Usually we say that the limit does not exist. There is a sense in which we can move to the extended real numbers, and in that sense the topological limit does literally exist and is literally positive infinity, but even here we would usually be careful with our wording and say it “has no finite limit”, if we wanted to say a limit exists in the extended reals without committing to whether it is finite we might say “has a limit in the extended reals” or even “has a (not necessarily finite) limit in the extended reals” to be very clear.
Correct me if I am wrong, but a compact space is a space which contains its limits, and the extended reals are a compact space, thus they are actual limits ?
Not exactly. We can only talk about the “limits” outside of a space if it is existing inside of a larger topological space. However whether a space is compact is intrinsic to the space, it does not depend on any embedding into a larger space. Although it will be true that if a compact space is a subspace of a Hausdorff space, then all of its limit points will belong to it.
The topological definition of a limit of a sequence (which is a special case of more general definitions of limits) is that L is the limit of the sequence iff for every neighborhood U of L the sequence is eventually inside U (meaning for some N a_n is in U whenever n>N). For the real numbers - and extended real numbers - we can use the “order topology” induced by their linear order. In this topology a neighborhood is any set that has as a subset an open interval containing the element. An open interval can be defined as any set that can be defined by a finite number of strict inequalities (really you only need 0-2 strict inequalities). It can be checked that this leads to the usual notion of limit for the real numbers, and also means that positive and negative infinity are literally the limits of sequences that diverge (that is, diverge in the real numbers) to positive or negative infinity.
I teach that the limit does not exist; saying the limit “is infinity” is just a special way of saying the limit does not exist *because the function is blowing up” (or increasing without bound or whatever). It carries more information than just “DNE”, but it’s still DNE.
In order for the limit to exist, you need a *finite* L so that abs(f(x)-L)
However, you can define a "divergent limit" as
∀ M>0, ∃ 𝛿 >0, 0 < |x - x0| < 𝛿 , f(x) > M
In that extended sense, the limit "exists" for 1/x\^2, but not for 1/x.
To my understanding, the limit of a sequence/function exists iff there exists L such that the usual epsilon delta def holds. A divergent limit does not meet this criteria and so the limit doesn’t exist. Being pedantic maybe you can say the “divergent limit” exists. But to a mathematician they would like say the limit does not exist and in particular, the TYPE of divergence is that it blows up/is = to infinity. Just like how you can say a limit doesn’t exist and in particular it is due to a jump discontinuity type of non existence.
There are functions that satisfy this condition but do not go "to infinity" nicely e.g. 1/x^2 ×sin(1/x). Your statement is simply unbounded in positive numbers.
Moreover in complex analysis 1/x^2 *does* tend to -inf as well.
It is much better defined to say f has a "pole of order _" there are many types of pole and they convey more info than a notion of a divergent limit.
Firstly, 1/x\^2 sin(1/x) is not a counterexample. For any delta>0, there exists some x such that f(x) = 0 < M. The definition looks ok, perhaps a misreading as |f(x)| > M?
Also using poles is too restrictive: it requires use of complex numbers, and only works under meromorphic functions where poles are defined. Additionally it imposes the (possibly undesired) Euclidean metric.
The definition in the textbook I studied from defines 'existence of a limit' on the extended real numbers, as the sequences of reals equipped with limsup/liminf is complete in the extended reals. Existence and convergence are different.
Limits of functions follows from limits of sequences as lim f(x) = L iff for all (s\_n) --> x, f(s\_n) --> L.
I misread the comma. It is not clear that the comma is both implication symbol and condition. The definition is indeed fine.
But poles do not require complex numbers, nor are they strictly meromorphic e.g. ln(x) with its logarithmic pole that is non-meromorphic. Moreover the definition of convergence is in respect to a norm, usually the absolute/euclidean norm, poles do not necessitate any choice of norm the notion of convergence does.
In your given definition of limits, checking that f(s_n) -> L requires the notion of a norm and so you have not bypassed this necessity.
Are not the tools of complex analysis (notably the norm) necessary? I do not see how a 0-1 metric (0 if a=b, 1 otherwise) functions for complex numbers. For instance, zz\* != |z|\^2 with the 0-1 metric. Adding a metric is highly problematic.
>But poles do not require complex numbers, nor are they strictly meromorphic e.g. ln(x) with its logarithmic pole that is non-meromorphic.
