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The following definition below is what I believe is the definition of a limit. However, there might be something subtle that I am missing.

Definition of a Limit: Let $f: X\to Y$ be defined on some open interval $I\subseteq X$ that contains $a$, except possibly at $a$ itself. Then, $\lim_{x\rightarrow a}f(x):=L$ if $\forall \epsilon \in \mathbb{R}^{>0}$, $\exists \delta\in \mathbb{R}^{>0}$, $\forall x\in I [\text{ if } 0 <|x-a|< \delta \text{, then } |f(x)-L|<\epsilon].$

Now, consider $f(x):=\sqrt{x}$ where $f$ is a function such that $f: [0, \infty)\rightarrow \mathbb{R}$. Why would $\lim_{x\to 0}f(x)=\lim_{x\to 0}\sqrt{x}=0$? Wouldn't it not make sense to figure out the limit here because we can't find some open interval $I$ that satisfies this? I'm looking for clarification on this definition to ensure this scenario makes sense. Any help would be greatly appreciated!

Definition: Let $A\subseteq X$ where $X$ is a metric space. Let $x\in X$. We say $x$ is a limit point of $A$ if $\forall B_x$ (i.e. an open ball with center x) $(B_x-\{x\})\cap A\neq \emptyset$.

Joe
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W. G.
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    That definition is a very restricted one. All that is needed for the definition of $\lim_{x\to a}f(x)$ is that $a$ is a limit point of $X$. There's no need for the existence of such an open interval. – José Carlos Santos Sep 12 '21 at 14:10
  • Related: https://math.stackexchange.com/questions/637280/limit-of-sqrt-x-as-x-approaches-0, https://math.stackexchange.com/questions/1914931/is-sqrt-x-continuous-at-0-because-it-is-not-defined-to-the-left-of-0 – Hans Lundmark Sep 12 '21 at 14:16
  • With the definition of the limit that you are using, it is indeed true that $\lim_{x\to0}\sqrt{x}$ does not exist. However, there is a more general definition of a limit that does state that $\lim_{x\to0}\sqrt{x}=0$:

    Suppose that $a$ is a limit point of the domain of $f$, i.e. every punctured neighbourhood of $a$ contains a point where $f$ is defined. Then, we write $\lim_{x\to a}f(x)=l$ if for every $\varepsilon>0$ there is a $\delta>0$ such that for all $x\in\operatorname{dom}(f)$, if $0<|x-a|<\delta$, then $|f(x)-l|<\varepsilon$.

     – Joe Sep 12 '21 at 14:48
  • @JoséCarlosSantos How would you define $a$ to be a limit point of $X$ applying to this definition here? I'm teaching a group of AP Calculus students who are pretty smart and trying to not make it too much topology if that makes sense. Can you please supply the definition of a limit that you are using exactly here as well as the definition of a limit point that is in relation to this definition? – W. G. Sep 12 '21 at 20:27
  • @W.G: In this context, the condition that $a$ is a limit point of the domain of $f$ is equivalent to requiring that for every $\delta>0$ there is a number $x\in(a-\delta,a+\delta)$ such that $x\neq a$, and $f(x)$ is defined. For example, if $f(x)=1$ for all rational $x$, and undefined otherwise, then every $a\in\Bbb R$ is a limit point of the domain of $f$. – Joe Sep 12 '21 at 20:35
  • @W.G.: Then, provided that $a$ is a limit point of the domain of $f$, we say that $\lim_{x\to a}f(x)=l$ if $$\forall\varepsilon>0:\exists\delta>0:\color{red}{\forall x\in\operatorname{dom}(f)}:0<|x-a|<\delta\implies|f(x)-l|<\varepsilon , .$$ – Joe Sep 12 '21 at 20:37
  • @W.G.: See here for further details. This definition of the limit looks a lot more complicated than it actually is! – Joe Sep 12 '21 at 20:40
  • What you are overlooking is that the domain is not $\mathbb R$ but is $[0,\infty)$. And although $[0,\epsilon)$ is not open in the space $\mathbb R$, it *IS* open in the space $[0,\infty)$. – fleablood Sep 12 '21 at 20:42
  • @fleablood I get the domain is $[0, \infty)$ here. Are you considering the metric space to be $[0, \infty)$ here? Is technically $I$ not an open interval than but rather an open set on a metric space $[0, \infty)$? I think that's what you are saying. How would you phrase the definition here? – W. G. Sep 12 '21 at 20:55
  • "Are you considering the metric space to be [0,∞) here?" Of course! The question distintly said $f: [0,\infty)\to \mathbb R$ so $[0,\infty)$ is your domain. And the definition you gave specifically says that $I$ is a subset of $X$, the domain of the function. .... so what is the defintion of an "open interval"? It is 1) an interval.... and $[0, \epsilon)$ is an interval, and 2) it is an open set... and $[0,\epsilon)$ is an open set in our domain..... so.... It is perfectly legitimate to claim $[0,\epsilon)$ is an open interval. – fleablood Sep 12 '21 at 21:20

