limit point

Mann

Aug 22, 2019 16:24

@Jasmine the function can be continuous on its domain and at the same time has a missing point discontinuity

This is what I am coming to here.

But for that I really need to tell you what a limit point is and what a isolated point is.

Jasmine

@Mann ohhhh !!

Mann

In our case 0 is the limit point of the domain $\mathbb{R}-0$

Here's how you characterizes it.

Jasmine

@Mann am I expected to answer this in JEE too

Mann

I am not sure, but I this concept is very fundamental to the definition of the limit. It's disastrous not knowing it

think*

And it's simple too

Jasmine

@Mann ok please enlighten

Mann

Aug 22, 2019 16:27

GIve me 10 min

Jasmine

ok

Mann

Brb

Mann

Aug 22, 2019 16:43

I am explaining something important to my friend. Please give me few minutes

i will be back

Jasmine

@Mann yea fine

Jasmine

Aug 22, 2019 17:12

@Mann I will have to go its raining heavily

please still write what you wanted to explain

I will see when I am back :)

Mann

@Jasmine I am back

For now, this definition is good enough for you.
A point $a \in \mathbb{R}$ is a limit point of a subset $D \subset R$ iff every open interval in $\mathbb{R}$ containing the point $a$ contains a point of $D$.

For example

$0 \in \mathbb{R}$

$\mathbb{R}-0 \subset \mathbb{R}$

Jasmine

@Mann ok

Mann

Any open interval contain $0$ is of the form $(-\epsilon,\epsilon)$ for arbitrary epsilon.

Every such open interval $(-\epsilon,\epsilon)$ contains a point of $\mathbb{R}-0$

Jasmine

@Mann go on

Mann

Therefore $0$ is a limit point of the subset $\mathbb{R}-0$

Understand this argument what I have laid properly Jasmine

So far

(The interval can also be non symmetric but it won't change anything)*

It will take some time, so think about it carefully.

@Jasmine

Mainly, I want you to visualize the definition I gave.

Jasmine

Aug 22, 2019 17:24

@Mann yes

Mann

You visualized it with the example?

Jasmine

I am reading now more carefully

Mann

Okayy

Jasmine

@Mann does contains a pont of D means contains at least a point of D

Mann

Yess

Atleast one point of D

Jasmine

Aug 22, 2019 17:27

@Mann ok

I think I understood what you mean by limit point

Mann

Now, let me give some counterexamples, you tell why they are so.

2 is not a limit point of the subset $(-1,1)$

You got it why?

Jasmine

@Mann yes because (1,3) doesnt contain anything from (-1,1)

Mann

YES!

Now one more

$(-8,-7) \cup \left\{0\right\} \cup (7,8)$

0 is not a limit point of this set.

I forgot to add one thing, I am sorry.

Contains a point of D other than $a$

A point a∈R is a limit point of a subset D⊂R iff every open interval in R containing the point a contains a point of D other than $a$

Any point in subset $D$ which is not a limit point is a isolated point of $D$

$0$ is a isolated point of $(-8,-7) \cup \left\{0\right\} \cup (7,8)$

THis is the last, thing and we are done with the base

Jasmine

@Mann (-5,5) doesnt contain D except for {0}

Mann

Yep!

Now limits are only for those values of $x$ which are limit points a domain $D$

We don't have any sense of what is a limit at a point which is not a limit point.

For example to talk about the limit at $x=0$ in the above is non sense

Jasmine

Aug 22, 2019 17:36

@Mann so essentially limit exists for xsin (2/x) at 0 as its the limit point

Mann

Yep!

$x=0$ is a limit point

even though it is not in domain but it doesn't matter

The reason for that is that, the very definition of limit is that we are approaching closer and closer to 0

but we never equal 0

So we do not need 0 to be in the domain as long as it can be "approached"

Jasmine

And what about the missing point discontinuity we are taught

Mann

And it can be approached exactly when it's a limit point.

Yeaa coming to that too.

BUt you understand until this right?

Jasmine

@Mann yes !

Mann

And why did I define limit point, you also understood right?

Here's one more example just to be safe

$f : ( -3,2) \cup \left\{0\right\} \cup (2,3) \to \mathbb{R} $ as $f(x)=x$

THere is no sense of talking about limit at $x=0$

But there is sense of talking about limits at $[-3,2] \cup [2,3]$

$0$ is "unapproachable" and the set I defined later on is approachable.

Jasmine

Aug 22, 2019 17:41

@Mann yes as 0 is not the limit point

Mann

Yep

Now, to the to the fun part.

$ f : (-2,3) \cup (4,6) \to \mathbb{R} $ given by $f(x)=x$

is continuous by definition

Can you see why?

Jasmine

@Mann one sec

Mann

Because if this is continuous, for the same reason. $f(x)=x\sin(2/x)$ was continuous on its domain $\mathbb{R} - 0$. This does need a little bit more proof, I haven't verified this exactly properly. But I think it's true most probably.

Jasmine

@Mann [-2,3] U [4,6] are limit points

Mann

Yep.

Jasmine

Aug 22, 2019 17:50

They are limit points so that means limit exists at those points

Mann

Yep.

BUt continuity is only checked at points in the domain $D$

Not at limit points.

which are outside domain $D$

Jasmine

@Mann ok so that would mean sin (2/x ) is discontinuous at 0

Mann

at what?

Nope.

$x=0$ is not a point of the domain.

Jasmine

@Mann but 0 is not the part of domain

Mann

Yes, continuity has no sense outside the domain

You can't talk about continuity of a function outside its domain.

No no.

Well yes in a sense

But

is different

That definition you gave me about discontinuity something something

See what it states.

Jasmine

Aug 22, 2019 17:54

@Mann missing point discontinuity you mean

Mann

Yess

THis new defintion

The discontinuity is said to be of missing point type if the limit of the function exists at point 'a' but the function is not defined at that point, i.e. lim x→a f(x) exists but f(a) is not defined."

It is defined for points specifically outside the domain.

It's not the same continuity.

Jasmine

@Mann ohhhhh

Mann

That's why I said, a function can be continuous and at the same time have a missing point discontinuity.

Jasmine

Coool !

Mann

The definition says that $f(a)$ is not defined.

therefore $a \notin D$

It's exclusively defined for points outside of function's domain.

You got it right?

Jasmine

Aug 22, 2019 17:57

@Mann one thing

Mann

Yes?

Jasmine

You said if say ${\alpha}$ is a limit point of any function then what is correct-

Mann

$\alpha \in \mathbb{R}$ can be a limit point of a $D \subset \mathbb{R}$ only.

Jasmine

a) we can talk about limit at ${\alpha}$

Mann

a) true.

Jasmine

Aug 22, 2019 18:00

b) limit exists at ${\alpha}$

I dint think b) is true

Mann

b) That will depend though, you will have to use the epsilon delta definition of limit.

Jasmine

If b) was true then limit would have existed at 0 in sin(1/x)

Mann

Yep!

Jasmine

Which is not true

Mann

b) is not necessarily true,

Jasmine

Aug 22, 2019 18:02

@Mann Thank you !! You are awesome !

Mann

Haha, no issues.

Jasmine

Got to learn a really nice thing today

$Mathematics$ JEE Maths Zone

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