help-26

[ruby tree]

Here's another way to look at it: imagine I've written a piece of software that simplifies logic expressions. It gives the correct answers and everything, but because of the way my parser works, you always need to put parentheses around binary operators and their inputs, otherwise your query is just invalid

I could write some more code to make the parse understand things like (a and b and c). I simply didn't because that's more effort

If you want to use my software, you need to input these "unnecessary parentheses"

Otherwise, go look for another program

These courses on logic begin with all the parentheses, because you need to understand exactly how these operators work (for example, on exactly two inputs)

Maybe later on you're allowed to omit these parentheses because you have the necessary understanding

That's the equivalent of me writing that additional piece of code to understand the new kinds of expressions where parentheses are absent

Or you still won't be allowed to do that, in which case I might call your course subpar but you have to live with it... and it's not as big a deal as you're making it out to be

[topaz sinew]

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[smoky sparrow]

recall the limit definition of the derivative

did you get tricked by the fact that the denominator is x - pi, not pi - x?

that could be why

[smoky sparrow]

so you need to multiply by -1 cause of this

[topaz sinew]

@hollow plover Has your question been resolved?

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[green dock]

Ok so I have to find the horizontal tangents but if f’(x)=0 equals to a root that’s not in the domain of f(x) then you don’t have a horizontal tangent there right?

[green dock]

For example the derivative of the function on x<-1 and x>1 is 2x-1 and the root is 1/2 but that’s not in the domain of f(x) for x<-1 and x>1

its ex 27

pm if u can help

[topaz sinew]

@green dock Has your question been resolved?

[mortal steeple]

Wlcmm bro

[topaz sinew]

@green dock Has your question been resolved?

[topaz sinew]

@green dock Has your question been resolved?

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@green dock Has your question been resolved?

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[mental trench]

How do I find the implicit differentiation of y = sin(xy)?

[sweet shard]

Chain rule and product rule on the right side

Might help to substitute y for f(x)

[mental trench]

ok thanks

After the chain rule and product rule I got dy/dx = (y+x*dy/dx)cos(xy). What do I do next?

[topaz sinew]

@mental trench Has your question been resolved?

[sweet shard]

Do next for what?

You already did the implicit differentiation

[mental trench]

To solve for dy/dx

[sweet shard]

Algebra

[topaz sinew]

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[graceful leaf]

so

i got 3.317 g

but

now i know ihave to plug it in m=mass substance Grams / amount substance Mol

they tell me use periodic table but like

do i look for an element with 3.317 mass?

but its in grams tho 😕

[raven field]

no if you want to look for an element you need molar mass

Convert the number of atoms into moles first

[graceful leaf]

ok

i got crazy number

1.024x10^29 divided by 6.02 x 10^23

right

[opal vault]

3g in the whole sample?

[graceful leaf]

like

i did d=m/v thing

[opal vault]

there's 21g in a cm^3

[opal vault]

so do you think there's less than that in more than a m^3?

[opal vault]

ok so density = mass/volume

so mass = ?

and then convert correctly

[graceful leaf]

based on what they gave theres no mass

But i have to find it

i found it but erased it one second 😭

[opal vault]

yeah you have to find it

[graceful leaf]

i will redo and show it

[opal vault]

you have density and volume. how do you find mass?

[graceful leaf]

m=v.d

[opal vault]

yes

[graceful leaf]

but the volume 1.550 m^3

should go to cm^3

right

[opal vault]

yes

how many cm^3 in a m^3

[graceful leaf]

can i do cm to m

and cube it

[opal vault]

sure, how many cm in a m

[graceful leaf]

so

0.01 m

then 10m^3

[opal vault]

?

[graceful leaf]

o

i see

if i cubed 0.01 not 10

[opal vault]

I still don't get where the 10 comes from

how many cm are there in a m

[graceful leaf]

1 meter is 100 cm

[opal vault]

yes

1m = 100cm

so

1m^3 = ?

[graceful leaf]

er

so would be

100 x 10^3

?

[opal vault]

???

[opal vault]

.

[graceful leaf]

oo

1m = 100cm

1m^3 = cm^3

it would be

100^3

100^3cm^3

[opal vault]

yes

so

1 m^3 is 1 million cm^3

1000000

[graceful leaf]

yes

[opal vault]

so how do you convert 1.550m^3 in cm^3?

