$x \mapsto \sqrt{x}$ on $[0, \infty)$
Terms
| Term | This example | Why |
|---|---|---|
| domain | $[0, \infty)$ | |
| continuous | ✓ | |
| uniformly continuous | ✓ | $x \mapsto \sqrt{x}$ is uniformly continuous, and so continuous, because $\abs{\sqrt{x} - \sqrt{y}} \leq \sqrt{\abs{x - y}}$, so the choice $\delta = \varepsilon^2$ works everywhere. |