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Today , me and my friend discussed on a question which is about whether there is limit in a given point.To answer this , we assumed that let $f$ be a function such that $f:{1} \rightarrow {4} ,f(1)=4$ ,i.e, its domain is only $1$ and image is only $4$ , but these values are isolated. Thanks to this function , we write the point $(1,4)$ in cartesian coordinate system.

I said that because of our $x$ value is an isolated number ,i.e $1$, we cannot close to $1$ because we do not have $f(x)$ values for those values. However , my friend said that we have a limit here and the limit value is $1$ by using $\epsilon - \delta$ definition.

After this discussion ,we went to our professors , one of them firstly said that there is no limit of a point , but after that she hesitated about it and wanted some time for thinking. The other professor said that we can find limit value for the given point using epsilon-delta.

I am confused here , because i learned that limit is the behavior of a function in given x value when we get close to (but not equal) this point. Here , our function has domain $1$ , so we cannot talk about $1.000000...1 $ or $0.9999..$.

Briefly , what do you think about whether a given point has a limit value there ? If it has limit value can you prove it using epsilon delta and for point $(1,4) $ ?

Addentum: By formal definition ; For the function f(x) defined on an interval that contains $x =a$. Then we say that,

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

If for every number epsilon($∈$) which greater than zero, there is some positive number delta $\delta$ such that,

$| f(x) – L | < ∈$ where $0 < |x – a| < \delta$

Source : https://www.geeksforgeeks.org/formal-definition-of-limits/

So, we see that to be able to use epsilon-delta , we must have a defined interval , but we do not have any defined interval except for $x=1$ isolated point. Then how dare can you use epsilon delta ?

