help-0

[alpine sable]

😢 ok

[oak chasm]

@alpine sable Simplify sqrt(12).

[alpine sable]

Ok...

[oak chasm]

Oh, never mind, the 12 · 8 is inside the square root.

[alpine sable]

No no See

Root is used only for 12

X 8 is different

[edgy breach]

So it's 8 sqrt(12), yes?

So simplify the root 12 first

[oak chasm]

Then none of the answers is correct.

[alpine sable]

Ok... How again?

[oak chasm]

Yeah, how can they get an integer times sqrt(6) with sqrt(12)?

[alpine sable]

Not possible

[oak chasm]

OK, how can they get an integer times sqrt(6) with sqrt(12)?

Since it's not possible, and all answers are an integer times sqrt(6), show how to do that.

[edgy breach]

Not possible mate

[alpine sable]

So none of the above?

[oak chasm]

The only way for it to work is if it's sqrt(12 · 8).

[alpine sable]

Hmm wait lemme try searching on Google

Omg sorry

I typed the q wrong

Mahn u ppl are genius

√12 x √8 is equal to?

A) 2√6
B) 3√6
C) 4√6
D) 6√6

Now it's correct

Check

[oak chasm]

OK, so factor 12 and 8.

Into primes.

What do you get?

[glass elm]

root 2 2 3 2 2 2

[alpine sable]

Ok lemme check

[glass elm]

@alpine sable You need to get the lowest factors and then add up the pairs together to form squares and the remaining single will be outside

[alpine sable]

Ohk

[glass elm]

when you get the lowest factors, 2 2 3 2 2 2

Pair them up

2 and 2

2 and 2

2 and 3

[alpine sable]

Got it

Thanks

[glass elm]

4root6

@oak chasm Could you perhaps give me a small intro to Limits?

I just want to understand the concept, I don't really understand what a Limit is despite checking so many source

Can please give a small intro?

[oak chasm]

@glass elm A function has an input and an output.

f(2) = 3 means the input is 2 and the output is 3.

[glass elm]

Oh, like, f(x) = 2 +x. If x is 2, output is 4?

[oak chasm]

Right.

So, if you get the input closer and closer to 2, like this:

f(1.9)
f(1.99)
f(1.999)

What happens to the output?

If it gets closer and closer to a number, that's the limit.

f(1.9) = 3.9
f(1.99) = 3.99
f(1.999) = 3.999

The input is getting closer to 2. The output is getting closer to 4.

[glass elm]

But, what's the point of it?

[glass lichen]

(keeping in mind the actual function value need not be the value of the limit)

[glass elm]

Like, if we take f(1) =3

[rustic rock]

yup hes right

[glass elm]

It's not close to anything

[oak chasm]

The point is that it's useful in calculus.

It's not about the function at one input.

[glass lichen]

yeah but that says nothing about the limit as x-> 1

[oak chasm]

It's about when you get the input closer and closer to something, does the output get closer and closer to something?

[glass elm]

Ohhh

So, for testing purposes?

[oak chasm]

Sort of.

[glass elm]

oh...

Okay, so, limit of f(1.6) is 4?

[oak chasm]

It's not about the value at f(1). It's about the value at f(0.9), f(0.99), f(0.999), and so on.

[glass lichen]

a function being defined at a point gives no information on what f does within the neighbourhood

[glass elm]

Since output is 3.6?

[glass lichen]

??

4 isnt 3.6

[glass elm]

Sorry, I'm new to this

[glass lichen]

the limit of a constant is itself

$\lim_{x\to a}c=c$

[ocean seal]

Mosh

[glass elm]

I don't get what a is here

What is a?

[glass lichen]

the number you base your neighbourhood off

the x value you want to inspect

[glass elm]

Can you explain with the help of a numerical, if possible?

[oak chasm]

x → what input is getting closer and closer to.

[glass elm]

Like, let's try to solve some question and I understand from that

[oak chasm]

So, x → a means x is getting closer and closer to a.

[glass elm]

Ohhhh

[oak chasm]

Does that make sense?

[glass elm]

So, a is the limit?

[oak chasm]

It never gets to a.

[glass lichen]

no

[oak chasm]

No, x never gets to a.

It just gets closer and closer to it.

[glass elm]

Then, limit is infinity but lesser than a?

[oak chasm]

Does that make sense?

Please stop guessing.

[glass elm]

Sorry

[oak chasm]

x → a means x gets closer and closer to a but never reaches a.

[glass lichen]

$\lim_{x\to a}f(x)=L$ means as I make x closer to a, f(x) gets closer to L

[ocean seal]

Mosh

[oak chasm]

So, if x → 2, then x might be 1.9, then 1.99, then 1.999, then 1.9999.

[glass elm]

Ohhh

[oak chasm]

See how x is getting closer and closer to 2?

[glass elm]

So, it keeps going closer to a

But never reachj

[oak chasm]

It doesn't reach 2, but it gets closer and closer.

Right.

[ocean seal]

Chai T. Rex

[oak chasm]

That's what that part means.

x is getting closer and closer to a without reaching it.

[glass elm]

Got it

1.9999999 -> 2

[ocean seal]

Chai T. Rex

[glass elm]

Is it like this?

[ocean seal]

Pig 2 Man

[oak chasm]

No, x is getting closer and closer to 2, not 1.9999999.

1.9999999 stays the same distance from 2 as it always was.

[glass elm]

Got it

[oak chasm]

x can change, but 1.9999999 can't.

So, let's look at this:

[ocean seal]

Chai T. Rex

[oak chasm]

We get x closer and closer to 2 without reaching it.

