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 $a \neq 0.$ Let $\delta=2\sqrt{a}\varepsilon.$ Then, $\forall x \in \mathit{dom},$ and $\left | x-x_0\right | < \delta \Rightarrow \left| \sqrt{x}-\sqrt{x_0}\right| = \left| \frac{x-x_0}{ \sqrt{x}+\sqrt{x_0}} \right| < \left|\frac{\delta}{2\sqrt{a}}\right|=\varepsilon.$
Can I do this?
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 $(0, \infty)$. $\square$
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 $0$.
Prove first the following classical theorem:
If $I$ is an interval (bounded or unbounded) and $f:I\to\mathbb{R}$ is one-to-one and continuous and $J = \{f(x): x\in I\}$, then the function $f^{-1}:J\to I$ is continuous
Hint to proof: observe that $f$ must be monotonic (it is a consequence of intermediate value theorem for continuous functions); the rest it is easy.
Later, apply this result to the function $f(x)=x^2$, restricted to the half-line $[0,+\infty)$.
I think you mean a≠0. As you have it written now, you still have to show $\sqrt{x}$ is continuous on $[0,a)$, but you are on the right track. Consider when $x=0$, let $\delta = \epsilon^2$ and it follows. Then, for $x>0$, at every point $x_0 \geq x$, $\sqrt{x}>0$, so $\frac{1}{\sqrt{x}+\sqrt{x_0}}<1$, then your inequality follows.
Prove that $\sqrt{x}$ is continuous on its domain $[0,\infty[$
I have used the following definition and therom:
A sequence of real numbers {$x_n$} is said to converge to a real number $a \in R$ if and only if for every $\epsilon>0$ there is an N $\in$ N such that
$n \geq N$ implies $|x_n-a|<\epsilon$
Suppose that E is a nonempty subset of R, that a $\in$ E, and that f: E$\rightarrow$ R. Then the following statemants are equivalent:
If $x_n \rightarrow a$ implies that $f(x_n) \rightarrow f(a)$ as $n\rightarrow \infty$ this means that the function f must be continuous. (This is just another way of stating the $\epsilon$ and $\delta$ statement)
Supose that as $n \rightarrow \infty$ then $x_n \rightarrow x$ (x plays the role of a)
We need to prove that if $x_n \rightarrow x$ then $\sqrt{x_n} \rightarrow \sqrt{x}$ as $n \rightarrow \infty$. If this is true then $\sqrt{x}$ is continuous.
(We cannot devide by zero, so we have to split up the proof, which will be clear in case 2.)
If x=0. We let $\epsilon > 0$ and choose N such that $n \geq N$ implies $|x_n-x|<\epsilon^2 \Leftrightarrow \sqrt{x_n-x} = \sqrt{x_n}<\epsilon$ for $n \geq N$.
So $\sqrt{x_n} \rightarrow 0$ when $n \rightarrow \infty$
(Remember that $\epsilon$ can be made aberitaryly small, but given a large enough N, $\sqrt{x_n}$ will be smaller thus it aproaches 0.)
$f(x_n)-f(x)=\sqrt{x_n}-\sqrt{x}=(\sqrt{x_n}-\sqrt{x})\cdot (\frac{\sqrt{x_n}+\sqrt{x}}{\sqrt{x_n}+\sqrt{x}})=\frac{|x_n-x|}{\sqrt{x_n}+\sqrt{x}}$
$\sqrt{x_n}\geq 0$ so we can write
$\sqrt{x_n}-\sqrt{x} \leq \frac{|x_n-x|}{\sqrt{x}}$
When $n \rightarrow 0$ then $x_n \rightarrow x$ meaning the righthandsite converges to 0.
$\sqrt{x_n}-\sqrt{x} \leq 0$
So it follows from the Squeeze Therom that $\sqrt{x_n} \rightarrow \sqrt{x}$ as $n \rightarrow \infty$
And therefore $\sqrt{x}$ is continuous on the domain [0,$\infty$[
