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This is sort of a follow up to my previous question

Say you have $$ \lim_{x\to +\infty} f(x) $$

where $f : \mathbb{R} \to \mathbb{R} , x \in \mathbb{R}$

What exactly does this mean? From the last question I asked I understand $+\infty$ to be a concept (I also read this) meaning "a number that is arbitrarily large" (taken from that page).

The definition of the limit of a function, as I pointed out in my last question, specifies that the point which $x$ approaches must be a limit point; however, again from my last question, I came to understand that $+\infty$ is not a limit point (considering how I described it above this makes sense, as its not a number but a concept), so how can we take the traditional limit of this?

I have thought about this and have come to the conclusion that a traditional $\epsilon , \delta$ proof wouldn't make sense, and that this simply means "what is the value of $f(x)$ as $x$ gets really really big. Is this correct thinking?

Thanks in advanced!

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DanZimm
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  • Does $\infty$ mean $+\infty$? If so, than it is less ambiguous to denote it that way. Otherwise, this could very well be the extra point of the one point compactification of $\mathbb{R}$, for instance. – Julien Mar 31 '13 at 02:33
  • yes - sorry, will edit to fix – DanZimm Mar 31 '13 at 02:35

2 Answers2

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The statement that

$$\lim_{x\to\infty}f(x)=L\tag{1}$$

means precisely this:

for each $\epsilon>0$ there is a real number $x_\epsilon$ such that $|f(x)-L|<\epsilon$ whenever $x\ge x_\epsilon$.

That’s the actual definition of $(1)$. As you can see, it’s very similar to the $\epsilon$-$\delta$ definition of $$\lim_{x\to a}f(x)=L\;,$$ but with $|x-a|<\delta$ replaced by $x\ge x_\epsilon$. You don’t have an actual notion of ‘distance from $x$ to $\infty$’, but just as making $\delta$ smaller makes $|x-a|<\delta$ a stronger statement about the closeness of $x$ to $a$, so making $x_\epsilon$ larger makes $x\ge x_\epsilon$ a stronger statement about the ‘closeness’ of $x$ to $\infty$, speaking informally. (As I said in comments on the earlier question, this can be made more formal in a more general topological setting.)

Brian M. Scott
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  • Aha! So doesn't this sort of describe mathematically what I said in words? $f(x)$ is as close to $L$ as you need provided $x$ is somehow arbitrarily large? – DanZimm Mar 31 '13 at 02:34
  • @Dan: Yes, at an informal level $-$ in the same sense that the $\epsilon$-$\delta$ definition tells you that you can get $f(x)$ arbitrarily close to $L$ by taking $x$ sufficiently close to $a$. – Brian M. Scott Mar 31 '13 at 02:37
  • Ok, is this then how the limit is taken of sequences? However for sequences $x_{\epsilon}$ is restricted to $\mathbb{N}$? And then you cannot take any other limits of sequences since there aren't any limit points? Or is there still something that I'm missing, from a topological perspective? – DanZimm Mar 31 '13 at 02:41
  • @Dan: Yes, for $\lim_{n\to\infty}x_n=L$ just replace $x_\epsilon\in\Bbb R$ by $m_\epsilon\in\Bbb N$, $|f(x)-L|<\epsilon$ by $|x_n-L|<\epsilon$, and $x\ge x_\epsilon$ by $n\ge m_\epsilon$. – Brian M. Scott Mar 31 '13 at 02:44
  • Again you have enlightened me, so, again, thank you very much! One last question, should I still consider the idea of the limit of a sequence as a closely but different concept to that of a function? Or does it suffice to understand that sequences inherently have no limit points so you can only take the limit as n increases without bound? – DanZimm Mar 31 '13 at 02:46
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    @Dan: The latter would do; you can then lump it together with the notion of $\lim_{x\to\infty}f(x)$. – Brian M. Scott Mar 31 '13 at 02:50
  • you have a fantastic night, and I'm jealous for all who you teach, if you teach this well over Math SE you must be phenomenal in person! – DanZimm Mar 31 '13 at 03:02
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    @Dan: Thanks! I’m retired these days, so MSE is virtually all of my teaching. – Brian M. Scott Mar 31 '13 at 03:05
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If you got the idea that $+\infty$ is the concept of arbitrarily large numbers, then you got the wrong idea. There are relationships, though; for example, $+\infty$ represents the 'limiting' behavior of "arbitrarily large positive numbers" in the same fashion that $0$ represents the 'limiting' behavior of "arbitrarily small positive numbers".

