Is $\lim_{x\rightarrow 0}\sqrt x$ well-defined? I have a limit, $\sqrt x$, in which I consider only values of $x\in\mathbb{R}$. If I then consider values of $x$ approaching zero, I know that:
$$\lim_{x\rightarrow 0^{-}}\sqrt x$$
does not exist, and that:
$$\lim_{x\rightarrow 0^{+}}\sqrt x$$
does exist, so I can then conclude that the limit does not exist and I understand that, but does this mean that the limit is not well defined as $x$ approaches $0$ as well? 
Thanks.
 A: For the function $f(x)=\sqrt{x}$, unless specified otherwise, we regard it as a function from $[0,\infty)$ to $\mathbb{R}$.

Since the domain is $[0,\infty)$, we can't approach $0$ from the left, hence in this context, since $0$ is a left endpoint of the domain, the phrase "$x$ approaches $0$" means "$x$ approaches $0$ from the right".

Hence we have
$$\lim_{x\to 0}\sqrt{x}=\lim_{x\to 0^{+}}\sqrt{x}=0$$.
A: If $f(x)=\sqrt{x}$, we consider $f(0)$ something called an endpoint discontinuity. How can it be discontinuous, though? Remember the limit definition of continuity:
$f(0)$ is continuous iff:
$$\lim_{x\to 0^-} \sqrt{x} = \lim_{x\to 0^+} \sqrt{x}$$
$$\lim_{x \to 0}=f(0)$$
$f(0)$ fails the first part of the test and the limit does not exist because you can't approach it from the left. If you want, you could prove this with the formal definition of a limit, but we could also use intuition. In your question, you said we are only looking at values of $x \in \mathbb R$, therefore we can have no value less than $0$ (for example, $\sqrt{-1}$) and the two-sided limit does not exist, therefore an endpoint discontinuity exists at $f(0)$.
