Let there be a function $f: \mathbb{Q} \to \mathbb{Q}$, such that $f(x) = 1$ for all rational numbers $x$.
What's the limit of this function if $x$ tends to $0$? Does it exist and is $1$? Does it not exist? Is asking about the limit invalid here?
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Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. – José Carlos Santos Sep 25 '19 at 09:11
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The limit might be non-existent. If we have for example $f(x)=0$ for all irrational $x$, then the limit does not exist. – Peter Sep 25 '19 at 09:19
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Assume that f is not defined for irrational. – Anshul Singhal Sep 25 '19 at 09:25
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Welcome, good question. I have edited your question to use math syntax (click "edit" if you want to see how) and to make a more helpful title of the question. – Caleb Stanford Sep 25 '19 at 09:42
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You are leaving "calculus" (your tag choice?) and moving to "general topology". That has a tag, too. – GEdgar Sep 25 '19 at 12:15
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What's the limit of this function if $x$ tends to $0$? Does it exist and is $1$? Does it not exist? Is asking about the limit invalid here?
The limit exists and equals $1$.
Values not in the domain of a function are not considered when taking the limit. For example, consider the following easier limit: $$ \lim_{x \to 0} \sqrt{x} $$
Now the function $\sqrt{x}$ implicitly has domain $[0, \infty)$. So, when we take this limit, we do not consider the values $x < 0$; we only consider $x > 0$. And so the answer is that the limit equals $0$. $$ \lim_{x \to 0} \sqrt{x} = 0. $$
Going back to your example, you asked us to consider $$ \lim_{x \to 0} f(x) $$ where $f: \mathbb{Q} \to \mathbb{Q}$ and $f(x) = 1$ for all $x \in \mathbb{Q}$.
Because the domain of $f$ is $\mathbb{Q}$, we do not consider irrational values of $x$ when taking the limit. So we only look at the values $x$ which are closer and closer to $0$, which are rational numbers. For such values of $x$, $f(x) = 1$. Therefore, the answer is $$ \lim_{x \to 0} f(x) = 1. $$
Caleb Stanford
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Isn't limit defined for a neighborhood? Neighborhood can not have points outside domain. No matter how small delta you take, you will still have points which are not in domain. As I understand, neighbourhood of a point is a set of points containing that point where one can move some amount in any direction away from that point without leaving the set. – Anshul Singhal Sep 25 '19 at 13:25
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2@AnshulSinghal You take relative neighbourhoods (neighbourhoods in the whole space, here $\Bbb R$, intersected with the domain). Those count as neighbourhoods in the domain (as a subspace). – Henno Brandsma Sep 25 '19 at 14:16
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@AnshulSinghal It's exactly as Henno Brandsma says. We only consider points that are both in a neighborhood of the limit point and also in the domain of the function. – Caleb Stanford Sep 25 '19 at 20:57
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The limit is $\lim_{x \to 0} f(x) = 1$ because whatever $\varepsilon$ you take, for any $\delta >0$ we have that $f[\mathbb{Q} \cap (-\delta, \delta)] =\{1\} \subseteq (1-\varepsilon, 1+\varepsilon)$, so we always get $1$ (so $\varepsilon$-close to $1$ ) for all $x$ nearby $0$ where $f$ is defined. It's irrelevant where $f$ is not defined, as long as we have domain points $\neq 0$ that are arbitarily close to $0$. Otherwise a limit to $0$ wouldn't make sense anyway.
So also $\lim_{x \to \pi} f(x)=1$ for similar reasons.
Henno Brandsma
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