Is $\sqrt{x}$ continuous at $0$? Is $f(x) = \sqrt{x}$ is continuous at $0$?

I see that people say that $\sqrt{x}$ is continuous at the interval $[0,\infty)$. Their proof is based on that: (continuous from the right):
$$\lim_{x \to 0^+}f(x) = f(0) = 0$$
At least from what I have seen: Is function continuous at 0?
But there is no left limit, namely $\lim_{x \to 0^-}$ doesnt exists. 
And from what I know the statement is that: $f$ is continuous at $x_0$ if and only if, it continuous from the left and from the right. 
But here, there is no limit from the left. 
So how $\sqrt{x}$ is continuous at $0$ so we say it continuous at $[0,\infty)?$
 A: Everyone will agree that the function $g: \Bbb R \to \Bbb R$
$$   
    g(x) = \left\{\begin{array}{lr}
        0, & \text{for } x \lt 0 \\
       \sqrt x , & \text{for } x \ge 0
        \end{array}\right\} 
$$
is continuous.
Restrict this function to the domain $[0,+\infty)$, defining
$\tag 1 \displaystyle{f = g_{\;|\,[0,+\infty)}}$
We leave it to OP to come up with sensible terminology/definitions concerning continuity for the function $f$. But if you want the restriction of a continuous function to also be continuous (seems reasonable), you won't have much 'leg room'.
A: Since the function is not defined on the left of 0 then the notion of continuity in 0 consists only of the continuity at the right of 0.
A: There is no thing as "continuity from the right (or left)".  Referring to limits, a function $f\colon G\to\mathbb R$ is continuous in $x\in G$ iff for every sequence $(x_n)$ with elements in $G$ that converges to $x$ we can conclude that $\lim_{n\to\infty}f(x_n)=f(x)$. Therefore, e.g., if $G=\mathbb Z$, we have that $f$ is continuous on $G$.
