# Different sizes of infinity

Correct me if I'm wrong, but this is what they taught us in precal: $$\lim_{x\rightarrow\infty}x=\infty$$ $$\lim_{x\rightarrow\infty}x^{2}=\infty$$ But, we also know that $n^{2}>n$ if $n\notin [0,1]$

Does that mean that some infinities are greater than others? Why don't we explicitly define infinity so that we can show differences in sizes?

• A quick remark, $x^2\leqslant x$ for $x\in [0,1]$. Jun 18, 2016 at 0:59
• It means some rates of approaching $\infty$ are greater than others. As far as some infinities being greater than others is concerned, there are many different concepts of infinity in mathematics that are quite different things from each other, and before talking about whether one infinity is greater than another, one must be precise about which of those various concepts is intended. $\qquad$ Jun 18, 2016 at 1:01
• @AdolfoHolguin Disagree that "uncountably infinite" even makes sense here. These aren't cardinalities! The right sense to compare them is as asymptotic growth rates, or elements of a Hardy field or something. I'll try and think of a good way to express it at a level I think is appropriate for the OP.
– user231101
Jun 18, 2016 at 1:04
• We cannot show infinities in different numerical sizes. infinity is a concept, not a number that has a position on the numberline. One should be very careful with using arithmetic operations on infinities, especially when it involves subtraction and division. A friend of mine, who is into economics, once compared the infinity idea with being rich. When you are a millionair, you are rich, but if you are a billionair you are richer, but that is still rich. Jun 18, 2016 at 1:18
• @imranfat: We can show infinite numbers of different sizes when such things exist (e.g. ordinal numbers and cardinal numbers). $+\infty$ is more geometric in origin and there is just the one, but you really should not confuse "big" with $+\infty$. At best you could say being a billionaire approximates having infinite money, but it's really not the same thing.
– user14972
Jun 18, 2016 at 4:23

The notation $\displaystyle \lim_{x\rightarrow\infty}f(x)=\infty$ where $f$ is any real-valued function of a real variable means exactly that as $x$ gets arbitrarily large, so does $f(x)$. That's all it means. This usage has no relation to any metaphysical ideas about infinity.

It's also the case that in the mathematical field of set theory, there is an entire elaborate theory of transfinite numbers. This is a very interesting area of math and the basics are accessible at an elementary level so do take a look at this if you're interested. This usage of infinity is not related to the infinity of the first paragraph.

The observation you made about $x$ and $x^2$ happens to lead to yet another interesting area of math: namely the study of the rate at which functions grow. For example as you noted, the functions $x$ and $x^2$ each go to infinity (meaning they grow without bound) as $x$ gets large. And yet, $x^2$ grows "faster" than $x$ in some way.

Growth rates of functions have been studied since the late nineteenth century. They've always been important in some areas of pure math; and today they are of extreme interest in computer science.

Suppose we have two algorithms whose running time increases as a function of the length of their input (sorting a list, say). Then as the input gets larger, the algorithm that grows faster starts to take an impractical amount of time. Computer scientists are always interested in finding algorithms that grow slowly.

The growth rate of a given function is described by its Big-O notation. The study of which functions have roughly the same growth rate is computational complexity theory.

• The $+\infty$ of the extended number line doesn't really bear any resemblance to the infinite numbers occurring in the various number systems studied in set theory.
– user14972
Jun 18, 2016 at 4:24
• @Hurkyl Agreed, I hope I didn't imply otherwise but perhaps I should have been more explicit. (ps) Added that point. Jun 18, 2016 at 5:28
• @ruakh Fixed. Thanks. Jun 19, 2016 at 5:46
• I disagree with the last sentence. If I want to learn "which functions have roughly the same growth rate" I do not look into a complexity book. Sure there likely are some results there on that (like the "master theorem") but after the preliminaries, in a complexity book I will find mostly algorithms and reductions.
– chi
Jun 19, 2016 at 7:59
• I can't provide a reference, but I'd look at calculus books, in general. Computational complexity is another topic, IMO -- where big-O notation is indeed frequently used, but is the means, not the aim.
– chi
Jun 19, 2016 at 15:00

I like this question a lot. The pedantic answer (and the right answer on an exam) is, of course, "They are the same. As limits, they are both equal to the formal symbol "$$+\infty$$"." That's pretty boring, though. There is a clear intuition that the behavior of $$x^2$$ at infinity is "bigger" than the behavior of $$x$$; the challenge is to try to figure out how to clarify that difference. In math, sometimes these intuitions go nowhere: when you try to understand the intuition well enough to express it clearly, it evaporates. In this case, you're actually really onto something.