[graceful leaf]

so

1.550m^3 times (1cm/0.01m)^3

[opal vault]

why go through 1/0.01 again

when you already know 1m^3 = 1000000cm^3

[graceful leaf]

oh

true

[opal vault]

1.550m^3 times 1000000cm^3/1m^3

[graceful leaf]

so just multiply

ok equal l to 1550000

cm^
3

[opal vault]

yes

so mass of the sample?

[graceful leaf]

m=v.d

so 1550000(21.40)

uhh equal

33,170,000g

[opal vault]

yes

and this is correct

because the unit is

cm^3 * (g/cm^3) = g

[graceful leaf]

yes

[opal vault]

ok

[graceful leaf]

so this is the mass of the substance?

[opal vault]

this is the mass of the whole sample

[graceful leaf]

oh

[opal vault]

on the other hand, the sample contains 1.024 * 10^29 atoms of said substance

so how many moles of substance does that make?

[graceful leaf]

so 1.024 * 10^29 divided by 6.02x10^23

[opal vault]

yes

[graceful leaf]

equal to 170,000 mol

[opal vault]

yep

so the sample contains 33,170,000g of substance

which is also 170000 mols of substance

[graceful leaf]

hm

[opal vault]

what's the molar mass of the substance?

[graceful leaf]

division then?

[opal vault]

yes

[graceful leaf]

M
is 195.23

[raven field]

Check periodic table

[graceful leaf]

ok

[opal vault]

yes, the periodic table gives molar mass of each element

find which one fits

[graceful leaf]

ok

brb

ok platinum

is closely

[opal vault]

yes it's the only one

[graceful leaf]

so answer is platinjum

thnx so much

.clos

.close

[topaz sinew]

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[sudden sorrel]

What went wrong here

[sudden sorrel]

I spend 1 hour

That’s the original question

[topaz sinew]

@sudden sorrel Has your question been resolved?

[sweet shard]

,w diff 5x(4-4x)^3

[thorny flame]

Results provided by WolframAlpha

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

Requested by rie.mann

[sudden sorrel]

I know the answer I wanna know how to do it haha

Halp

🥲

[sudden sorrel]

That’s not the answer btw anyway cuz you need to find the equation of the line at x=2

It’s a long ass question

[sudden sorrel]

<@&286206848099549185>

[scarlet silo]

Uhhhh

[rigid cloak]

Yo

[scarlet silo]

y-f(a)= f’(a)(x-a)

Plug 2 into f

[rigid cloak]

@sudden sorrel

[scarlet silo]

Differentiate, then plug 2 in again

[sudden sorrel]

I did that

What did I do wrong

[sudden sorrel]

Here

[scarlet silo]

Uhhhh

I don’t know where your work is

I can do it myself when i get the chance tho

[sudden sorrel]

I pinged it

[scarlet silo]

Ik but I don’t know your thought process

[sudden sorrel]

Oh

Then what can I do🥲

I don’t know what I did wrong

[sudden sorrel]

Yes please

[scarlet silo]

Ok

Gimme a few mins

if this doesn’t answer your question, i am sorry ): but i must go to bed now

[sudden sorrel]

I’m so confused🥲 how did you get -320(1-x)^2

But that’s fine, good night😅

[topaz sinew]

@sudden sorrel Has your question been resolved?

[sudden sorrel]

<@&286206848099549185>

[strange kindle]

There's a factor of (4x-1) too

I'll write out on paper

Wait a sec

@sudden sorrel

[sudden sorrel]

Wait what

You simplified the expression?

We were told not to simplify so we can factor

[strange kindle]

This is factored form

[sudden sorrel]

How do I know how to do this

We learned factoring but not at that level

How do you know when to simplify and when to factor

[strange kindle]

You see something common multiplied by the terms

So factor it out

[strange kindle]

I mean if you can see something is common in every term

Then just factor THAT thing out

[sudden sorrel]

Where did you see something common with (1-x)^3?

[strange kindle]

(1-x)^2 * (1-x)

Then (1-x)^2 is common

Factor it out

[sudden sorrel]

I’m so confused

But it’s 4x-4

So why not factor (4x-4)^2 out

And then multiply it by (4x-4)

To get ^3

[strange kindle]

?

