<s>[INST] Proof Verification: Prove $\sqrt{x}$ is uniformly continuous on [0, $\infty$) Proof:
Fix $\epsilon \gt 0$. We want to find $\delta \gt 0$ such that:
$$|x-a| \lt \delta \Rightarrow |\sqrt{x} - \sqrt{a}| \lt \epsilon$$
$$\phantom{2000i11111}\Rightarrow |\frac{x-a}{\sqrt{x}+\sqrt{a}}| \lt \epsilon$$
If $|x-a|\lt \delta$ then $|\frac{x-a}{\sqrt{x}+\sqrt{a}}| \lt \frac{\delta}{\sqrt{a+1}+\sqrt{a}}$ for $\delta \lt 1$
Thus, choose $\delta = min\{1, (\sqrt{a+1}+\sqrt{a})\epsilon\}$. Then we have:
$$|x-a| \lt \delta \Rightarrow |\sqrt{x} - \sqrt{a}|=|\frac{x-a}{\sqrt{x}+\sqrt{a}}| \lt \frac{\delta}{\sqrt{a+1}+\sqrt{a}}=\frac{(\sqrt{a+1}+\sqrt{a})\epsilon}{\sqrt{a+1}+\sqrt{a}}=\epsilon$$
 [/INST] Simply, if $|x - a| < \delta(\epsilon) = \epsilon^2$, then, since $|\sqrt{x} - \sqrt{a}| \leqslant |\sqrt{x} + \sqrt{a}|$, we have  
$$|\sqrt{x} - \sqrt{a}|^2 = |\sqrt{x}-\sqrt{a}|\, |\sqrt{x}-\sqrt{a}|\leqslant |\sqrt{x}-\sqrt{a}|\, |\sqrt{x}+\sqrt{a}| = |x - a|< \epsilon^2\\ \implies|\sqrt{x} - \sqrt{a}|< \epsilon$$
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