As I see it, there are three ways to talk about what's going on. First, there's the pedantic way. There it quite a good reason why, if we want to call what we're doing a "limit," we shouldn't distinguish between the behavior of the two functions. Next, there is the idea of "asymptotic growth rate:" even though both of these functions tend to $$+\infty$$, we can compare how fast that tendency is. Computer scientists like that view, because it is useful when analyzing the performance of algorithms on large inputs. Finally, and my personal favorite, is the idea of comparing "germs at infinity." That method allows us to make much finer comparison.

Limits

First of all, limits are about continuity. When we say, for example, $$\lim_{x \to 0} \frac{\sin(x)}{x} = 1 \text{,}$$ what we really mean is that, if we want to consider $$\frac{\sin(x)}{x}$$ as a continuous function, its value at 0 would have to be 1. $$\frac{\sin(x)}{x}$$ isn't actually defined at 0 (since we'd have to divide by 0), so this is about extending the domain of the function. So what does that mean for limits involving infinity?

Imagine for a moment that on the real line, sitting out past all the positive real numbers, there was a single extra point called "$$+\infty$$." Similarly, past all the negative real numbers might lie a point called "$$-\infty$$." The real line doesn't actually look like this (those points don't exist); what I've described instead is something else we call the Extended Real Line, sometimes denoted $$\overline{\mathbb{R}}$$.

Now, our original functions $$x$$ and $$x^2$$ go from $$\mathbb{R}$$ to $$\mathbb{R}$$. As it turns out, it makes sense to talk about what continuity would mean for functions involving $$\overline{\mathbb{R}}$$, not just $$\mathbb{R}$$. So we can ask: if we wanted to extend the domain of these functions to $$\overline{\mathbb{R}}$$ (and maybe extend the range to $$\overline{\mathbb{R}}$$, too, if we have to), what should the value of these functions be at $$\pm \infty$$? Our limits tell us the answer.

When talking about the behavior of our functions at infinity in this sense, it doesn't make sense to distinguish between them: we're only adding one point on either end of the real number line.

Asymptotic growth rate

As you know, both $$x$$ and $$x^2$$ approach $$+\infty$$ as $$x$$ approaches $$+\infty$$. That statement, however, just tells us where they're going; we can still ask about how quickly they do so. As your intuition tells you, $$x^2$$ is heading there a lot faster than $$x$$ is. We can quantify that, if we want:

Definition. Given two functions $$f(x)$$ and $$g(x)$$ from $$\mathbb{R}$$ to $$\mathbb{R}^{>0}$$, say $$f$$ grows asymptotically faster than $$g$$ at $$+\infty$$ if $$\lim_{x \to +\infty} \frac{f(x)}{g(x)} = +\infty\text{.}$$ In that case, write $$f(x) \gg g(x)\text{.}$$

In your specific example, we clearly have that $$x^2 \gg x$$ in this sense.

This notion is useful when trying to analyze the qualitative behavior of a compound system. Roughly, we know that if $$f \gg g$$, the contribution of the parts that look like $$f$$ will eventually dominate the contribution of the parts that look like $$g$$.

Germs at infinity

Germs are a precise way to talk about "how a function behaves at infinity." Unlike asymptotic growth rates, these germs capture all the behavior of the function at infinity, while ignoring all the other stuff.

Definition. Functions $$f(x)$$ and $$g(x)$$ have the same germ at $$+\infty$$ if there is some number $$a$$ such that $$f(x) = g(x)$$ for all $$x > a$$.

Similarly, say (the germ of) $$f$$ is greater than (the germ of) $$g$$ at $$+\infty$$ if there is some number $$a$$ such that $$f(x) > g(x)$$ for all $$x > a$$.

For example, consider the functions $$|x|$$, $$|x-1|+1$$ and $$|x-2|+2$$. These are all different functions as a whole, but they are identical once we get past $$x = 2$$. (See graphs of these functions here.) Consequently, we say the have the same germ at $$+\infty$$.

In this view, $$x^2 > x$$ at $$+\infty$$ merely because $$x^2$$ is eventually greater than $$x$$. Note that this notion is much finer than asymptotic growth rates, which involve the limits of quotients: $$x + 1 > x$$ at $$+\infty$$, even though neither grows asymptotically faster than the other ($$\lim_{x \to +\infty} \frac{x+1}{x} = 1$$).