That's already factored

Other than the constant 4

[sudden sorrel]

So where do you see (1-x)^2?🫠🥲

[strange kindle]

After differentiating it

We have two terms

[sudden sorrel]

After applying the product rule f’g+g’f
I got
(5)(4-4x)^2 + (-12(4-4x)^2)(5x)

Is this correct?^

Then the next step is to find factors.
All I’m seeing in common is
5 and 5x
And
(4-4x)^2 and another one

[strange kindle]

Wht

[sudden sorrel]

That’s incorrect?🥲

[strange kindle]

Yeah

[sudden sorrel]

Damn it

Okay a step back then

The original function

5x is f

[strange kindle]

In the first term you'll have power 3

[sudden sorrel]

And (4x-4)^3 is g

Oh right right my bad it’s 3 I looked at the wrong thing

Yes that’s what I did with power of 3

The 2 is after I took the factor

Cuz you take the smallest power as factor

[strange kindle]

Idk what you meant but ok

[sudden sorrel]

This is what I got

[sudden sorrel]

Like when you factor something with a power you take the smallest one

Like if you had x^2+x^3

[strange kindle]

Oh I think I got it

[sudden sorrel]

You’d factor x^2

[strange kindle]

So you do know about taking factors

[sudden sorrel]

So the same here we have (4-4x)^3 and one ^2

[strange kindle]

What are common in both terms

[sudden sorrel]

So I factored the ^2

[sudden sorrel]

This and the 5 with 5x

[strange kindle]

Yeah

[strange kindle]

Yeah

So what do you have

[sudden sorrel]

This?

Wait no

This(I’ll write it)

[strange kindle]

Wht

[sudden sorrel]

(5)(4-4x)^2 + (-12(4-4x))(x)
The factors^

[strange kindle]

Heh?

[sudden sorrel]

No?🥲

[strange kindle]

Idk

[sudden sorrel]

Wait are the factors correct at least?

[strange kindle]

I don't understand this

[strange kindle]

Yeah

[sudden sorrel]

It’s this

[strange kindle]

Uh no? When you take 5(4-4x)^2 as a factor

Then it's MULTIPLIED by the remaining thing

Here you have just added it

To something?

[sudden sorrel]

Wait not divided by the remaining things???

OHHHHH

It is multiplied oh my god

You gotta find like what you multiply it by to get the original answer before factoring

[strange kindle]

Yeah..?

[sudden sorrel]

Damn it

Thank you lemme look at it again real quick

[sudden sorrel]

Well there is a + here so what do I do with it then

(5)(4-4x)^3 + (-12(4-4x)^2)(5x)

Here I meant^

There is a plus

So after taking the factors idk what to do

[strange kindle]

Uh didn't I tell you that is wrong?

[strange kindle]

This is wrong..

[sudden sorrel]

I meant this

(5)(4-4x)^3 + (-12(4-4x)^2)(5x)

This^^

That’s before factoring

Next step I take (5)(4-4x)^2 as factors

But then what do I do with that plus sign

[strange kindle]

Oh

[sudden sorrel]

This is before factoring, this is the step when you apply product rule

The plus is from the product rule formula

[strange kindle]

I say first simplify the first term

Wait

What's the original

Question even

[sudden sorrel]

It’s a long ass question

After that I need to find y and the line

[strange kindle]

That's short bruh

[sudden sorrel]

But first this haha

[sudden sorrel]

I’m on it for the past 4 hours😭

[strange kindle]

You know if you want to find the tangent just plug the numbers for slope

[sudden sorrel]

Yes but first you need to solve the derivative

Which is the step I’m stuck on

[strange kindle]

Plug x=2 in this

[sudden sorrel]

I know you can but I need to know how to simplify that term in general for the exam on Monday

Because we are gonna have a question where it asks just for the derivative

And it’s multiple choice

[strange kindle]

Ok

[sudden sorrel]

So you’ll need to make it look like one of the choices

Well what is the next step😅

How do I factor that

[strange kindle]

Factor out 5(4-4x)^2

[sudden sorrel]

I did but then what

Cuz there is a plus sign

[strange kindle]

So what

[sudden sorrel]

Do I keep it?

There?

When I did you said the terms are being added and then it’s wrong

So what do I do with it

[strange kindle]

, rotate

[strange kindle]

@sudden sorrel

[sudden sorrel]

Thank you, lemme digest this real quick haha

I see so you rewrote it and put the plus in the middle and negative multiplied by the plus

Then added the x from the 5x to the 12?

[strange kindle]

Not added but multiplied but yeah

[sudden sorrel]

And then the second expression you’d have:
(4-4x) -12x
?

