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This post says that right-hand limit, or similarly left-hand limit, is unique when it exists by the uniqueness of limit. That can be asserted because $$\lim_{x\to c} f(x) = L \Longleftrightarrow \lim_{x\to c^-} f(x) = L = \lim_{x\to c^+} f(x).$$

Now let $f : [0, \infty) \to \mathbb{R}$ be defined as $f(x) = \sqrt{x}$. This post says that $$\lim_{x\to 0} \sqrt{x} = 0 \text{ because } \lim_{x\to 0^-} \sqrt{x} = 0 = \lim_{x\to 0^+} \sqrt{x}$$ asserting that the left-hand limit exists by the vacuous truth of an empty domain, which is confirmed in this post.

But, by the vacuous truth of an empty domain, the left-hand limit is no longer unique because $$\begin{align*} &\forall z \in \mathbb{R} : \lim_{x\to 0^-} \sqrt{x} = z \\ \Longleftrightarrow\;&\forall z \in \mathbb{R} : \forall \varepsilon\in\mathbb{R}:\exists\delta\in\mathbb{R}:\forall x\in \varnothing : -\delta < x < 0 \longrightarrow \left|\sqrt{x} - z\right| < \varepsilon\;\ldots\;\text{LH limit definition} \\ \Longleftrightarrow\;&\top\;\ldots\;\text{vacuously true by }\forall x \in \varnothing\text{ for all }z \end{align*}$$

But, it must be unique because $$\lim_{x\to 0} \sqrt{x} = 0$$ is unique.

What is my logical fallacy such that I can show that the left-hand limit is no longer unique by the vacuous truth of an empty domain?

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Tadeus Prastowo
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  • From more general definition of limit, it is incorrect to say that $\exists \lim_{x \to 0^-}\sqrt{x} = 0$ (because function must be defined on corresponding filter elements). Likewise, there's certain "handwaving" when one says that $\exists \lim_{x \to 0}\sqrt{x} = 0$, since actual statement is ($\exists \lim_{x \to 0^+}\sqrt{x} = 0$ and $\sqrt{x}$ is undefined for $x<0$). – Abstraction Nov 25 '15 at 10:29
  • @Abstraction: "Likewise, there's certain 'handwaving' ..." is exactly what I mean by being a bit sloppy in my comment to http://math.stackexchange.com/a/1543993/137540 But, the guy insists that there is no sloppiness (i.e. handwaving) in writing $\lim_{x\to 0} \sqrt{x} = 0$ because the intention is 100% $\lim_{x\to 0} \sqrt{x} = 0$, not $\lim_{x\to 0^+} \sqrt{x} = 0$ as I suggested. So, if I accept your answer, then the other guy is being a bit sloppy. But, if he is not sloppy, then you are incorrect. Bottom line: how to resolve this contradiction? What is the most general definition of limit? – Tadeus Prastowo Nov 25 '15 at 10:55
  • "Most general" is a bit ambiguous. Definition of limit via filters covers most general case (for $h \in {X \to Y}, F$ being filter on $X$, $Y$ being topological space, $y \in Y$, $\lim_Fh=y$ is short for $\forall N_y$ being neighbourhood of $y$, $\exists f \in F : h(f) \subset N_y$). When applying this definition to $h(x)=\sqrt{x}$, our $X$ is $[0;+\infty)$, not $\mathbb{R}$. Thus, filter $F$ in question is filter of neighbourhoods of $0$ in $[0; +\infty)$. "Handwaving" here is that we may call this filter $x \to 0$, omitting "but only where $h(x)$ is defined" part. – Abstraction Nov 25 '15 at 11:09
  • @Abstraction: One more question, so for $\sqrt{x}$, left-hand limit $\lim_{x\to 0^-}\sqrt{x}$ exists or not? – Tadeus Prastowo Nov 25 '15 at 11:21
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    As per given definition it doesn't. Effectively for $x \to 0^-$ $h(x)$ is meaningless, you reduce its domain to $\emptyset$ (other way to put it is to say you use values outside its domain). For function with domain $\emptyset$, there can be no (proper) filter and thus no limit. – Abstraction Nov 25 '15 at 13:08
  • @Abstraction: As per another recent comment to this post http://math.stackexchange.com/q/1544273/137540 I think now I see the light to answer my own question in this post http://math.stackexchange.com/q/1545193/137540 – Tadeus Prastowo Nov 25 '15 at 15:46

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