There are a lot of neat things we can do with these germs at infinity. We can add, subtract and multiply them. We can sometimes take their derivatives and divide by them. How they work, and when they behave especially nicely, is an active area of research math. See, for example, Hardy fields at Wikipedia.

• Nice answer! Asymptotic notation can also be used for a classification in between the two types you've given, since we can choose to keep as many terms as we wish. For example $\sqrt{x^2+2} \in x + \frac{1}{x} + O(\frac{1}{x^3})$ as $x \to \infty$. (See my answer for more examples.) So we can easily distinguish between $x+1$ and $x$, while identifying $x$ and $|x-1|+1$ and ignoring higher-order terms. Jun 18, 2016 at 11:30
• @user21820 The frameworks are closely related. Take any nice family of germs at infinity, such as those of rational functions. (Nice means contains the constants and forms an ordered field.) In any such case, you automatically get an Archimedean valuation that tells you the "order of growth" of the germ. The possible values form a group. For rational functions, the valuation is the exponent of the highest order term, and the group is the integers. What you're suggesting is exactly that, even when germs have the same valuation, we can look at the valuation of their difference.
– user231101
Jun 18, 2016 at 18:09
• "There it quite a good reason why, if we want to call what we're doing a "limit," we shouldn't distinguish between the behavior of the two functions." << When I read this sentence I thought the reason would be "subsequences". Typically, for a function $f$ we want $\lim f(x_n) = \lim f(y_n)$ if the two sequences $(x_n)_{n \in \mathbb N}$ and $(y_n)_{n \in \mathbb N}$ have the same limit, regardless of how fast $(x_n)_{n \in \mathbb N}$ and $(y_n)_{n \in \mathbb N}$ go to their limit.
– Stef
Jan 4 at 8:49
• Well, at least we want $\lim f(x_n) = \lim f(y_n)$ when $\lim x_n = \lim y_n$ and $f$ is continuous at $\lim x_n$. And in particular in the case where $y_n = x_{k(n)}$ for some increasing function $k:\mathbb N \to \mathbb N$.
– Stef
Jan 4 at 9:29

If that seems puzzling to you, this will surely blow your mind:

$$\lim_{x \to \infty} \frac{1}{x} = \lim_{x \to \infty} \frac{1}{x^2} = 0$$

even though $$\dfrac{1}{x} > \dfrac{1}{x^2}$$ for all $$x > 1$$. Are there different sizes of $$0$$, too?

Actually, there's nothing special about $$\infty$$ or $$0$$ in this respect. It's perfectly possible for two functions to approach the same limit (any limit!) from the same direction, but for one function to be consistently closer to the limit than the other. This doesn't make the limits any different — it just means that one of the functions approaches the limit faster than the other.

(Of course, $$\infty$$ as a limit is somewhat special in other ways: when we write $$\lim_{x \to a} f(x) = \infty$$, what we really mean is that $$f(x)$$ diverges without bound as $$x$$ approaches $$a$$ — or as $$x$$ grows without bound, if $$a$$ itself is $$\infty$$. But all these special definitions are really only necessary because the real number line with the usual metric we use on it is not topologically compact, and does not even include $$\pm \infty$$ as proper points. Instead, we can easily define a metric on the extended real number line, e.g. by homeomorphically mapping it to a closed interval, such that limits at (and to) infinity work just like normal limits. I go into more detail on that in another answer here.)

• +1 for pointing out that "faster convergence" makes sense for finite limits as well as for infinite limits. In fact, when using iterative methods to approach the solution of an equation $f(x) = 0$, we typically consider different methods (simple dichotomy, Newton's method, the secant method, iterative fixed-point methods...) and compare their rate of convergence. One simple way to measure convergence is to count how many digits are correct after n iterations.
– Stef
Jan 4 at 8:45

The notation $\lim _{x\to \infty}f(x)=\infty$ is just an abbreviation for $$\forall y\in \mathbb R\;\exists z\in \mathbb R\;\forall x\in \mathbb R\; (x>z\implies f(x)>y).$$

Literally.Nothing more. It does not assume the existence of a "number" $\infty.$ It is useful, in pure math, as well as in applied math, to intuit $x$ as an object in motion, with $f(x)$ varying over time as $x$ does. But the logic of it is that numbers don't move or change. The ( abbreviated ) sentence $\lim_{x\to \infty}f(x)=\infty$ is about a property that is shared by every set of the form $\{(x,f(x)):x>r\}$.

