \begin{cases} \frac{\sqrt{1 - \cos 2x}}{x}, & x \ne 0 \\ \sqrt{2}, & x = 0 \end{cases}$ behave at $x = 0$?

Explanation

1. 01

   $1 - \cos 2x = 2\sin^{2} x$, so $\sqrt{1 - \cos 2x} = \sqrt{2}\lvert \sin x \rvert$ and $f(x) = \frac{\sqrt{2}\lvert \sin x \rvert}{x}$.

2. 02

   From the right this tends to $\sqrt{2}$, from the left to $-\sqrt{2}$: the one-sided limits differ.