We see what x² gets closer and closer to.

[glass elm]

Ohhhh, wait, I understand now

We need to check x^2

And find out what x^2 is 'closse to' as well

But what do we take x as?

[oak chasm]

x = 1.9, x² = 3.61
x = 1.99, x² = 3.9601
x = 1.999, x² = 3.996001

We take x getting closer and closer to 2 and see what x² is getting closer and closer to.

[glass elm]

4?

[oak chasm]

So, x = 1.9, then x = 1.99, then x = 1.999.

See how I got x² for each of those xs?

[glass elm]

Yeah, 3.61 and the others

[oak chasm]

Right, and it is getting closer to 4.

[glass elm]

And they're all getting closer to 4 but not reach

[oak chasm]

Right, so the limit is 4.

[glass elm]

So, do we always take it as whole number?

[ocean seal]

Chai T. Rex

[oak chasm]

No, it can be something else.

[glass elm]

Like, we could say 3.9

But we say 4 instead

[glass lichen]

cause you'll get past 3.9 eventually

[oak chasm]

No, there are ways of getting the correct value, which may or may not be a whole number.

[glass lichen]

which happened with x=1.99

[glass elm]

ahh

[oak chasm]

But first, do you understand what a limit is?

[glass elm]

Yes, it's to check the result of a function for which value it never reaches but gets closer to for any value of x lesser than the tending value 'a'

[oak chasm]

OK, so that's almost right, I just have to show you one more thing.

That's called the limit from the left.

[glass elm]

oh...

[oak chasm]

@boreal whale Sorry, we can't help with tests.

<@&268886789983436800>

@glass elm So, there's also a limit from the right.

[glass elm]

Limit from Right is like 4, right?

And limit from left is a negative number?

[oak chasm]

That's where we go x = 2.1, x = 2.01, x = 2.001.

No.

See how we're getting closer and closer to x = 2 from the right?

[glass elm]

Ahh, yes

Value is getting lesser

and closer

[oak chasm]

From the left, we do x = 1.9, x = 1.99, x = 1.999, ....

From the right, we do x = 2.1, x = 2.01, x = 2.001, ....

Those are the left and right side limits.

[glass elm]

For which function though?

[oak chasm]

For any function.

[glass elm]

Ahhh

So, is it mean that left and right lims are 2?

[oak chasm]

No, those are how the inputs to the functions go.

[glass elm]

Oh, got it

the inputs need to be closer and closer

[oak chasm]

If the function is x², x² will get closer and closer to 4 as x gets closer to 2.

But what I was talking about was just what x gets closer to.

[glass elm]

And same goes from right side?
2.1, 2.01, the square value come near 4 again?

from right

[oak chasm]

Right.

And if the left and right side limits agree, that's the limit.

[glass elm]

Okay, so we have two method to solve limit?

[glass lichen]

$\lim_{x\to a}f(x)=L\iff \lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=L$

[ocean seal]

Mosh

[oak chasm]

Right.

[glass elm]

Do we need to have both limit same or can we just conclude with one side?

[oak chasm]

So, if the left and right side limits give us the same value, that value is called the limit.

No, we need both side limits to get the limit.

They have to agree for the limit to exist.

And if they agree, the value they agree on is the limit.

[glass elm]

Got it, both sides need same limit for it to be a limit

[oak chasm]

If they don't agree, there is no limit.

[glass elm]

How come there is no limit btw?

[glass lichen]

example $\lim_{x\to 0}\frac{1}{x}$ DNE since the sided limits don't agree

[ocean seal]

Mosh

[oak chasm]

Because the limit is what the left side and right side limit agree on.

If they don't agree, there is no limit.

[glass elm]

But, which function don't have limit?

[oak chasm]

Well, 1/x doesn't.

[glass elm]

Can give an example pls?

[oak chasm]

,w plot 1/x

[glass lichen]

I just did

[ocean seal]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Chai T. Rex#7323

[glass elm]

f(x) = 1/x?

Is it because it is constant?

[glass lichen]

?????

1/x is clearly not a constant

[glass elm]

Oh, it's not?

[oak chasm]

No, look at the graph.

See how if x → 0, it goes to -∞ from the left and ∞ from the right?

[glass elm]

yeah

[oak chasm]

So, the left limit is -∞. The right limit is ∞.

They don't agree, so there's no limit.

[glass elm]

Ohhh

Infinity basically means no limit anyway

Cuz it keeps going on

[oak chasm]

No, limits can have infinite results.

[glass elm]

And they also don't have same from both sides, so we can conclude that not have limit

[glass elm]

ohhh

[oak chasm]

Here, the limit from the left is -∞ and the limit from the right is ∞, so they're not the same, so the limit doesn't exist.

[ocean seal]

Chai T. Rex

Chai T. Rex

[oak chasm]

That's how we write the left side limit.

[glass elm]

Left Side Limit means negative above the 'a'?

And Right Side with + i guess?

lim x->0+?
\

Right side?

[oak chasm]

Right.

Because the left side limit is where you subtract a little something from 0 and the right side limit is where you add a little something to 0.

[glass elm]

damn, that's smart notation

and easy to understands

[oak chasm]

Also, the - and + are written like exponents. They're raised a little bit.

[glass elm]

mhm

[glass lichen]

(superscripts)

[alpine sable]

i have a dumb question
y*dy/dx=0 its not linear right??

[oak chasm]

@alpine sable Sorry, this channel is busy.

[alpine sable]

ok sorry

[oak chasm]

#help-5 looks open.