You misremember how limits interact with inequalities:

Theorem: If

• $f(x) < g(x)$ for all $x$ sufficiently near $a$
• $\lim_{x \to a} f(x)$ exists
• $\lim_{x \to a} g(x)$ exists

Then

• $\lim_{x \to a} f(x) \leq \lim_{x \to a} g(x)$

Note that "less than" becomes "less than or equal to" after taking the limits; the point being that it is possible for $f(x)$ and $g(x)$ to both converge to the same limit, but with $g(x)$ doing so more quickly.

For a finite example, observe that $x^2 < x$ for all $x \in (0,1)$, however $$\lim_{x \to 1^-} x^2 = \lim_{x \to 1^-} x$$ or if you want to see it for limtis at $+\infty$, $0 < \frac{1}{x}$ for all positive $x$, however $$\lim_{x \to +\infty} 0 = \lim_{x \to +\infty} \frac{1}{x}$$

One way to think about how $+\infty$ and $-\infty$ work is to observe that the number line is an interval; however it is an interval without endpoints, like $(0,1)$. The extended number line is formed by adding the two endpoints; so the number line can be written as $(-\infty, +\infty)$, and the extended number line is $[-\infty, +\infty]$.

As $x \to +\infty$, both $x$ and $x^2$ approach the right endpoint $+\infty$ without reaching it; it's just that $x^2$ approaches "faster".

Sometimes, we do want to talk about the rate at which a function approaches $+\infty$ (or other values), in which case we would point out that $x^2$ grows asymptotically faster than $x$ as $x \to +\infty$. Things like this are the subject of asymptotic analysis.

Infinity is not quite a number. If you must make it one, you have to allow for $\bar {\Bbb R}$, the extended real line. But even in the sense of $\bar{\Bbb R}$, the "number" $+\infty$ is a single point, so the second infinity is no different than the first one because they are just the same point.

The gist here, however, is not to compare the two resulting infinities in the "larger-smaller" sense, but the realisation that $x^2$ approaches infinity faster than $x$.

The fact that $x^2 > x$ eventually should not necessarily mean that we could have defined different kind of infinities i.e..

$$\lim _{x\rightarrow \infty }x^2> \lim _{x\rightarrow \infty} x$$

This is intuitive since when restricting taking limit whose limits point are real numbers does not suggest such a property.

That is even if $f(x)>g(x), x<x_0$ then it is still possible that $\lim_{ x\rightarrow x_0}f(x)= \lim_{ x\rightarrow x_0}g(x)$.

Now in our example $x^2 >x$, but notice that for every $x_1$ there is $x_2$ such that $x_1^2<x_2$. So the function $x$ can outgrow every value that the function $x^2$ achieves. So both limits should indeed talk about the same symbol.

Another way to think about the uniqueness of $\infty$ is the unique up the isomorphism of the one compactification of the real numbers, and if you think of how the topology is defined there the symbol $\infty$ should be the same element.

When we say "infinite" we don't mean some specific number. It is more along the lines of, "without bound." Or, more formally, "pick a number, bigger than that."

$x= \infty \implies \forall M>0, x>M$

• I am deleting the bit about cardinality Jun 18, 2016 at 1:30
• Don't confuse the specific point $+\infty$ with different kinds of concepts like "approaching $+\infty$" or "a set whose least upper bound is $+\infty$".
– user14972
Jun 18, 2016 at 4:30

It seems to me that you have more of a misunderstanding regarding the limit concept rather than anything having to do with infinity.

One can have two functions $f,g$ such that $f(x)>g(x)$ for all $x$ but

$\lim_{x \to \infty} f(x) = \lim_{x \to \infty} g(x) = L$. That is, both $f$ and $g$ tend to the same limit. Take, for example $f,g:\mathbb{R}_{\geq 0} \to \mathbb{R}$ respectively given by $f(x) = 1$ and $g(x) =\frac{x}{x+1}$. Generally speaking, note that $L$ can be anything, including infinity.

As a side note, note that you never want to think of $\infty$ as a number. In set theory, there is a notion of some infinities being "bigger" or "smaller" than others, but,when discussing the limiting behaviour of real functions, this is essentially irrelevant.

Big $O$ notation is a way of comparing asymptotic behaviours no matter the expression is convergent or not.