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    Your confusion comes from simple thing: what is limit point? Once you properly and precisely define it you will have no problem. Note that limit point can mean multiple, non equivalent things, depending on context. For example limit of sequence and limit of set are different things. Continuity at a point is another. Your example may satisfy some but not other definitions. – freakish Mar 09 '23 at 13:07
  • @freakish i dont say limit point , it is limit of a given single point in cartesian coordinate according to its x value –  Mar 09 '23 at 13:14
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    There's a limit of sequence, limit of set, etc. There is no such thing as limit of a single point, or rather: define what limit is and then you'll have your answer. Seriously, it's all about definitions, nothing else. – freakish Mar 09 '23 at 13:15
  • @freakish but my prof say that there is a limit of single point , it is the problem –  Mar 09 '23 at 13:17
  • I don't care what your prof says. Maybe they thought about different definitions. That's irrelevant. Stick to maths: definitions, theorems and proofs. That's what matters in maths. And that's what allows us to be objective and not depend on someone's opinion. Again: pick the definition and you'll have your answer. – freakish Mar 09 '23 at 13:17
  • This seems relevant: https://math.stackexchange.com/q/2732987/139123 – David K Mar 09 '23 at 13:18
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    Also relevant: https://math.stackexchange.com/q/2216560/139123 – David K Mar 09 '23 at 13:20
  • Your friend cites the definition. If his argument uses mathematical logic correctly then it is true. – David K Mar 09 '23 at 13:21
  • @DavidK we talked about this I thought like nameless –  Mar 09 '23 at 13:28
  • That answer uses a definition that requires a limit point. It immediately says that your function has no limit at $1$. But as you see from other answers, not everyone uses that definition of limit. This situation (where some people define it one way, some another) may be tolerated because nobody really cares about the limit at an isolated point. – David K Mar 09 '23 at 13:36
  • @DavidK please correct me , you says that if we use one definition ,then there is not limit at $1$, but if we use epsilon delta , then there is limit at $1$ –  Mar 09 '23 at 13:44
  • @DavidK "nobody really cares about the limit at an isolated point" , we were about to fight because of this discussion –  Mar 09 '23 at 13:44
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    That's not epsilon. See https://math.stackexchange.com/questions/2926005/how-can-%e2%88%88-be-less-than-greater-than-something – mr_e_man Mar 09 '23 at 14:28
  • True, I should not have said "nobody"! Clearly you care. I also find the question interesting. It might be better to say that in the course of doing mathematics, mathematicians rarely if ever encounter a situation where one definition or the other prevents them from accomplishing their goals. – David K Mar 09 '23 at 14:31
  • The use of "where" in your definition is very strange. Usually, we say "where" when we have introduced a new symbol (or new value of the symbol) and need to say what the symbol means (or what the value is). For the definition of a limit, the wording would usually be "$| f(x) – L | < \epsilon$ if $0 < |x – a| < \delta$," and as one of the linked questions says, when the "if" part of a conditional is false (which it is for every $x$ in your case) then the conditional is true. – David K Mar 09 '23 at 14:38
  • @DavidK - It is unusual, but "where" actually makes sense here. It refers to locations on the $x$-axis. Though, it might be better to use "wherever", or "whenever". – mr_e_man Mar 09 '23 at 14:48
  • @mr_e_man "When" or "whenever" would be fine. The point is that this part of the definition is a conditional, which can be vacuously true. The use of "where" makes this feature of the definition obscure (if indeed it still holds at all). – David K Mar 09 '23 at 15:45
  • I just noticed that your definition of "limit" says that $\lim_{x\to0} \frac xx$ is undefined. We usually would like to say that the limit is $1$ even though $\frac xx$ itself (not the limit) is undefined when $x=0.$ – David K Mar 09 '23 at 15:48
  • @DavidK it is the formal definiton which can be found in internet , i did not understand your comment –  Mar 09 '23 at 15:52
  • Your definition says $\lim_{x\to0} \frac xx$ does not exist because $\frac xx$ is not defined on an interval that includes $x=0.$ – David K Mar 09 '23 at 15:58
  • Compare with this definition: https://web.mit.edu/wwmath/calculus/limits/formal.html ... In the discussion it is noted that $f(x)$ need not be defined at $x=a$ and you can still take a limit at $a$ (according to that definition). – David K Mar 09 '23 at 16:01
  • @DavidK yes , we restricted our domain in only $x=1$ and it does not even allow an open ball around it if we talk in language of topology. We are just talking about a single point in cartesian cordinate whose coordinate is $(1,4)$ and to work over it we used a function using given domain and range –  Mar 09 '23 at 16:02
  • We understand all that, but the answer to "does the limit exist" still depends on the precise definition you use. Various respected mathematicians use definitions that imply the answer is "yes". Others don't. – David K Mar 09 '23 at 16:03
  • So if you are just going to go hunting for definitions, you can find either answer. A better approach might be to look up the definition given in your last calculus course, if both of you took the same course. – David K Mar 09 '23 at 16:05
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    Just to emphasize, there is no "the formal definiton which can be found in internet". There are many formal definitions, not all equivalent, that can be found on the internet. – David K Mar 09 '23 at 16:13
  • @DavidK i now think that there cannot be thought about limit using any epsilon-delta ,because in my question , there would no be any $x$ value for any given epsilon value , because my function has only one element in its domain. So it does not matter the type of epsilon -delta , there is no limit process in a isolated point such that it is represented by a function like above. Am I right here ? –  Mar 09 '23 at 17:35
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    That is not completely correct. Most definitions rely on a logical implication or conditional statement. A conditional such as "if $x\in A$ then $Q$" is vacuously true when $A$ is the empty set. That is, a very simple epsilon-delta definition could apply precisely because you cannot find any $x$ in the domain such that $0<|x|<\delta$ and therefore the epsilon-delta condition is vacuously true. But as noted in https://math.stackexchange.com/a/3166117/139123, that means the limit could be any number, so it's not a very useful definition in that case. – David K Mar 09 '23 at 17:46
  • @DavidK THANK YOU VERY MUCH :) –  Mar 09 '23 at 18:08
  • @boun: You might find this answer helpful which cites different references for both versions of limit. – Markus Scheuer Mar 09 '23 at 21:34

2 Answers2

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It all depends on precisely what your $\epsilon$-$\delta$ definition of limits is, namely whether you use $0<|x-a|<\delta$ (and require $a$ to be an accumulation point of the function's domain) or just $|x-a|<\delta$. (Yes, both definitions are used; see for example this question.) With the first definition, the limit doesn't exist, but with the second definition, the limit is 4.

