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I know this question has been asked and answered before and I found this answer Limit of $\sqrt x$ as $x$ approaches $0$ helpful, in particular the response from @pie, and I am convinced that $f(x)=\sqrt(x)$ is continuous at $x=0$. However, this raises another question about how we define one-sided limits.

I have always understood that $\lim_{x\to c} f(x)$ exists and $\lim_{x\to c} f(x)=L$ if and only if $\lim_{x\to c^{+}} f(x)$ and $\lim_{x\to c^{-}} f(x)$ exist and $\lim_{x\to c^{+}} f(x)=\lim_{x\to c^{-}} f(x)$. However, when $f(x)=\sqrt(x)$, is $\lim_{x\to 0^{-}} f(x)$ defined?

Based on @pie's response, we can define the left-hand limit as such:

Let $A \subseteq \mathbb{R}$ and let $c$ be a limit point of $A$. For a function $f : A \to \mathbb{R}$, a real number $L$ is said to be a limit of $f$ at $c$ if, given any $\epsilon > 0$, there exists a $\delta > 0$ such that if $x \in A$ and $0<c-x<\delta$ then $|f(x)-L|<\epsilon$.

In the case $f(x)=\sqrt(x)$ and $A=[0,\infty)$, $c=0$ and $L=0$ it is vacuously true that given any $\epsilon > 0$, there exists a $\delta > 0$ such that if $x \in A$ and $0<c-x<\delta$ then $|f(x)-L|<\epsilon$ as there are no $x\in A$ such that $0<c-x<\delta$ as this requires $x<0$, so it appears by this definition that $\lim_{x\to 0^{-}} f(x)=0$. But the same is true when $L=1$ or any other real number, so the left-hand limit is not unique.

I suppose we could change the definition to stipulate that the set $(x \in A : 0<c-x<\delta)$ is non-empty, but then we conclude that $\lim_{x\to 0^{-}} f(x)=0$ does not exist, which contradicts $\lim_{x\to c} f(x)$ exists and $\lim_{x\to c} f(x)=L$ if and only if $\lim_{x\to c^{+}} f(x)$ and $\lim_{x\to c^{-}} f(x)$ exist and $\lim_{x\to c^{+}} f(x)=\lim_{x\to c^{-}} f(x)$.

The problem here presumably is the fact that zero is boundary point of the domain? It appears that if $c$ (in the closure of $A$) is not a boundary point then it is true that $\lim_{x\to c} f(x)$ exists and $\lim_{x\to c} f(x)=L$ if and only if $\lim_{x\to c^{+}} f(x)$ and $\lim_{x\to c^{-}} f(x)$ exist and $\lim_{x\to c^{+}} f(x)=\lim_{x\to c^{-}} f(x)$, but if $c$ is a boundary point of $A$ then $\lim_{x\to c} f(x)=\lim_{x\to c^{-}} f(x)$ or $\lim_{x\to c} f(x)=\lim_{x\to c^{+}} f(x)$, whichever exists, but this seems a bit ad hoc?

Also, what about the case when $A$ is a single real number (eg. $f : \{0\}\to\mathbb{R}, f(x)=x$). Does $\lim_{x\to 0} f(x)=0$ but neither the left-hand nor right-hand limit exist?