[glass elm]

Okay, so, what do we do next?

okay one sec i try to solve

[ocean seal]

Chai T. Rex

[glass elm]

first we do left side limit

[oak chasm]

Correct.

[glass elm]

1.9 1.99 1.999

It become 1

No wait

One min

[stark trail]

eh

[glass elm]

-0.1/-0.1?

[glass lichen]

do the simplification first

[stark trail]

It’s always 1 lol

[oak chasm]

No, it's not always 1.

Try it with x = 2.

[devout sigil]

No

It is my favourite rule

[glass elm]

No Limit is answer?

[stark trail]

Yeah u meant it’s undefined at x=2 so it’s not always 1?

[oak chasm]

@glass elm Right, so what's -0.1/-0.1?

[glass elm]

1

[oak chasm]

Right.

[glass elm]

But that's weird, it reached 1

instead of being close

[oak chasm]

Right.

[glass elm]

so I think No Limit(guess)

not sure though

[oak chasm]

No, we're doing the left side limit.

[glass elm]

Ohhhh

[oak chasm]

Not the whole limit.

[glass elm]

Okay, right side then

2.1 2.01 2.001

[stark trail]

I meant if x\neq2

[glass elm]

Wait, it come 1 as well

[oak chasm]

Right, so the left and right side limits are both 1.

So, what's the limit?

[glass elm]

1

[oak chasm]

Right.

[glass elm]

But, this is confusing since how it is constant?

Shouldn't it be near something?

like 3.9 and 4.1

near 4

[oak chasm]

No, the limit doesn't care.

[glass elm]

ohh

So it just have to be same?

[oak chasm]

The limit is about what the output goes towards.

[glass elm]

not matter if near or far

got it

[oak chasm]

If it's always 1, the output is going towards 1.

[glass elm]

So, in constant function like this, it always be easy?

Here, everything become constant 1

So 1 is limit?

[oak chasm]

Well, it's not a constant function.

Because when x = 2, you get (2 - 2)/(2 - 2) = 0/0.

0/0 is undefined.

[glass elm]

oh right

But that not matter since 2 cannot be input

Right?

that's why we take limit 1 here

And no undefined problems

[oak chasm]

Right, as we get closer and closer to 2 without touching it, we never actually use 2.

[glass elm]

got it

[oak chasm]

So, that's the basics of what a limit is.

[glass elm]

Can we also go further if possible?

if u dont mind

[oak chasm]

Well, the next part would be the limit laws.

[glass elm]

it's easier to understand with you

as compare to website/video

[glass lichen]

time to break out e-d definition \s

[glass elm]

oh, it has laws?

[oak chasm]

@glass cave Sorry, this channel is busy.

[glass cave]

sorry\

[oak chasm]

@glass elm Yes, for example:

[ocean seal]

Chai T. Rex

[glass elm]

we are adding functions too...?

[oak chasm]

If you have a limit of two things added together, you can take the limits of each term and then add those limits together.

[glass elm]

this is getting complex, damn...

[glass lichen]

...

[oak chasm]

No, it's not that complex.

[glass elm]

oh, sorry, yeah it's just that first time im learning this

[ocean seal]

Chai T. Rex

[oak chasm]

This is an example.

[glass elm]

actually was supposed to last year but due to covid not much focus and I found it hard and never bothered

[oak chasm]

See how if you have the limit of things added together, you can take the limit of each part and then add those limits together?

[glass elm]

and now I need to learn derivatives and stuff and first basics, so back to limits

[glass elm]

yeah

[oak chasm]

That's all it means.

[glass elm]

it's just taking common, right?

[oak chasm]

You can do each part separately, which makes it easier.

I don't know what you mean by common.

[glass elm]

Like

2(5+3) = 10 + 6

[oak chasm]

No, it's not the distributive property.

[glass elm]

oh, sorry

[oak chasm]

It means if you have the limit of something + something, you can do the limits of the somethings and then add them together.

[glass elm]

Okay, so, if it's together, you can do it separately and just add?>

[ocean seal]

Chai T. Rex

[oak chasm]

Do the limit of x. Do the limit of 2. Add them.

The limit of x is easier to do than the limit of (x + 2).

The limit of 2 is easier to do than the limit of (x + 2).

So, you do the two easier limits.

[glass elm]

okay, one minute

[oak chasm]

Then you add the limits together.

[glass elm]

Can you give value for a?

[oak chasm]

Let's say 2.

[alpine sable]

Hey chai i got a question for yah if you got time after this 😛

[glass elm]

limit is 2 for x

And for lim 2

also 2

4 is total limit?

[oak chasm]

Right.

You add the limits together.

[glass elm]

So, this is a kind of Limit Distributive Law but not exactly distributive?

[oak chasm]

No, thinking of it that way won't help.

[glass elm]

Ahh, okay, won't associate the two

I understand the law though yeah

[oak chasm]

Because there are similar laws for addition, subtraction, multiplication, division, and exponents.

[glass elm]

lim(a+z-5) is lim(a) + lim z + lim -5

[oak chasm]

Right.

[glass elm]

its okay?

[oak chasm]

Yes.

[glass elm]

but sometime when its easier to do all together, we do all together, yes?

[oak chasm]

Right.

[ocean seal]

Chai T. Rex

[glass elm]

okay i try one min

no wait, this look too hard

[oak chasm]

If you have the limit of a power, you take the limit of the base and the limit of the exponent and then you.

[glass elm]

oh, this is new rule?

[oak chasm]

Yes.

It's like the sum law.

[alpine sable]

i learned something new. cool.