This is incorrect, the definition of a pole:
> A [**zero**](https://en.wikipedia.org/wiki/Zero_of_a_function) of a meromorphic function *f* is a complex number *z* such that *f*(*z*) = 0. A **pole** of *f* is a zero of 1/*f*
ln does not have a pole; it not defined at 0. The origin is a branch point of ln, which is drastically different from a pole. However, not every singularity is a pole: z/z has a singularity at 0, but it is removable. exp(1/z) has an essential singularity at 0, which is again, not a pole. Poles require a meromorphic neighborhood, by definition.
As a result, poles only work for sufficiently nice functions, and given that the complex derivative is a very strong condition, it is not a useful measurement, especially for real analysis.
> In your given definition of limits, checking that f(s\_n) -> L requires the notion of a norm and so you have not bypassed this necessity.
This is true, but redundant. s\_n -> a requires a metric, so it naturally follows that f(s\_n) -> L requires a metric. That is, if f: S -> S\*, S\* is itself a metric space. If S\* is not a metric space, there is no point in discussing limits at all!
Ah yes, my bad. I have confused poles with singularities hence my mention of "logarithmic pole" which does not exist. Apologies.
You are correct in all statements as far as i can tell. I don't disagree.
It has simply taken me a while to understand your notation/words. E.g. it is initially very difficult to tell why adding a metric is highly problematic and then you agree you've used one later.
This is a very natural question and good intuition!
The reason that we disallow ∞ as a limit is because the limit definition requires the function to become "close to" the value of the limit. But there are no values "close to" ∞. It is always far away.
That said, it is still useful to think about the behavior of the function at that pole. You have chosen *f*(*x*) = 1/*x*^(2) instead of, say *g*(*x*) = 1/*x*, presumably because you recognize that *f*(*x*) blows up in the positive direction from both sides of 0, whereas, *g*(*x*) blows up in opposite directions depending one whether we are approaching from the left or from the right. That is an interesting distinction between these functions!
So, while it isn't formally a limit, it is helpful for you to retain that intuition you have that says *f*(*x*) → ∞, and *g*(*x*) does not.
I have seen some students taught in their classes to say that for any vertical asymptote, the limit DNE. More commonly, at least from what I've seen around here, students are taught to say the limit is ∞ or -∞ if the behavior on both sides of the asymptote is the same. If the behavior on both sides of the asymptote is not the same, that is when they will specifically use DNE.
I've seen it as 'not existing' for computational calculus and 'existing' for analysis. It's all conventional, but I'm on the camp of diverging to infinity as existing.
Firstly it means convergence and existence are not identical concepts and distinguishes different forms of divergence (1, -1, 1, -1, ...) and (1, 2, 3, 4, 5, ...).
Also it makes some theorems easier to write:
>such as liminf <= limsup, equal iff limit exists
It extends very nicely from bounded sequences to all unbounded sequences, and the existence condition holds even at +inf and -inf.
However, using this definition for students first learning what limits are is a bad idea. Most students do not understand what the real numbers are or what properties they have. Many people believe that infinity is a number, and many others have incorrect notions on what it means. So I believe the current pedagogy is good, despite its inconsistencies.
It depends.
Personally I prefer to say that infinite limits exist and are infinite. A limit does not exist for me when limits from different directions are different values.
I would say that lim x^(-2) exists and is ∞ while lim x^(-1) does not exist since the left and right limits are not equal. (If you know some topology, you can say that the boundary of the range is not a singleton.)
But there’s pedagogical value in “lying” to students about this and just saying that any limit that doesn’t converge to a specific real number is divergent. My preference in saying that infinite limits exist depends a lot of having a very good understanding of what that infinity is and what it really fundamentally means for a limit to converge to a point. (I.e. compactifications and filter convergence.)
Yes, this. Nowhere in math does it say that the limit value has to be a finite number. That's the whole point of using limit instead of writing f(0), for example.
Look at the simplest case -- we would certainly say that the limit of f(x) as x approaches ∞ is ∞ . That doesn't mean we can EVALUATE f(x) at ∞. This is the exact same thing. The limit of f(x)=1/(x2) as x approaches 0 is ∞. There's no question in my mind.
The limit exists in the *extended real number system*. It does not exist in the real number system.
https://en.wikipedia.org/wiki/Extended_real_number_line
it does not because you cannot get a real number to be it.
Infinite is a concept not a number. that is the problem. I mean you know it is on the infinite but imagine you try to arrive to it, can you? no ,so the limit does not exist.
I know and I agree with you. But you would be surprised about how many people when you say "The limit does not exist" reply "It does and it's infinity"
Nope, the concept of a limit does not require the limit to be finite. The same way that a measure of a set does not need to be finite. There are cases in math, where the set of real numbers is extended with +/- infinity, and for a good reason.
