# Does infinity have a limit?? [closed]

Infinity being the far extent that the numerical system can stretch,can we say that infinity is actually a limit or infiity has another limit?

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## closed as unclear what you're asking by Asaf Karagila, Daniel Rust, Lord_Farin, Dennis Gulko, Dominic MichaelisNov 6 '13 at 12:11

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I may be missing the point here, but I'm fairly sure we can say that infinity is indeed a limit. How else would we answer something like $\lim_{x \rightarrow 0} \frac{1}{x}$? –  ymbirtt Nov 6 '13 at 11:00
Infinity is something reached in a limit, but I don't think it would be proper to say that it has a limit. –  Jaycob Coleman Nov 6 '13 at 11:08
What does "have a limit" even mean? –  Asaf Karagila Nov 6 '13 at 11:28
Well @ymbirtt, we would answer to what you ask saying the limit does not exist, as it definitely is not infinity. If you take the limit from the right, though, then it indeed is infinity. –  DonAntonio Nov 6 '13 at 11:31

If with Infinity being the far extent that the numerical system can stretchyou mean something like the extended real line (the extended real line, denoted by $\overline {\Bbb R}$, is by definition $\Bbb R\cup \{-\infty, +\infty\}$ - where $+\infty$ is an object we define as being greather than all real numbers and dually for $-\infty$), then yes, you can say that $+\infty$ is an actual limit, because $+\infty$ is something that exists, it is an object. For instance $\lim \limits_{x\to 0^+}\left(\dfrac 1 x\right)=+\infty$.

On the other end, over the real numbers, something like $\lim \limits_{x\to 0^+}\left(\dfrac 1 x\right)=+\infty$ is simply shorthand notation for the formula $$\forall \varepsilon >0\exists \delta >0\forall x\neq 0\left(|x|<\delta \implies \left|\dfrac 1 x\right|>\dfrac 1\varepsilon\right).$$

In this sense $+\infty$ is not an actual entity, it's not an object and you can't say something equals $+\infty$. You can write $\lim \limits_{x\to 0^+}\left(\dfrac 1 x\right)=+\infty$, but you shouldn't read this as an equality. It's simply a string of symbols which happens to have a $=$ in the middle. You can even say that the limit is $+\infty$, but only as short for the above formula.
More generally, given a function $f$, let $D_f$ be its domain. Given any $a\in \Bbb R$, we say that the limit of $f$ at $a$ is $+\infty$ and we write $\lim \limits_{x\to a}\left(f(x)\right)=+\infty$ (not an equality, just a meaningless string of symbols which abbreviate the formula below), if, and only if, $$\forall \varepsilon >0\exists \delta >0\forall x\in D_f\left(|x-a|<\delta \implies \left|f(x)\right|>\dfrac 1\varepsilon\right).$$

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The answer above is meant to answer the question in the question box. As for the question in the title, I can't give a meaning to it and I think the OP is interested in the question in the question box. –  Git Gud Nov 6 '13 at 11:30
You mean $|1/f(x)|$ instead of $|1/x|$ in your last line, don't you? –  Michael Hoppe Nov 6 '13 at 11:43
@MichaelHoppe Thanks for noticing the typos. Fixed it. –  Git Gud Nov 6 '13 at 11:50

A suggestion for a slight improvement of Git Gud's answer: for $\epsilon>0$ and $a\in\mathbb R$ we define as usual $B_{\epsilon}(a):=\{x\in \mathbb R\mid |x-a|<\epsilon\}$. We may safely define $B_{\epsilon}(\infty)=]\epsilon,\infty[$ and then apply the usual definition of a limit to know what $\lim \limits_{x\to a}f(x)=\infty$ means.

That is: we don't have to alter the definition of limit, but declare what a neighborhood of $\infty$ should be instead.

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