Prime counting function $$\pi(x) = O \left(\frac{x}{\ln x} \right)$$

$n$-th prime number $$p_{n}=n \left[ \ln n+\ln \ln n-1+O\left( \frac{\ln \ln n}{\ln n} \right) \right]$$

Stirling Series $$n! = \sqrt{2\pi n}\left( \frac{n}{e}\right)^n \left[1 +\frac{1}{12n}+O\left( \frac{1}{n^2} \right) \right]$$

Exponential integral $$\int_{x}^{\infty} \frac{e^{-t}}{t} \, dt = e^{-x} \left[ \frac{1}{x}-\frac{1}{x^2}+\ldots+(-1)^{n-1} \frac{(n-1)!}{x^n}+ O\left( \frac{n!}{x^{n+1}} \right) \right]$$

See more examples here.

Infinite limits and limits at infinity have standard ε-δ definitions using real numbers, and $$\infty$$ is not a real number. It is possible to use the affinely-extended real line and extend our definitions for limits to cater for these, but they will still not satisfy the properties of limits, such as:

? $$\lim_{x \to a} ( f(x) + g(x) ) = \lim_{x \to a} f(x) + \lim_{x \to a} g(x)$$.

because $$\lim_{x \to \infty} ( x+(1-x) ) = 1$$ while $$\lim_{x \to \infty} x = \infty$$ and $$\lim_{x \to \infty} (1-x) = -\infty$$, and there is clearly no consistent way to assign $$\infty + (-\infty)$$ to a value.

So if one wishes to extend the usual definitions, there will be plenty of exceptions to the usual limit properties.

But here you want finer control over what you say about an expression when it has an infinite limit. This is certainly useful, just as knowing runners' speeds (convergence rate) gives you more information than just where they end up (the limit). In that case you're looking for asymptotic expansions (using Landau notation), which usually exist for reasonably nice expressions. For example:

As $$n \to \infty$$:

$$\frac{1}{n} - \frac{1}{n+1} = \frac{1}{n^2+n} \in \frac{1}{n^2} - \frac{1}{n^3} + O(\frac{1}{n^4})$$.

Note that this gives more information than just saying $$\frac{1}{n} - \frac{1}{n+1}$$ does not tend to a limit.

And we can from this obtain related limits such as $$n^2 ( \frac{1}{n} - \frac{1}{n+1} ) \in 1 - \frac{1}{n} + O(\frac{1}{n^2}) \to 1$$.

I gave some more involved examples at:

I personally find it better to always use asymptotic expansion instead of limit laws, because it is not only systematic but also yields much more information. One can always extract the limit from the asymptotic expansion but not necessarily the other way around. The exception is when it is not a concrete expression, in which case one might have to use abstract methods including the fundamental theorem of calculus and L'Hopital's rule.

There is even an algorithm to deterministically compute limits (as long as the expression uses only elementary functions; no summation, product, ...) using asymptotic expansions, which is what modern computer algebra systems do.

Back to the example in your question, we have $$x^2$$ and $$x$$, which tend to $$\infty$$ as $$x \to \infty$$, but if we are interested in their asymptotic behaviour we would simply leave them as they are, since they are already simple polynomials. On the other hand, we would usually 'simplify' more complicated expressions like "$$\ln(\exp(x)+1)$$" as follows:

As $$x \to \infty$$:

$$\exp(-x) \to 0$$.

$$\ln(\exp(x)+1) = \ln(\exp(x)(1+\exp(-x))) = x + \ln(1+\exp(-x))$$

$$\ \in x + \exp(-x) - \frac12 \exp(-2x) + O( \exp(-3x) )$$.

Functions may grow in quite complicated manner – it is not that simple as discerning between $x$ and $x^2$. For example $$f(x) = \frac{x(x+1)}2 + \frac{x(x-1)}2\cos x$$ oscillates between $x$ and $x^2$:
for $x=2n\pi$ with natural $n$, that is whenever $\cos x = 1$, we get $f(x)=x^2$,
but for each $x=(2n+1)\pi$ we have $\cos x=-1$ and $f(x)=x$.

The function obviously satisfies $$\lim_{x\to\infty}f(x) = \infty$$ because is gets arbitrarily large when $x$ gets large enough ($f(x)\ge x$ for positive $x$), however it would be very hard to tell if it grows 'like $x$' or 'like $x$ squared' (whatever meaning you would like to assign to these expressions).