[oak chasm]

But this time, it's for powers.

[glass elm]

Oh, I understand, so, we can do it easier method by taking the limit of exponent itself and exponenting that vakue to the main lim?

[oak chasm]

Right.

[glass elm]

but not always necessary, yea? only when can do easier

[oak chasm]

This breaking it into parts and combining the limits of the parts works for addition, subtraction, multiplication, division, and powers.

Well, it should be your first way of trying it.

[glass elm]

got it

so priority

[midnight skiff]

@alpine sable Try in another channel. This one’s busy, as far as I see.

[oak chasm]

There's one more limit law and one more thing to know.

The other limit law is the constant multiple law.

[glass elm]

ooh

[ocean seal]

Chai T. Rex

[glass elm]

Ohhh, that's interesting

[alpine sable]

same as a sum

[glass elm]

But, it's kinda pointless too, right?

[oak chasm]

If you have the limit of a constant multiple times something, that's the same as the constant multiple times the limit of the something.

No, it simplifies your work.

[glass elm]

Oh, if it simplify then its good

[oak chasm]

Do you know what a continuous function is?

[glass elm]

sorry, i don't know what functions are properly except for the basic thing like it has input and give output

but not the different types or anythiong else

[oak chasm]

OK, you know the graph of a function?

[glass elm]

not so properly

But, what I know:

f(2) = 5

[stark trail]

Linear functions?

[glass elm]

y is 5, x is 2

And then we draw the graph something, but I don't remember properly

[oak chasm]

So, if you have f(x) = x + 2, can you graph it?

[glass elm]

let me try, 1min

[stark trail]

Ganbatte cookie chan

[alpine sable]

lol😆

[glass elm]

5, 7
7, 9

mark those points

draw line

in cart plane

Like this?

[oak chasm]

Yes.

[glass elm]

ohh so, line graph?

i don't know so properly for complex function but if easy like this then yes i can

[oak chasm]

Now you can draw that graph without lifiting your pen from the paper since it's a straight line forever, right?

[glass elm]

yeah it infinite straight line

[oak chasm]

So, that's what continuous means.

You can graph it without ever lifting your pen from the paper.

Now, let's talk about x/x.

[glass elm]

So, to know if something is continous, we need to dreaw and graph and check?

[oak chasm]

Well, there are easier ways, but that's just what continuous means.

[glass elm]

okay, got it

[oak chasm]

So, with x/x, what happens when x = 0?

[alpine sable]

draw a line up and down, it shouldnt cross that line more than once.

[glass elm]

uhh

undefine?

[oak chasm]

Right.

So, you can't draw that graph without lifting your pen.

[glass elm]

so, graph not exist?

[oak chasm]

No, the graph exists.

You just have to lift your pen to not draw anything for x = 0.

Because the function doesn't have a value at x = 0.

But everywhere else, it has a value.

Like x = 2, it's 2/2.

[glass elm]

oh but it has value at others, righttt

[oak chasm]

Right.

[glass elm]

but that would be a weird graph, right?

[oak chasm]

So, the other parts are continuous.

[glass elm]

So, it's a line segment?

no wait, ray?

[oak chasm]

Well, it's two rays.

[glass elm]

or two lines?

two rays*

[oak chasm]

One ray going left from x = 0, One going right.

[glass elm]

and middle part is empty at 0

[oak chasm]

Right.

[glass elm]

got it

[oak chasm]

It's continuous everywhere except at x = 0.

[glass elm]

so, it's basically discontinous as whole?

[oak chasm]

Now with limits, here's a trick.

Yes, but it's continuous everywhere except x = 0.

[glass elm]

understood

[oak chasm]

If the function is continuous where your limit input is going towards, you just fill in the thing you're going towards into the function.

[ocean seal]

Chai T. Rex

[glass elm]

Wait, i'm confuse

What's f(2)?

or do you mean f(x) = 2?

[oak chasm]

No.

Let's say f(x) = x/x.

What's f(2)?

[glass elm]

1

[oak chasm]

Right.

You fill in x with 2 on the right side of the =.

f(2) = 2/2

f(2) = 1

[glass elm]

yeah

But i dont understand f(2) = f(1)

what the other f do?

[oak chasm]

I didn't say f(2) = f(1).

[glass elm]

in this

what the other f means?

[oak chasm]

I didn't say it there either.

[glass elm]

but, there, f(x) = f(2)

[oak chasm]

No, I didn't say that.

[glass elm]

Oh wait, f(2) is not y?

[oak chasm]

No.

[glass elm]

I actually think f(2) is y thats why I confuse sorry

[stark trail]

There’s a limit on the left side

[glass elm]

You mean that x is 2, right?

[ocean seal]

Chai T. Rex

[oak chasm]

That's not f(x) = f(2).

[glass elm]

ohhhh

Okay thats why Im confuse

What is the thing in the righr side? i not understand

[oak chasm]

OK, let's go over functions.

f(x) = x + 2

That means that you have a function called f.

[glass elm]

yes

[oak chasm]

Its input is written x.

f(x)

[glass elm]

yep

[oak chasm]

It's output is x + 2.

[alpine sable]

chai would be an outstanding pre calc teacher 😆

[glass elm]

mhm

[oak chasm]

So, let's find out f(2).

[glass elm]

4

[oak chasm]

Right.

So, let's go over that.

[stark trail]

The Legend of Chai

His name is cool

[oak chasm]

f(x) = x + 2
f(2) is a lot like f(x).

It has x replaced with 2, right?

[glass elm]

but f(2) not have output

[oak chasm]

No, hold on.