Like the next line

How is it 16x 🥲

Plus times minus is minus so you are left with just -12x

Then multiplied by (4-4x)

It’s -48x+48x^2

So how did we get 4-16x

[strange kindle]

-4x -12x

=

-16x

[sudden sorrel]

Isn’t it being multiplied??

[sudden sorrel]

Before the factoring they are being multiplied

So how can you just change it to them being added

It’s originally -12(4-4x)

[strange kindle]

Try to factor it

Or

[strange kindle]

distribute in the first line

And you'll see why

[sudden sorrel]

How do you simplify 4-4x

It’s simplified

Oh wait

You mean with the 1-x

No I don’t get before that how you changed -12times(4-4x)
To (4-4x) plus -12x

How can you change something that is being multiplied to something that is being added

[strange kindle]

bro do what I said

Either

[strange kindle]

..

Or

[strange kindle]

..

[sudden sorrel]

(5)(4-4x)^2 ((+)-12(4-4x)^2(5x)

?

That’s what I would have done

Cuz the plus is being multiplied but also the -12 is being multiplied

[strange kindle]

That's wrong

Lemme ask you

When you factor

5x^2 + 5(-12)x^3

What do you get

[sudden sorrel]

Well I would rewrite it as:
5x^2 + 5x^3(-12)
Then factor out 5x^2
And get
(5x)^2 (5x)(-12)

[strange kindle]

That is so wrong

[sudden sorrel]

How🥲

Then I don’t know

I just know normal factoring

[strange kindle]

First of all the 5 isn't in bracket

And you'd have 1

[sudden sorrel]

But it’s being multiplied

[strange kindle]

In addition with 12x

[sudden sorrel]

Bracket means multiply no??

[strange kindle]

https://youtu.be/9MUDAr1PnlE?feature=shared

[sudden sorrel]

If all three terms are being multiplied can’t they be written the same??

[strange kindle]

Watch this

[lost hare]

(8x10^3)+(90x-32^4)

[strange kindle]

???????

[lost hare]

Its in the book

grade 11

[strange kindle]

Oh it's someone else

[sudden sorrel]

3x(-12)5 = 3x(5)(-12)
Right??

[strange kindle]

Yeah?

[sudden sorrel]

So why can’t I rewrite
5(-12)x^3
As
5x^3(-12)

[strange kindle]

Let's take a simpler example

[sudden sorrel]

All three are being multiplied tho

I don’t see the difference

[strange kindle]

Wait this is right

I thought you said something else

[strange kindle]

.......

[sudden sorrel]

So then this is the factor?

[sudden sorrel]

Yes please

[strange kindle]

...

Let's take

X^2 + x^3

Can you factor

[sudden sorrel]

x^2(x)

[strange kindle]

Wth

[strange kindle]

Watch this bro

[sudden sorrel]

???

But that gives you x^3🥲

Then you have both x^2 the factor

And the answer

[strange kindle]

Yeah

[sudden sorrel]

x^3

[strange kindle]

Oh

Write the whole damn thing bro

[sudden sorrel]

This is the whole thing

[strange kindle]

No it's not

The factored form IS

[plush delta]

I got a question

[strange kindle]

! occupied

[topaz sinew]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[sudden sorrel]

Wait the factor doesn’t stay after you multiply it with something??

[strange kindle]

??

[plush delta]

Ok so this is Kinda of a grade 7 question
8927382118.8274212341 in exponential form.
I don't remember this
I need help.

[strange kindle]

Watch the video bro

[sudden sorrel]

Oh…

[strange kindle]

! occupied

[topaz sinew]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[sudden sorrel]

I thought it stays😭

[strange kindle]

Idk what you're talking about

[sudden sorrel]

So it’s x^2(1+x)???

[strange kindle]

Yeah

[sudden sorrel]

Wtf

[strange kindle]

Finally

[strange kindle]

Wdym

[sudden sorrel]

Omg that makes so much sense actually

Ohhhh

[strange kindle]

If you distribute the x^2

[sudden sorrel]

Wait hold on lemme try the previous one

[strange kindle]

You get back the same thing

[strange kindle]

K

[sudden sorrel]

5x^2((-12)+x)

?

Correct?

[strange kindle]

No

[sudden sorrel]

Ffs😫

Close??

[strange kindle]

Yeah... No..?

