Negative $x$ values are not in the domain. Shall we say that it does not exist because left hand limit does not exist or it is $0$ because (-ve) values are not in the domain, then do not talk about it?
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$\lim_{x\to 0}\sqrt{x}$ is indeed not well defined on $\mathbb R$, but it is well defined on $[0,\infty )$. – Surb Oct 26 '20 at 10:18
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1See this question. While the title addresses a different question, your question is also addressed there. – N. F. Taussig Oct 26 '20 at 10:20
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See also the related OP. – user Oct 26 '20 at 10:41
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Implicitly, $x$ tends to $0$ while being in the domain of the function. – Bernard Oct 26 '20 at 10:50
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The limit is equal to zero, from what I gather. The question seems to be coming from the common introductory calculus technique to prove a limit exists, namely, if:
$$\lim_{x\rightarrow 0^-} \sqrt{x} = \lim_{x\rightarrow 0^+} \sqrt{x}$$
and they exist, then so will $$\lim_{x\rightarrow 0} \sqrt{x} $$
The confusion arises when you try to take the limit from the left, as there is no "left" side of the function. On the other hand, someone with an analysis background will tell you that the square root function isn't defined in the negative domain and therefore, you cannot take the limit in that domain. This is not to say that the limit does not exist there, it just means that you cannot take a limit there to begin with. In other words, you cannot talk about a "left" $\epsilon$-neighborhood centered at $0$, so it doesn't make sense to discuss the limit approaching from the left. In otherwords, we should only consider the function only in its defined domain. Someone with a background in analysis will tell you that because $0$ is a limit point, then the limit exists at that point, but as I'm assuming this is not in your background, I will leave it to you to reserach more on that topic.
You can also check this by $\epsilon$ analyses. Given $\epsilon > 0$, we have $\sqrt{x} < \epsilon$ $\Rightarrow$ $0 < x < \varepsilon^{2}$.
Cheers.