[glass elm]

just f(2)

Oh sorry

[glass elm]

yes

[oak chasm]

f(2) is f(x) with x replaced by 2.

So, if f(x) = x + 2, then f(2) is where you replace the xs with 2s.

f(x) = x + 2
f(2) = 2 + 2

[glass elm]

So, it's the same function?

But with input as 2?

[oak chasm]

No, hold on.

[glass elm]

But then, where is the main function? like, where is it defined?

[oak chasm]

f(x) = x + 2 is the definition.

[glass elm]

Yeah

[oak chasm]

If you have f(2), you're filling in x with 2.

[glass elm]

mhm

[oak chasm]

You had f(x). You filled in x with 2. Now you have f(2).

[glass elm]

ahh

[oak chasm]

So, you do that to both sides.

f(x) = x + 2
f(2) = 2 + 2

See how I filled in x with 2 on both sides?

[stark trail]

Well I don’t think you’re supposed to learn about continuity rn since you’re new at Limit.

[glass elm]

yeah

[oak chasm]

OK, so that's how you find f(2).

[glass elm]

Chai is teaching me with proper sequence its fine

[oak chasm]

Let's say we have this:

g(x) = x/x.

[stark trail]

Okay then

[oak chasm]

What's g(2)?

[glass elm]

1

[oak chasm]

Right.

g(x) = x/x
g(2) = 2/2

If we replace x with 2, we have to do the same thing to both sides.

[glass elm]

yeah

[oak chasm]

OK, so let's do a limit without f(x).

[glass elm]

okay

[ocean seal]

Chai T. Rex

[glass elm]

2

[oak chasm]

Now, x/x is continuous everywhere except at x = 0, right?

[glass elm]

Oh wait

The limit is 1

[oak chasm]

No, don't solve it.

[glass elm]

sorry

[glass elm]

yeah

Since 0/0 is not define

[oak chasm]

OK, so if x = 2, it's continuous there, right?

[glass elm]

but, x can't be 2, right?

since x -> 2

[oak chasm]

Why not?

Right.

But hold on.

Not with the limit.

With x/x without the limit.

[glass elm]

oh, so,, the functionm

yes, it continous except at 0/0

[oak chasm]

OK, so x/x is continuous at x = 2.

[glass elm]

yes

[oak chasm]

So, if the the thing you're taking a limit of (x/x) is continuous at where x is going to (x → 2), then you can just fill in 2 to x/x.

x/x is continuous at x = 2.

So, you can just fill in 2 to x/x.

[ocean seal]

Chai T. Rex

[glass elm]

I'm confused about the statement

wouldn't we take 1.9/1.9 and 2.1/2.1?

[oak chasm]

We could.

[glass elm]

But since it's 1, we represent it as 2/2?

[oak chasm]

But there's something special about functions where they're continuous.

[glass elm]

Or hmm, sin^2+ cos^2

Since it give value as 1 too?

[oak chasm]

No, it has nothing to do with 1

[glass elm]

limx->2 x/x = sin^2 90degree + cos^2 90degree

Like this?

[oak chasm]

No.

It has nothing to do with 1.

[glass elm]

?

ohh, sorry

so, it's not about representing with 1

[oak chasm]

No.

It's about whether the function is continuous.

[glass elm]

okay, wait, i got it

The 'a' value

its tending to

[oak chasm]

Right.

[glass elm]

its continous at that ppint

[oak chasm]

Right.

[glass elm]

So we can just fill it in with 2?

[oak chasm]

Yes.

[glass elm]

2/2

But, what the point of it?

We already get limit as 1

[oak chasm]

It makes the limit easier sometimes.

[glass elm]

ohhh

okay i get it now

its just utility feature if tro make easier

[cyan grail]

if a function is continous at that point then you can just plug it. But sometimes it isn't continous at the point which is why we use limits.

[oak chasm]

If you can just get rid of the limit and do it with algebra from before, that's easier than the limit.

[glass elm]

okay so only whern continous

[oak chasm]

Right.

[glass elm]

If not continous then cant

but how to know if continous?

From graph?

That take longer time though but?

Draw graph

[ocean seal]

Chai T. Rex

[cyan grail]

to know if something is continous it will either state it. But in most cases it is the domain, if something is not able to be represented there like a fraction or a logarithm

[glass elm]

Limit not exist here, right?

[oak chasm]

Because it's not continuous at zero, so you can't use that trick.

It exists.

[glass elm]

for x-> 0?

[oak chasm]

But you can't use our continuous-there trick.

The limit is 1.

[glass elm]

oh wait, yeah, approaching

[oak chasm]

But you can't use our continuous-there trick to get the limit.

[glass elm]

but we cant directly plug in

[oak chasm]

Right.

We have to get it a different way.

[glass elm]

i understand now

And for tends to 0

0.1

and -0.1

which give 1 both siudes

[oak chasm]

Well, be careful.

It's about 0.1, 0.01, 0.001, and so on.

Not just 0.1.

[glass elm]

we cant directly check with 0.1?

[oak chasm]

Nope.

[glass elm]

okay, so, i need to use upto 3 decimal point?

only .001?

[oak chasm]

Limits are about how the output changes as the input gets infinitely close to something.

No, you need to get infinitely close.

[glass elm]

.00001?

[oak chasm]

That's not infinitely close.

That's 0.00001 away.

[dusk peak]

How to solve -1,3,7,11 (explicit formula

[glass elm]

but we can't calculate infinitely close, right?

[oak chasm]

@dusk peak Sorry, this channel is busy.

[midnight skiff]

Please try another channel.