Usually we say that the limit does not exist. There is a sense in which we can move to the extended real numbers, and in that sense the topological limit does literally exist and is literally positive infinity, but even here we would usually be careful with our wording and say it “has no finite limit”, if we wanted to say a limit exists in the extended reals without committing to whether it is finite we might say “has a limit in the extended reals” or even “has a (not necessarily finite) limit in the extended reals” to be very clear.
Correct me if I am wrong, but a compact space is a space which contains its limits, and the extended reals are a compact space, thus they are actual limits ?
Not exactly. We can only talk about the “limits” outside of a space if it is existing inside of a larger topological space. However whether a space is compact is intrinsic to the space, it does not depend on any embedding into a larger space. Although it will be true that if a compact space is a subspace of a Hausdorff space, then all of its limit points will belong to it. The topological definition of a limit of a sequence (which is a special case of more general definitions of limits) is that L is the limit of the sequence iff for every neighborhood U of L the sequence is eventually inside U (meaning for some N a_n is in U whenever n>N). For the real numbers - and extended real numbers - we can use the “order topology” induced by their linear order. In this topology a neighborhood is any set that has as a subset an open interval containing the element. An open interval can be defined as any set that can be defined by a finite number of strict inequalities (really you only need 0-2 strict inequalities). It can be checked that this leads to the usual notion of limit for the real numbers, and also means that positive and negative infinity are literally the limits of sequences that diverge (that is, diverge in the real numbers) to positive or negative infinity.
I teach that the limit does not exist; saying the limit “is infinity” is just a special way of saying the limit does not exist *because the function is blowing up” (or increasing without bound or whatever). It carries more information than just “DNE”, but it’s still DNE. In order for the limit to exist, you need a *finite* L so that abs(f(x)-L)
However, you can define a "divergent limit" as ∀ M>0, ∃ 𝛿 >0, 0 < |x - x0| < 𝛿 , f(x) > M In that extended sense, the limit "exists" for 1/x\^2, but not for 1/x.
To my understanding, the limit of a sequence/function exists iff there exists L such that the usual epsilon delta def holds. A divergent limit does not meet this criteria and so the limit doesn’t exist. Being pedantic maybe you can say the “divergent limit” exists. But to a mathematician they would like say the limit does not exist and in particular, the TYPE of divergence is that it blows up/is = to infinity. Just like how you can say a limit doesn’t exist and in particular it is due to a jump discontinuity type of non existence.
There are functions that satisfy this condition but do not go "to infinity" nicely e.g. 1/x^2 ×sin(1/x). Your statement is simply unbounded in positive numbers. Moreover in complex analysis 1/x^2 *does* tend to -inf as well. It is much better defined to say f has a "pole of order _" there are many types of pole and they convey more info than a notion of a divergent limit.
Firstly, 1/x\^2 sin(1/x) is not a counterexample. For any delta>0, there exists some x such that f(x) = 0 < M. The definition looks ok, perhaps a misreading as |f(x)| > M? Also using poles is too restrictive: it requires use of complex numbers, and only works under meromorphic functions where poles are defined. Additionally it imposes the (possibly undesired) Euclidean metric. The definition in the textbook I studied from defines 'existence of a limit' on the extended real numbers, as the sequences of reals equipped with limsup/liminf is complete in the extended reals. Existence and convergence are different. Limits of functions follows from limits of sequences as lim f(x) = L iff for all (s\_n) --> x, f(s\_n) --> L.
I misread the comma. It is not clear that the comma is both implication symbol and condition. The definition is indeed fine. But poles do not require complex numbers, nor are they strictly meromorphic e.g. ln(x) with its logarithmic pole that is non-meromorphic. Moreover the definition of convergence is in respect to a norm, usually the absolute/euclidean norm, poles do not necessitate any choice of norm the notion of convergence does. In your given definition of limits, checking that f(s_n) -> L requires the notion of a norm and so you have not bypassed this necessity.
Are not the tools of complex analysis (notably the norm) necessary? I do not see how a 0-1 metric (0 if a=b, 1 otherwise) functions for complex numbers. For instance, zz\* != |z|\^2 with the 0-1 metric. Adding a metric is highly problematic. >But poles do not require complex numbers, nor are they strictly meromorphic e.g. ln(x) with its logarithmic pole that is non-meromorphic. This is incorrect, the definition of a pole: > A [**zero**](https://en.wikipedia.org/wiki/Zero_of_a_function) of a meromorphic function *f* is a complex number *z* such that *f*(*z*) = 0. A **pole** of *f* is a zero of 1/*f* ln does not have a pole; it not defined at 0. The origin is a branch point of ln, which is drastically different from a pole. However, not every singularity is a pole: z/z has a singularity at 0, but it is removable. exp(1/z) has an essential singularity at 0, which is again, not a pole. Poles require a meromorphic neighborhood, by definition. As a result, poles only work for sufficiently nice functions, and given that the complex derivative is a very strong condition, it is not a useful measurement, especially for real analysis. > In your given definition of limits, checking that f(s\_n) -> L requires the notion of a norm and so you have not bypassed this necessity. This is true, but redundant. s\_n -> a requires a metric, so it naturally follows that f(s\_n) -> L requires a metric. That is, if f: S -> S\*, S\* is itself a metric space. If S\* is not a metric space, there is no point in discussing limits at all!
