# When $\lim_{n \to \infty} x_n = \infty$, is infinity viewed as the limit of the sequence $x_n$?

If I understand correctly, the limit of a function when approaching a limit point of its domain may not be achieved by the function, i.e. the limit may not be the function value at some domain point. So I guess positive and negative $\infty$ can be limits of real-valued functions?

A sequence is also a function with domain being $\mathbb{N}$. I wonder if for $\lim_{n \to \infty} x_n = \infty$, is the sequence viewed as converges? Can the infinity be viewed as the limit of the sequence? I remember some book said $\infty$ couldn't be the limit of any sequence.

Thanks and regards!

-
You can do two things: consider the extended reals, $\mathbb{R}\cup\{-\infty,\infty\}$, which has certain rules for operations (and certain forbidden operations, such as $\infty-\infty$); in which case $\infty$ really "is" the limit. Or, as is usual in calculus, you can view $\lim x_n=\infty$ as a statement that says the limit does not exist, but which also explains why the limit does not exist (the values $x_n$ grow without bound). Same goes for limits of functions: either they are the actual value in the extended reals, or they are a description of why the limit does not exist. – Arturo Magidin Feb 19 '12 at 6:42
@ArturoMagidin That looks like an answer to me. – Alex Becker Feb 19 '12 at 6:43
In addition to the two things Arturo Magidin described, there are at least two other good ways of looking at this question. One is to use big O notation en.wikipedia.org/wiki/Big-o_notation . Another is non-standard analysis (NSA) en.wikipedia.org/wiki/Non-standard_analysis , in which the real number system is expanded to the hyperreal numbers. One way of describing the hyperreals is that a hyperreal number is an equivalence class of sequences $\{x_n\}$ that agree almost everywhere. In neither of these approaches is it possible to define an infinite limit in the way you want. – Ben Crowell Feb 19 '12 at 6:57

The standard (calculus-level and beginning real analysis) interpretation is that when we write $$\lim_{x\to a}f(x) = \infty$$ or $$\lim_{n\to\infty} x_n = \infty$$ we are saying that the limit does not exist, but we are also specifying why it doesn't exist: because the values (of the function, as $x$ gets closer and closer to $a$; or of the sequence, as $n$ gets larger) "grow without bound." In this sense, neither limit exists, so it is incorrect to say that $\infty$ "is" the limit of the sequence/function (because that implies that the sequence/function converges, and that is not the case). Similar comments apply to the case of $\lim\limits_{x\to a}f(x)=-\infty$ or $\lim\limits_{n\to\infty}x_n = -\infty$.
An alternative approach is to consider the "extended reals", $\mathbb{R}^{\#}=\mathbb{R}\cup\{-\infty,\infty\}$. We need to extend some of the operations to include these new symbols (e.g., $a+\infty = \infty$ for all $a\in\mathbb{R}$, $a\infty =\infty$ if $a\gt 0$, $a\infty=-\infty$ if $a\lt 0$, etc), and explicitly "forbid" other operations (e.g., $0\infty$, $\infty-\infty$, $\frac{\infty}{\infty}$, etc.) This is often done, for example, in Measure Theory, where we allow functions and sequences that take values in the "extended reals". We also need to extend the topology of $\mathbb{R}$ to $\mathbb{R}^{\#}$ (intuitively, create an analogue of "open interval around a point" for $\infty$ and for $-\infty$); this is achieved by making sets of the form $(a,\infty]$ and $[-\infty,b)$ the "neighborhoods of $\infty$ and $-\infty$", respectively. If you do all of this preparation and groundwork, then you could say that the limit does exist and is equal to $\infty$ or to $-\infty$, as appropriate.