[oak chasm]

Right, so we use nice techniques.

[dusk peak]

I accidentally send it sorry

[oak chasm]

That's why the limit laws are important.

That's why the continuous-there trick is important.

[midnight skiff]

All good, np.

[glass elm]

got it

[oak chasm]

That way we can get the limit without having to get infinitely close.

To do infinitely close, we can do that without our techniques so far.

[glass elm]

mhm

[oak chasm]

But it's somewhat complicated.

[glass elm]

okay i understand these laws now

[oak chasm]

What we do is this.

We show that once the input gets within a certain distance to a, the output is always within a certain distance to the limit value.

So, maybe when we get within 0.1 of a, the output is always within 0.01 of the limit value.

[glass elm]

so, we always have to show three times?

like always always?

[oak chasm]

What do you mean by three times?

[glass elm]

1x10^-1, then ^-2, then ^-3?

[oak chasm]

Nope.

Those were only to show you what limits mean.

[glass elm]

ahh

[oak chasm]

To get the limit by going infinitely close, we have to do an infinite number of times.

[glass elm]

wait so the way we've been solving the limits is wrong and we can't take 0.1 and stuff like that?

[oak chasm]

Right, that was just to show you what limits mean.

To get the limit that way, we need an infinite number of tries.

[glass elm]

got it

[oak chasm]

So, what we do is this.

We show that if the input gets within 0.1 of a, the output is stuck somewhere within 0.01 of the limit value.

Or within 0.00001 or within 3 or whatever.

It doesn't matter how far away from the limit value, as long as the outputs are stuck within a certain distance of the limit value.

Does that make sense so far?

[glass elm]

i dont understand the statements

gets within?

[oak chasm]

Like if I have less than 1 gallon of fuel, I'm stuck within 15 miles of my house.

Does that make sense?

[glass elm]

that does, yes

[oak chasm]

OK, so 1 gallon is the input and 15 miles is the output.

[glass elm]

yep

[stark trail]

Bro you’re amazing at giving examples

[glass elm]

I swear ^

chai is amazing

[oak chasm]

So if the input is within 1 of a.

Like a + 0.5 is within 1 of a.

a - 0.75 is as well.

Then the output is stuck 15 miles from my house.

Or the output is within 15 of the limit value.

[glass elm]

i didn't get that part

[oak chasm]

Like the limit value + 7 is fine.

[glass elm]

if we missing 2 gallons

30 miles, right?

[oak chasm]

If we have 2 gallons, yes.

We can go 30 miles.

[alpine sable]

Please tag me when I can ask a question. Pretty easy one but the wolfram system won’t give me the breakdown of steps (even with premium)

[midnight skiff]

@alpine sable You can ask in another channel and if someone can answer, they’ll respond.

[glass elm]

got it

[oak chasm]

@glass elm So, let's say we have x → 2 and the limit = 5.

Does it make sense how we could have a limit like that?

[alpine sable]

@midnight skiff I have, no answer there so I’ll wait in the queue here 🙂 if it gets answered in the other channel I’ll lyk

[ocean seal]

Chai T. Rex

[glass elm]

that's because 1.9 + 3 is 4.9 and comes to 5
And same for 2.1

[oak chasm]

Right.

[glass elm]

but you said we can't solve that way

So, im not sure what other method

[oak chasm]

No, I didn't.

@shrewd quail Sorry, this channel is busy.

[shrewd quail]

my bad

[oak chasm]

I said we can't do it with a finite number of tries.

So we can't do it with 3 tries.

Does that make sense?

@midnight skiff Sorry, this channel is busy.

[glass elm]

yeppp

[oak chasm]

OK, so we can do it with infinite tries.

We're doing it that way.

Here's our first try.

Then I'll show you how to use that to get infinite tries.

[glass elm]

Ahh,

okayy

[ocean seal]

Chai T. Rex

[oak chasm]

So, if the input is within 0.1 of 2, the output is within 0.1 of 5, right?

Like if the input is 1.99, the output is 4.99, which is within 0.1 of 5.

Does that make sense?

[glass elm]

i don't get 'within'

do you mean

1.9

and 4.9?

you mean both have the same difference?

[oak chasm]

I mean the distance away from.

[glass elm]

of 0.1?

ahh okay i get it now'

distance

[oak chasm]

Like 8 and 8.1 are 0.1 away from each other.

[glass elm]

yep

[oak chasm]

So, 1.99 is within 0.1 of 2.

Because 1.99 is 0.01 away from 2.

And 0.01 distance doesn't go higher than 0.1 distance.

[glass elm]

yeah

So like

if x is 5 km far away from y

it is inside 10 km range?

[oak chasm]

Right.

So, let's work out how to get the distances.

[glass elm]

yeah

[oak chasm]

For the input distances, the formula is |x - a|.

Like if a = 2 like in our example.

And our test value is 1.99.

Then the distance from 2 is |1.99 - 2|, which is |-0.1|, which is 0.1.

Does that make sense?

[glass elm]

yeah

[oak chasm]

OK, let's work out how to get the distance of the output from the limit value.

Our limit value is 5, right?

[glass elm]

yres

[oak chasm]

And our function output is x + 3.

So, the distance is |(x + 3) - 5|.

Does that make sense?

[glass elm]

x+3 - Limit?

[oak chasm]

Right |output - limit value|.

Just like |input - a value|.

[glass elm]

got it

[oak chasm]

The distance between two values is just |q - r|.

So, let's simplify our input and output distances.

|x - 2| is the input distance, which is already simplified.

|(x + 3) - 5| is the output distance.

|x - 2| is that simplified.

