<s>[INST] Prove that $\sqrt{x}$ is continuous on its domain $[0, \infty).$ Prove that the function $\sqrt{x}$ is continuous on its domain $[0,\infty)$.
Proof.
Since $\sqrt{0} = 0, $ we consider the function $\sqrt{x}$ on $[a,\infty)$ where $a$ is real number and $s \neq 0.$ Let $\delta=2\sqrt{a}\epsilon.$ Then, $\forall x \in dom,$ and $\left | x-x_0\right | < \delta \Rightarrow \left| \sqrt{x}-\sqrt{x_0}\right| = \left| \frac{x-x_0}{ \sqrt{x}+\sqrt{x_0}} \right| < \left|\frac{\delta}{2\sqrt{a}}\right|=\epsilon.$
Can I do this?
 [/INST] We need to prove that for any point $a \in (0, \infty)$, for every $\varepsilon>0$ there exists a $\delta > 0$ such that $$|x-a|<\delta\implies|\sqrt{x}-\sqrt{a}|<\varepsilon.$$
So, to find a $\delta$, we turn to the inequality $|\sqrt{x}-\sqrt{a}|<\varepsilon.$ Since we want an expression involving $|x-a|$, multiply by the conjugate to remove the square roots.
$$|\sqrt{x}-\sqrt{a}|<\varepsilon\implies|\sqrt{x}-\sqrt{a}|\cdot|\sqrt{x}+\sqrt{a}|<\varepsilon\cdot|\sqrt{x}+\sqrt{a}|$$
$$|x-a|<\varepsilon\cdot |\sqrt{x}+\sqrt{a}|. \tag{1}$$
Now, if you require that $|x-a|<1$, then it follows that $x-a<1$, so $a - 1<x<a+1$, and therefore that $\sqrt{x}<\sqrt{a+1}.$ Therefore, $\sqrt{x}+\sqrt{a}<\sqrt{a+1}+\sqrt{a},$ which combined with $(1)$ tells us that $$|x-a|<\varepsilon(\sqrt{a+1}+\sqrt{a}).$$
So, let $\delta = \mathrm{min}(1,\ \varepsilon(\sqrt{a+1}+\sqrt{a}))$. This proves that $f(x) = \sqrt{x}$ is continuous on $(0, \infty)$. $\square$
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