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Note that $X$ is a space that does not have to be equal to $\mathbb R$.

The definition of an interval in the space $X$ is any set $I$ so that for any $a, b \in I$ then the all the elements $w\in X$ so that $\min(a,b) \le w \le \max(a,b)$ will be $w\in I$.

So although the $S=(1,2)\cup \{3\}\cup (4, 5)$ will not be an inteval in $\mathbb R$, $S$ will be an interval in the space $X= (-\infty, 2)\cup \{3\} \cup (4,\infty)$.

The definition of an open interval is an interval $I$ so that for every $a \in I$ there exists a $\delta > 0$ so that $x\in S$ so that $|x-a| < \delta$ will be that $x\in I$.

By that definition $[0, 1)$ is not an open interval in $\mathbb R$. It fails because $0\in I$ but for any $\delta> 0$ we can find an $x: -\delta < x < 0$ and $|x-0| < \delta$ but $x \in [0,1)$ so $[0,1)$ can not be an open interval in $\mathbb R$.

But $[0,1)$ is an open interval in $[0,\infty)$. Why? Consider $0 \in [0,1)$. Let $\delta = \frac 12$. If $|x- 0| < \frac 12$ and $x\in [0,\infty)$ then $0\le x < \frac 12$. And so $x \in [0,\infty)$. So $0$ being an endpoint of $[0, 1)$ with nothing to the left of it does not cause $[0,1)$ to fail to be an open interval. It's okay that there is nothing to the left of the point $0$ in $[0,1)$ because..... there is nothing to left of $0$ in our entire universe space of $[0, \infty)$.

.....

So according to you definition $\lim_{x\to 0} f(x) = 0$ where $f:[0,\infty)\to \mathbb R: f: x\to \sqrt x$ works out just fine.

"Let $f:[0,\infty)→\mathbb R$ be defined on some open interval $I\subset [0,\infty)$ that contains $0$ except possibly at a itself".

Okay... $[0,1)$ is such an open interval. $[0,1)$ is open in $[0,1)$ and $\sqrt x$ is defined on $[0,1)$.

"The $\lim_{x\to 0} f(x) = 0$ if $\forall \epsilon > 0$ there exists a $\delta >0$"

Okay.... let $\delta = \epsilon^2$.

"So that for all $x\in [0, 1)$ if $0< |x-0| < \delta$ then $|f(x) -0| < \epsilon$".

That follows.

If $x \in [0,1)$ and of $0 < x-0 < \delta$ then $-\delta < x < \delta$ and as $0 \le x \le 1$ we have $0 \le x < \delta$.

And as $\delta =\epsilon^2$ we have $0 \le x < \epsilon^2$ so $0 \le \sqrt x < \epsilon$.

So $0 \le f(x) < \epsilon$ so $|f(x) -0| = f(x) < \epsilon$.

The definition is satisfied.

W. G.
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fleablood
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