Does that make sense?

[glass elm]

so distances are equalified

[oak chasm]

In this case, yes, but not always.

[glass elm]

and when its not?

means no limit?

[oak chasm]

No, it just means the distance formula is different.

It can still work.

[glass elm]

gotit

[oak chasm]

OK, now we want to show that for any output distance > 0, we can find an input distance that works.

Like we can say the output distance is 3.

3 > 0, so we can use that.

Is there an input distance that works?

Does that make sense?

[glass elm]

uhh

could and could not be?

[oak chasm]

What do you mean?

[glass elm]

some case not?

and some case yes?

that output distamnce = input distance?

[oak chasm]

Yes, if the limit value is the wrong limit value, it won't work.

No, it doesn't matter that the input and output distances are equal.

As I said earlier.

[glass elm]

ahh

okaay

[oak chasm]

So, let's set the output distance to 3.

|x - 2| ≤ 3.

[glass elm]

mhm

[oak chasm]

The distance between the output and the limit value has to be within 3.

[glass elm]

yep

[oak chasm]

What is the distance between the input and a when that happens?

Well, in this case, the distance expressions are the same, so |x - 2| ≤ 3 for the inputs as well.

Does that make sense?

[glass elm]

mhm it dos

[oak chasm]

OK, so when the input is within 3 of 2, the output is within 3 of 5.

When the input is stuck there, the output is stuck there.

[glass elm]

mhm so both dist are within a certain area, equal or not equal

[oak chasm]

Right.

So, let's choose a smaller distance.

Let's say we want outputs within 0.000000000000000000001 of the limit value.

Well, the same thing.

|x - 2| ≤ 0.000000000000000000001 for the outputs.

|x - 2| ≤ 0.000000000000000000001 for the inputs.

Does that make sense?

[glass elm]

yeah

|x-2| <_ infinity but not reaching limit for both inp and outp

[oak chasm]

OK, so no matter how small we choose the distance from the limit value, we can find a distance from a that leaves the outputs stuck there.

Does that make sense?

[glass elm]

yeah

[oak chasm]

So, we have infinite tries.

[glass elm]

mhm

[oak chasm]

The distance between the limit value and the output can be any number that's greater than 0 and we'll find an input distance that works.

So, for all those infinite numbers (infinite tries) that get closer and closer to zero distance, we find that it narrows our outputs to be stuck closer and closer and closer to our limit value.

Does that make sense?

[glass elm]

it does, yeah

[oak chasm]

Let me rephrase that.

As we move the output distance closer and closer to zero, we can find an input distance that works and it gets closer and closer to zero as well.

[glass elm]

yep

[oak chasm]

And that works for all output distances greater than zero.

That's how we do limits when we can't use easier methods.

[glass elm]

mhm

[ocean seal]

Chai T. Rex

[glass elm]

ok wait

[oak chasm]

OK.

[glass elm]

first calculate lim 3

no wait

lim already given

what to do here then?

[oak chasm]

No, we're not using the nice limit laws.

We're using the harder method just to get a feel for it.

So, what's the input distance formula?

[glass elm]

ohhh

mod(2-x)

[oak chasm]

Right.

What's the output distance formula?

[glass elm]

mod(7-x^2+3)

i think?

[oak chasm]

Almost.

|(7) - (x² + 3)|

[glass elm]

oh yes

[oak chasm]

|7 - x² - 3|
|-x² + 4|
|x² - 4|

[glass elm]

mhm

[oak chasm]

So, we have:
δ = |x - 2|
ε = |x² - 4|

δ is our input distance.

ε is our output distance.

[glass elm]

sigma and eigenvalue?

[oak chasm]

Delta and epsilon.

[glass elm]

uh lets just use input and output dist

😅

not know any of this greek letters

[oak chasm]

OK, so let's say we want to get our output distance within d.

Like d might be 3 or 0.00001 or whatever.

[glass elm]

yep

[oak chasm]

But we'll call it d to represent whatever it is.

|x² - 4| ≤ d

Now solve for x.

[glass elm]

mod(x^2 - 2^2)

=>

one min solving

|x+2| <_ d; |x-2| <_ d

[oak chasm]

No, the zero product property only works with 0, not d.

[glass elm]

mod(x+2) <_ d / mod(x+2)

mod(x-2) <_ d / mod(x-2)

[oak chasm]

That's not x.

[glass elm]

i dont know how to solve, sry

[oak chasm]

That's |x + 2| and |x - 2| without x gone on the other side.

OK, so:

|a| = a if a ≥ 0
|a| = -a if a < 0

[glass elm]

dont understand

|a| is +a in all conditions, right?

[oak chasm]

No.

What if a is -5?

[glass elm]

|-5| is 5

[oak chasm]

|-5| = 5, right?

[glass elm]

right?

yeah

[oak chasm]

Right, so |a| = -a in that case.

[glass elm]

mhm

[oak chasm]

|-5| = -(-5)

[glass elm]

yeah

[oak chasm]

|a| = -a

Because a = -5.

[glass elm]

But wont that not work?

|5| is 5

[oak chasm]

What's a?

[glass elm]

-5

[oak chasm]

|a| = |-5| = 5, right?

[glass elm]

ahh...

yeah

got it

for a = -5

[oak chasm]

Right,

|a| = a if a ≥ 0
|a| = -a if a < 0

[glass elm]

pls another channel, soprry

[oak chasm]

@tropic sail Sorry, this channel is busy.

So, there are two cases for |x² - 4|.