DTt
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  • Yes, it will be continuous at zero. (You're overthinking it.) The reason is that the epsilon condition need not be satisfied on inputs which do not exist in the domain $[0,\infty).$ Does this help? – J.G.131 Jun 21 '24 at 17:22
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    Your answer depends on the definition. State the exact definition of "continuous" to be used. – GEdgar Jun 21 '24 at 17:46
  • @GEdgar Judging by his definition of “left-hand limit” as well OPs fixation on limits, would it not be reasonable to assume his definition involves showing $f(c)=\lim_{x\to c}f(x)?$ (So his definition of continuity would just be the standard epsilon-delta definition, again judging by his definition of limit.) – J.G.131 Jun 21 '24 at 17:55
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    This is a very FAQ. https://math.stackexchange.com/search?q=one-sided+limit+definition We should probably close as a duplicate, but which target to choose? – Anne Bauval Jun 21 '24 at 17:58
  • Yes, sorry, I'm defining continuous as $f(c) = \lim_{x \to c}f(x)$. Really, the question is whether $\lim_{x \to 0}\sqrt(x)$ exists. I'm happy that it does, using the definition on the post I link to, but if our definition requires that $c \in I$ where $I$ is some open set and $I\setminus {0}$ is a subset of the domain then I see that the limit does not exist, as there is no such $I$. – DTt Jun 21 '24 at 18:12
  • My problem is that it appears that $\lim_{x\to 0}\sqrt{x}=0$ but $\lim_{x\to 0^{-}}\sqrt{x}$ does not exist and so $\lim_{x\to 0}\sqrt{x}\neq\lim_{x\to 0^{-}}\sqrt{x}=0$. – DTt Jun 21 '24 at 18:15
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    All of these comments depend on the definition. Just saying "usual epsilon-delta definition" isn't enough. We need to know exactly what the quantifiers are. – GEdgar Jun 21 '24 at 19:46
  • The definition of the left-hand limit I am using is this: Let $A\subset\mathbb{R}$ and let $c$ be a limit point of $A$. For a function $f:A\to\mathbb{R}$, a real number $L$ is a left-hand limit of $f$ at $c$ if and only, given any $\epsilon>0$ there exists $\delta>0$ such that if $x\in A$ and $0<c-x<\delta$ then $|f(x)-L|<\epsilon$. – DTt Jun 21 '24 at 20:52
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    The definition of the limit I am using is this: Let $A\subset\mathbb{R}$ and let $c$ be a limit point of $A$. For a function $f:A\to\mathbb{R}$, a real number $L$ is a limit of $f$ at $c$ if and only, given any $\epsilon>0$ there exists $\delta>0$ such that if $x\in A$ and $0<|x-c|<\delta$ then $|f(x)-L|<\epsilon$. – DTt Jun 21 '24 at 20:53
  • By my definitions, $\lim_{x\to 0}\sqrt{x}=0$ but $\lim_{x\to 0^{-}}\sqrt{x}$ is not unique. If in the definition of the left-hand limit we insist that $c$ is a limit point of $A\cap(-\infty,c)$ then $\lim_{x\to 0^{-}}\sqrt{x}$ does not exist. Either way $\lim_{x\to 0}\sqrt{x}\neq\lim_{x\to 0^{-}}\sqrt{x}$. Is the correct statement, according to my defintions, that, if $c$ is not a boundary point of $A$ then $\lim_{x\to 0}f(x)=L$ iff $\lim_{x\to 0^{-}}f(x)=\lim_{x\to 0^{+}}f(x)=L$? – DTt Jun 21 '24 at 21:15
  • In Germany , I learnt it different to the "official" definition. We did not consider this function continuous since it is not defined for $x<0$. For the "official" definition , this is irrelevant and therefore this function is considered to be continuous. I am glad to have learnt the officially wrong way since it describes more intuitive when a function is continous. We only would have called this "right-continous". – Peter Jun 21 '24 at 22:20
  • Yes, and right-differentiable. I guess my question is which is the authoritative epsilon-delta definition that I should use? I know that there is not one “correct” definition, but which is the most widely accepted definition? – DTt Jun 22 '24 at 18:15
  • Welcome to MSE! <> The definition in your comment would best be edited into the question body; otherwise, future passersby must read the comments to understand the question. <> Given the definition: "Yes, the limit exists." Your understanding about one-sided limits turns out not to apply here, i.e., is not equivalent to the definition. <> You haven't given your definition of continuity but in the circumstances it would be perverse not to call the square root continuous at $0$. – Andrew D. Hwang Jun 23 '24 at 00:01

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