Continuity of square root function at the first point

Question

So, as you know the continuity of a function at a point has some conditions. One of them is that the limit of this function at that point has to exist and be the same as the function's output of that point. Consider the function sqrt(x), which is defined over [0, +inf), if we take the limit of the function at x = 0, the limit would exist from the right side but not the left side. Thus, the full limit wouldn't even exist and yet it's considered continuous at x = 0. How's that?

Answer 1

The limit of a function at a point is defined with respect to its domain. If $D\subseteq\mathbb{R}$, $f:D\to\mathbb{R}$, and $a\in\mathbb{R}$, then we say $L=\lim_{x\to a}f(x)$ if for all $\epsilon>0$ there exists $\delta>0$ such that all $x\color{red}{\in D}$ with $0<|x-a|<\delta$, $|f(x)-L|<\epsilon$. Notice that this definition only involves values of $x$ that actually are in the domain $D$ of the function. So if $f(x)=\sqrt{x}$ with domain $[0,\infty)$, then $\lim_{x\to a}f(x)$ does exist. It's no problem that $f$ is not defined to the left of $0$, since the definition of the limit only cares about inputs that are in the domain of $f$.

(In fact, I would say it's not even correct to say that the limit from the left does not exist. To define the limit from the left, you would use the same definition as above but replace $0<|x-a|<\delta$ by $0<a-x<\delta$. For the case of the square root function, the condition would then hold vacuously, since there are no $x\in D$ such that $0<a-x<\delta$. So actually, it's not that the limit from the left doesn't exist--instead, $L=\lim_{x\to 0^-}\sqrt{x}$ is simultaneously true for all real numbers $L$. In particular, it is true that $0=\lim_{x\to 0^-}\sqrt{x}$, which is what continuity from the left would require. See my answer at The definition of continuity and functions defined at a single point for some related discussion.)