Confusion about having limit value an given point

Question

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 ?

Answer 1

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.

Answer 2

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.