@Mann xsin(2/x) has missing point discontinuity at x=0 right

What is missing point discontinuity? @Jasmine

@Mann its discontinuous there right

missing point is that discontinuity which is removable

Can you define the function?

@Jasmine

It really depends on what the function is

@Mann function is xsin(2/x)

the question doesnt provide more information

ill-posed question

The question is bad.

See

ok if we take domain as R - {0}

and range as R

On that domain function is continuous

then

and if nothing is given then

Question is wrong

The function is not properly defined

Domain and range are a part of the definition of function

is it not obvious that the domain is R - {0}

Well, I can assume that.

Or maybe the question wants me to assume that/

In that case, the function is continuous on its domain

if were given another question like what is the domain of xsin(2/x) we would have said its R - {0}

I can say that such functions are also defined for complex numbers.

Then?

Areyy, I don't understand the concept of "missing point discontinuity" though.

@Mann we are working in cartesian plane

Yeaa, but why should I still say the domain is R-0

Why not [-1,1] - 0

this seems confusing

Moreoever, I just realizes -1 < 2/x < 1

d*

See

I don't know what you had been taught.

But I can definitely assure you that a proper definition of functions means that domain and range is specified.

Secondly,

Maybe before college, you are taught under the assumption that real number is everything. Therefore domain is everything that lies in real number where the function is defined

Sure, you can use that definition to construct the domain for certain questions, no issue.

But you have to remember that this is still a definition.

Yeaa, this definition makes sense.

But wait.

Oh wait my badd

Yea

forget the 2/x wala part

THat was a bad mistake on my part

then

So

In the current situation

"you are taught under the assumption that real number is everything. Therefore domain is everything that lies in real number where the function is defined"

Use this definition to construct a domain for xsin(2/x)

@Mann yes

Please remember that this is still an assumption.

It's not a given truth.

Going on with this

We get the domain that function has domain R-0

@Mann yes

Now 0 is a limit point of this domain

Do you know the concept of a limit point though?

@Mann no

Consider the space of whole real number R and in it you have a set R-0

ok

You see that the set R-0 look as some disjoint line segments in R

With open intervals at 0

Okay?

You got this @Jasmine?

$\mathbb{R} = (-\infty,\infty)$ and $\mathbb{R}-0=(-\infty,0) \cup (0,\infty) $

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

I will wait until you get back.

sorry was afk

@Mann yes

so now since we have defined a domain so can we say the function is continuous

but what about this

@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.

@Mann ohhhh !!

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

Here's how you characterizes it.

@Mann am I expected to answer this in JEE too

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

@Mann ok please enlighten

GIve me 10 min

ok

Brb

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

i will be back

@Mann yea fine

@Mann I will have to go its raining heavily

please still write what you wanted to explain

I will see when I am back :)

@Jasmine I am back

For example

$0 \in \mathbb{R}$

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

@Mann ok

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

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

@Mann go on

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)*

@Jasmine

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

@Mann yes

You visualized it with the example?

I am reading now more carefully

Okayy

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

Yess

Atleast one point of D

@Mann ok

I think I understood what you mean by limit point

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?

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

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

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

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

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

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"

And what about the missing point discontinuity we are taught

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

Yeaa coming to that too.

BUt you understand until this right?

@Mann yes !

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.

@Mann yes as 0 is not the limit point

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?

@Mann one sec

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.

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

Yep.

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

Yep.

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

Not at limit points.

which are outside domain $D$

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

at what?

Nope.

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

@Mann but 0 is not the part of domain

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.

@Mann missing point discontinuity you mean

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.

@Mann ohhhhh

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

Coool !

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?

@Mann one thing

Yes?

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

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

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

a) true.

b) limit exists at ${\alpha}$

I dint think b) is true

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

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

Yep!

Which is not true

b) is not necessarily true,

@Mann Thank you !! You are awesome !

Haha, no issues.

Got to learn a really nice thing today

I have been through those times too.

I really don't understand why they don't teach these things.

It's not like it's too difficult either when applied at specific cases.

@Mann sad but its not something too difficult I guess

Exactly!

Anyway, I gotta go now though. Need to prepare for GRE

See you.

Goodnight

Goodnight