Show that $g(x)=\sqrt{x}$ is uniformly continuous on $[0, \infty)$

Question

I must show that $g(x)=\sqrt{x}$ is uniformly continuous on $[0, \infty)$. I know we have to show that $\sqrt{a-b}\ge \sqrt{a} - \sqrt{b}$ and $\sqrt{a + b} \le \sqrt{a} + \sqrt{b}$ but I don't know how to do this.

Answer 1

Suppose $M(>0)\in\mathbb{R}$, then clearly $\sqrt{x}$ is continuous $[0,M]$ which is a closed and bounded set. Then $\sqrt{x}$ is uniform continuous on $[0,M]$.

Also $g'(x)$ is bounded on $[M,\infty)$.(by what?) Hence $g$ is uniform continuous on $[M,\infty)$. Club the two statements.

Answer 2

Since $|\sqrt{x} - \sqrt{y}| \leqslant|\sqrt{x} + \sqrt{y}| $, we have

$$|\sqrt{x} - \sqrt{y}|^2 \leqslant |\sqrt{x} - \sqrt{y}||\sqrt{x} + \sqrt{y}|= |(\sqrt{x})^2 - (\sqrt{y})^2|=|x - y|.$$

Hence, $|x-y| < \delta = \epsilon^2 \implies |\sqrt{x} - \sqrt{y}| < \epsilon$, for all $x,y \in [0,\infty).$