The right (IMHO) way treat $+\infty$ in the way it's used in calculus is by the extended real numbers. The extended real numbers comprise all ordinary real numbers, and two additional numbers which we call $+\infty$ and $-\infty$.

It turns out that, in this setting, $\lim_{x \to +\infty} f(x) = L$ or $\lim_{x \to a} f(x) = +\infty$ means exactly the same thing as any other limit... once we pass to the general topological notion of limit.

In the more general setting, rather than having an $\epsilon$ or a $\delta$ that says how far some number can be from some other number, we instead consider the idea of an open set or an open neighborhood. The definition of a limit is:

$\lim_{x \to a} f(x) = L$ if and only if, for every open neighborhood $U$ containing $L$, there exists an open neighborhood $V$ containing $a$, such that for every $x \in V$ such that $x \neq a$, we have $f(x) \in U$.

In the standard real numbers, we can choose to define "open neighborhood" to mean "open interval". So every neighborhood looks like $(a,b)$: e.g. $(L - \epsilon, L + \epsilon)$ is an open neighborhood when $\epsilon > 0$. Hopefully it's clear how the general notion of limit reduces to the version you learned in elementary calculus!

When using the extended real numbers, we also take $(a, +\infty]$ and $[-\infty, a)$ as open neighborhoods (where $a$ is an ordinary real number). So,

$$\lim_{x \to +\infty} f(x) = L$$

Means, after a little bit of rewriting and simplification,

For every $\epsilon > 0$, there exists an $N$ such that for every (ordinary) real number $x > N$ we have $|f(x) - L| < \epsilon$

which is exactly what you learned in elementary calculus!

  • Warning: I think I have some subtle error in my definition of limit; it should be correct in 'most' cases, but I think it gets some weird cases wrong. (e.g. when you have open neighborhoods consisting of only a single point) –  Mar 31 '13 at 03:06
  • I believe with regards to the definition of a limit you mean $f(x) \in U$ not $x \in U$ - this makes sense except for the fact that I'm not talking about the extended real number line, I'm talking about strictly $\mathbb{R}$. So then with regard to your definition of a limit, $\infty$ would have to be quantified in some sort of way in order to have the neightborhood of $\infty$, wouldnt it? – DanZimm Mar 31 '13 at 03:08
  • @DanZimm: Okay, so I had a blatant error too! Fixed. Yes, you're talking about the real number line, but the real number line is a subspace of the extended real number line, and your function $f(x) : \mathbb{R} \to \mathbb{R}$ automatically becomes a (partial) function on the extended real number line, defined everywhere except at $\pm \infty$... and being undefined at $+\infty$ doesn't get in the way of the definition of a limit at $+\infty$ (or at any other point). –  Mar 31 '13 at 04:23
  • It is also practical and useful to do "continuous extension" at $\pm \infty$ as you would do in other cases (e.g. extending $x/x$ to be defined at $0$). I wouldn't expect any of my colleagues to even bat at an eye if I wrote something like $\arctan(+\infty) = \pi/2$ -- and would only give slight pause (if any) if I did the same with any other function that had a limit at $+\infty$. –  Mar 31 '13 at 04:28
  • Ok, so this sort of unifies all types of limits with regard to neighborhoods, right? If so (as long as I understand this properly) I like how this is phrased all the same – DanZimm Mar 31 '13 at 04:34