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?

Formal Proof:
Let $\epsilon > 0$, $a = 0$. Choose $\delta = \epsilon^2$. Then $\left |x \right| < \delta$ implies $\left | \sqrt{x} \right | < \epsilon$, as desired.
Let $\epsilon > 0$, $a \in (0,\infty)$. Choose $\delta = \sqrt{a}\epsilon$. Then $\left |x - a \right | < \delta$ implies $\left | \sqrt{x} - \sqrt{a} \right | < \frac{\delta}{\left | \sqrt{x} + \sqrt{a} \right |} \leq \epsilon$, as desired.
Thus $f$ is continuous at $x = a$ for all $a \in [0, \infty)$.
Explanation:
The first part is more obvious in its design. We would like to arrive at the form $\left | \sqrt{x} \right | < \epsilon$, so we notice we may square both sides to reach $\left | x \right | < \epsilon^2$, meaning we may choose $\delta = \epsilon^2$, as these steps may be undone.
For the second part, we would like to arrive at the form $\left | \sqrt{x} - \sqrt{a} \right | < \epsilon$ from $\left |x - a\right| < \delta$. We note $x - a = (\sqrt{x} - \sqrt{a})(\sqrt{x}+\sqrt{a})$, so whatever $\delta$ we choose, we will arrive at $\left | \sqrt{x} - \sqrt{a} \right | < \frac{\delta}{\left | \sqrt{x} + \sqrt{a} \right |}$, and we wish this to be less than or equal to $\epsilon$. We cannot choose $\delta$ immediately because our expression still depends on $x$, but we may easily remedy this. Noting that $\sqrt{a} \neq 0$, $\left | \sqrt{x} + \sqrt{a} \right | \geq \left | \sqrt{a} \right |$, meaning $\frac{\delta}{\left | \sqrt{x} + \sqrt{a} \right |} \leq \frac{\delta}{\left | \sqrt{a} \right |}$. This means $\delta = \sqrt{a}\epsilon$ is a good choice, as it is only in terms of the given point $a$ and $\epsilon$.
The proof was split into two cases, $a = 0$ and $a > 0$, because $a = 0$ is rather easy to prove, and the proof of the second part would have been problematic if $a$ were allowed to be $0$.