[glass elm]

oui

[glass elm]

its just tgat

i undertand word equation better

than with symbol

like if you say

|a| = a if a is posiitv

|a| = -a if a is negative

[oak chasm]

OK, then:

|a| = a if a is nonnegative
|a| = -a if a is negative

[glass elm]

got it thats better

[oak chasm]

OK, so we have two cases. For the first case x² - 4 is nonnegative.

[glass elm]

yes

-(x^2-4)

[oak chasm]

No.

[glass elm]

(x^2-4)

[oak chasm]

It's nonnegative.

[glass elm]

these two case?

(x^2-4)

-(x^2-4)

[oak chasm]

Yes, x² - 4 can be nonnegative or negative.

We'll deal with nonnegative first.

[glass elm]

okaye

[oak chasm]

|x² - 4| ≤ d
x² - 4 ≤ d
x² ≤ d + 4
|x| ≤ sqrt(d + 4)

x is around 2 (a = 2) when it's narrowing in on a, so it's positive:

x ≤ sqrt(d + 4)

No, we do x² - 4 is negative.

[glass elm]

yeah

|x| ≤ sqrt(d + 4)

why mod here again

x^2 always be positive, rightr?

[oak chasm]

Because the square root of a squared value is the absolute value of the value.

sqrt((-3)²) = sqrt(9) = 3
sqrt((3)²) = sqrt(9) = 3

See how the value that you square initially has its absolute value as the result?

[glass elm]

oh ok

i just confuse because | | come again

i think modulus again

but yeah we can just write x i guess?

x ≤ sqrt(d + 4)

x is positive here anyway, right? cuz we know sqrt makes positive

[oak chasm]

Right.

[glass elm]

got it

[oak chasm]

Well, no.

x isn't necessarily positive.

It could be x = -2.

[glass elm]

oh wait, square makes positive not sqrt sorry

[oak chasm]

But we're taking the limit x → 2, so x is around 2, which is positive.

[glass elm]

okay I understand now why we take mod again

let's go negative part now

[oak chasm]

OK, now here's the thing.

[glass elm]

mhm

[oak chasm]

|x² - 4| ≤ d
-(x² - 4) ≤ d
x² - 4 ≥ -d
x² ≥ 4 - d
|x| ≥ sqrt(4 - d)

x is around 2, so x is positive.

x ≥ sqrt(4 - d)

[glass elm]

yep

[oak chasm]

So, sqrt(4 - d) ≤ x ≤ sqrt(d + 4)

[glass elm]

x ≤ sqrt(d + 4)
x ≥ sqrt(4 - d)

mhm

[oak chasm]

OK, so that's where is x is stuck when the output values are stuck within d of the limit value.

[glass elm]

yeah

[oak chasm]

And the input distance is |x - 2|.

[glass elm]

mhm

[oak chasm]

So, let's take our extreme x values.

When x = sqrt(4 - d) (minimum x can get):

@alpine sable Sorry, this channel is busy.

The input distance is:

|x - 2|
|sqrt(4 - d) - 2|

x will be less than 2 since this is the minimum for x.

x will be less than 2 because d is greater than 0 and sqrt(4 - d) will be less than sqrt(4), which is 2.

[oak chasm]

@sturdy lake Sorry, this channel is busy.

[sturdy lake]

q3

ok

btw chai means tea

[oak chasm]

|sqrt(4 - d) - 2|
-(sqrt(4 - d) - 2)
2 - sqrt(4 - d)

Now let's take the maximum x value.

|x - 2|
|sqrt(d + 4) - 2|

d is positive, so sqrt(d + 4) will be greater than sqrt(4), so sqrt(d + 4) is greater than 2.

|sqrt(d + 4) - 2|
sqrt(d + 4) - 2

[glass elm]

yep

[oak chasm]

So the input distance is between 2 - sqrt(4 - d) and sqrt(d + 4) - 2.

[glass elm]

yeah

though i dont know how to evaluate that

[oak chasm]

OK, now d is our output distance.

[glass elm]

mhm

[oak chasm]

We're going to make it go closer and closer to zero, right?

[glass elm]

So, to a?

[oak chasm]

a is our input.

[glass elm]

x?

[oak chasm]

Why would we make the output go to an input?

[glass elm]

a is the tending value right?

[oak chasm]

What?

[glass elm]

x -> a

[oak chasm]

Functions have an input and an output, right?

[glass elm]

yeah

[oak chasm]

They're different things.

[glass elm]

oh sorry

so you mean input

and not tending vaklue

[oak chasm]

The output distance is about the outputs.

So, no the output distance won't go towards one of the inputs.

[glass elm]

mhm

[oak chasm]

d is our output distance.

We're trying to narrow in on the limit value, so the output distance will go towards zero as we narrow in.

[glass elm]

So, limit will be 0?

[oak chasm]

Sorry?

[glass elm]

like, going towards zero

[oak chasm]

Why do you think that?

[glass elm]

towards the limit?

Or is it going towards zero and the limit both?

[oak chasm]

How can something go towards two different numbers?

[glass elm]

oh waitt

sorry i confuse

I get it now

the output distance

will get lesser and lesser

[oak chasm]

Right.

[glass elm]

I thought you were meaning the function goes near zero lol sorry

[oak chasm]

So the output distance goes towards zero and the output distance is d.

[glass elm]

yes

[oak chasm]

So, d will be close to zero.

So, sqrt(4 - d) will be close to 2.

And 4 - d will be close to 4.

Does that make sense?

[glass elm]

another channel pls

[glass elm]

yeah

cuz

taking d as negligible