Hans Lundmark
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  • when we discuss the question , we think over the first definition ,i.e $0<|x-a|< \epsilon$. However , even though the first definition , they still insist on that limit is $4$. It is the point where i try to explain them. –  Mar 09 '23 at 15:37
  • Well, with the first definition it is usually stated explicitly that $a$ must be an accumulation point of $D_f$, since otherwise it would be true, when $a$ is an isolated point of $D_f$, that $f(x) \to L$ as $x\to a$ for any number $L$. (The condition $0<|x-a|<\delta \implies |f(x)-L|<\epsilon$ becomes vacuously true, regardless of the value of $L$.) And then the notation $\lim_{x\to a} f(x)$ would be meaningless since you can't have $\lim_{x\to a} f(x) = L$ for all $L$. So unless your friends are prepared to accept such weirdness, they had better require $a$ to be an accumulation point. – Hans Lundmark Mar 09 '23 at 15:43
  • And then, since an isolated point of $D_f$ is not an accumulation point of $D_f$, the limit (by definition) does not exist at an isolated point. – Hans Lundmark Mar 09 '23 at 15:45
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    @boun As I just observed in a comment under the question, your definition is actually more restrictive than even the first definition, since you require not only that $a$ is a limit point but actually it must be in the domain and the domain must include an interval about $a.$ A limit point would not necessarily meet those conditions. – David K Mar 09 '23 at 15:51
  • @DavidK: That's a good point. I didn't notice that. So with that definition, the immediate answer would be “we haven't even defined what it means to have a limit at an isolated point, so it doesn't make any sense to claim that the limit exists”. – Hans Lundmark Mar 09 '23 at 15:55
  • @HansLundmark i now think that there cannot be thought about limit using any epsilon-delta ,because in my question , there would no be any $x$ value for any given epsilon value , because my function has only one element in its domain. So it does not matter the type of epsilon -delta , there is no limit process in a isolated point such that it is represented by a function like above. Am I right here ? –  Mar 09 '23 at 17:35
  • Yes, with the $0<|x-a|<\delta$ definition, that condition is satisfied for no $x \in D_f$. And since $P \implies Q$ is always true when $P$ is false, no matter what $Q$ is, the implication $(\forall x \in D_f) , (0<|x-a|<\delta \implies |f(x)-L|<\epsilon)$ becomes true, no matter what $L$ is. So the $\epsilon$-$\delta$ condition holds for any $L$ when $a$ is an isolated point in $D_f$ (as I wrote in my previous comment). Hence the extra condition that $a$, to begin with, needs to be an accumulation point of $D_a$; otherwise we don't even consider talking about $\lim_{x\to a} f(x)$. – Hans Lundmark Mar 09 '23 at 19:54
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First things first. Let us state your definition properly.

Let $f:I \rightarrow \mathbb{R}$ be a function defined in an interval $I=[a,b]$ (possibly degenerated). We say that $f$ has limit equals to $L$ at $x_0 \in I'$, where $I'=\{x\in \mathbb{R}: \forall \varepsilon>0\exists x'\in I\setminus\{x\}\mbox{ such that} |x-x'|<\varepsilon\}$ ($x$ points that can be aproximated by points of $I$ other than $x$), if for all $\varepsilon>0$ there is $\delta>0$ such that all $x\in I\setminus\{x_0\}$ that satisfies $|x-x_0|<\delta$ we have $|f(x)-L|<\varepsilon$.

By using your definition (Notice that you ask $0<|x-a|<\delta$) it makes no sense to ask for limit if your interval is degenerated. There is no point that can be approximated by a point of $I$ other than itself.

Notice that if we define the limit "allowing" $0=|x-a|$, then you always will have limit, because you always will have that the only point in $I$ satisfies the conditions for $L=f(a)$.

I hope that is clear that is a matter of definition you are using.

Ygor Arthur
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  • when we discuss the question , we think over $0<|x-a|< \epsilon$. However , even though we use this definition , they still says that limit is $4$. It is the point where i try to explain them. –  Mar 09 '23 at 15:37
  • In this case it make no sense to ask if the limit exists, because there is no $x$ in the domain of $f$ that satisfies this condition. The point is that you are not in the context to ask this. For example, supose that you work with the function that sends any natural $n$ to $2n$. It makes no sense to ask what is the image of $\pi$ in this context. – Ygor Arthur Mar 10 '23 at 21:16