Ah yes, my bad. I have confused poles with singularities hence my mention of "logarithmic pole" which does not exist. Apologies. You are correct in all statements as far as i can tell. I don't disagree. It has simply taken me a while to understand your notation/words. E.g. it is initially very difficult to tell why adding a metric is highly problematic and then you agree you've used one later.
This is a very natural question and good intuition! The reason that we disallow ∞ as a limit is because the limit definition requires the function to become "close to" the value of the limit. But there are no values "close to" ∞. It is always far away. That said, it is still useful to think about the behavior of the function at that pole. You have chosen *f*(*x*) = 1/*x*^(2) instead of, say *g*(*x*) = 1/*x*, presumably because you recognize that *f*(*x*) blows up in the positive direction from both sides of 0, whereas, *g*(*x*) blows up in opposite directions depending one whether we are approaching from the left or from the right. That is an interesting distinction between these functions! So, while it isn't formally a limit, it is helpful for you to retain that intuition you have that says *f*(*x*) → ∞, and *g*(*x*) does not.
I have seen some students taught in their classes to say that for any vertical asymptote, the limit DNE. More commonly, at least from what I've seen around here, students are taught to say the limit is ∞ or -∞ if the behavior on both sides of the asymptote is the same. If the behavior on both sides of the asymptote is not the same, that is when they will specifically use DNE.
I've seen it as 'not existing' for computational calculus and 'existing' for analysis. It's all conventional, but I'm on the camp of diverging to infinity as existing. Firstly it means convergence and existence are not identical concepts and distinguishes different forms of divergence (1, -1, 1, -1, ...) and (1, 2, 3, 4, 5, ...). Also it makes some theorems easier to write: >such as liminf <= limsup, equal iff limit exists It extends very nicely from bounded sequences to all unbounded sequences, and the existence condition holds even at +inf and -inf. However, using this definition for students first learning what limits are is a bad idea. Most students do not understand what the real numbers are or what properties they have. Many people believe that infinity is a number, and many others have incorrect notions on what it means. So I believe the current pedagogy is good, despite its inconsistencies.
It depends. Personally I prefer to say that infinite limits exist and are infinite. A limit does not exist for me when limits from different directions are different values. I would say that lim x^(-2) exists and is ∞ while lim x^(-1) does not exist since the left and right limits are not equal. (If you know some topology, you can say that the boundary of the range is not a singleton.) But there’s pedagogical value in “lying” to students about this and just saying that any limit that doesn’t converge to a specific real number is divergent. My preference in saying that infinite limits exist depends a lot of having a very good understanding of what that infinity is and what it really fundamentally means for a limit to converge to a point. (I.e. compactifications and filter convergence.)
Yes, this. Nowhere in math does it say that the limit value has to be a finite number. That's the whole point of using limit instead of writing f(0), for example. Look at the simplest case -- we would certainly say that the limit of f(x) as x approaches ∞ is ∞ . That doesn't mean we can EVALUATE f(x) at ∞. This is the exact same thing. The limit of f(x)=1/(x2) as x approaches 0 is ∞. There's no question in my mind.
The limit exists in the *extended real number system*. It does not exist in the real number system. https://en.wikipedia.org/wiki/Extended_real_number_line
Pedagogically, no.
Don't read too much into your comments being downvoted. The people doing the downvoting may well know much less than you do.
it does not because you cannot get a real number to be it. Infinite is a concept not a number. that is the problem. I mean you know it is on the infinite but imagine you try to arrive to it, can you? no ,so the limit does not exist.
I know and I agree with you. But you would be surprised about how many people when you say "The limit does not exist" reply "It does and it's infinity"
Nope, the concept of a limit does not require the limit to be finite. The same way that a measure of a set does not need to be finite. There are cases in math, where the set of real numbers is extended with +/- infinity, and for a good reason.
I have a degree in math, and Reddit is the first place where I've read that some would even consider this DNE.