It kinda is

Just watch the video bro

I can't explain very well on chat

[sudden sorrel]

Okay I’ll go watch it, I like that YouTuber too haha

[strange kindle]

He speaks and explains by doing examples

[strange kindle]

👍

[topaz sinew]

@sudden sorrel Has your question been resolved?

[topaz sinew]

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[ivory sorrel]

Problem 7

[ivory sorrel]

so I started by flipping the graph

so $y =- \sqrt{3x-x^2}$
I then shift the graph to the left by 4 units so
\
$y = -\sqrt{3(x+4}-(x+4)^2}$
\
I then shift it down by one unit so $y+1 = -\sqrt{(3(x+4) - (x+4)^2}$

[thorny flame]

Veni, vidi, perii is not f(wai)
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[torpid matrix]

so whats your question?

that does match the graph

[ivory sorrel]

Just wanted confirmation

As I messed it up the first time

Thanks

.close

[topaz sinew]

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Use .reopen if this was a mistake.

[topaz sinew]

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[feral forge]

i got the answer but I was wondering why I need to subtract it instead of adding it since it want the total charge for that month

[mortal steeple]

Let's say you have a scale

But the start part is broken

So you measure a pencil from 2cm mark to 6cm mark

What will be the length of the pencil

@feral forge

[feral forge]

4 cm

[mortal steeple]

Exactly

[mortal steeple]

Same can be applied here

[feral forge]

ohhhhh

shshhsjsjs

I feel dumb now haha

thank youuu

😭

[neon iron]

Ur handwriting is so neat thats crazy

[mortal steeple]

Wlcm

[mortal steeple]

Yeah

[feral forge]

AWHHHH THANKIUU 🥹🥹

[neon iron]

The charge for that month means end - initial then you’ll get the usage during the month:)

[mortal steeple]

Yes .close if your doubt is fine

Done*

[feral forge]

hehe I get it noww thanks guys!!

.close

[topaz sinew]

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[topaz sinew]

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[ivory sorrel]

I'm trying to find the limit of $\sqrt{x}e^{sin(\frac{\pi}{x}})$ at $x=0^+$

[thorny flame]

Veni, vidi, perii is not f(wai)

[ivory sorrel]

I was thinking $\pm \sqrt{x} e^x$

[swift jackal]

Tried L hospital?

[ivory sorrel]

for the squueze theorm

nvm

It's 0

[thorny flame]

Veni, vidi, perii is not f(wai)

[swift jackal]

I think there is no need to think cause you can directly put the limit.

[ivory sorrel]

I've been asked to use the squeeze theorm here

[swift jackal]

Ok got it

[swift jackal]

Is this function?

[golden blade]

yea apply it since sine is bounded

between -1 and 1

[ivory sorrel]

cool

[ivory sorrel]

to bound the function, this is what I wanted to use

[golden blade]

e^-1 <= e^sin(pi/x) <= e^1

[ivory sorrel]

nor e^sin(pi/x)

$\sqrt{x} e^{\sin ( \pi /x)}$

[thorny flame]

Veni, vidi, perii is not f(wai)

[golden blade]

sqrt(x) is increasing and positive you can multiply the inequalities by it

[golden blade]

.

[ivory sorrel]

Cool

thanks

[golden blade]

i was just making a point

[ivory sorrel]

.close

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[rain sluice]

for these why does it matter what order t he sin and cos is if they are multiplying?

[ruby tree]

It matters whether A or B is in the sin (with the other one in the cos)

It's really the exact same formula because
-sin(A-B) = sin(B-A)

[rain sluice]

oh ok thanks

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[violet sundial]

Is this right?

It's Function

<@&286206848099549185>

[neon iron]

You forgot that x had a square in the gx

[violet sundial]

Its on the parenthesis

+3 should be -3 mybad

[neon iron]

The denominator is supposed to be (x^2 -3)^2 if I’m not wrong

[violet sundial]

So its 4/x^2 -6x +6

[neon iron]

Should I show you what I got

[violet sundial]

Yes

[neon iron]

I could be wrong so confirm from someone else 😶

[topaz sinew]

@violet sundial Has your question been resolved?

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[jolly agate]

2. a

[neon iron]

,rotate

[jolly agate]

Is it -4?

[neon iron]

you can just substitute yeah

[fallow igloo]

Correct

Do you know why?

[jolly agate]

No

I put in x = 0

Will be -4

[fallow igloo]

That’s the solution

You can simply plug it in, because the function is consecutive at x = 0

[jolly agate]

So for example e)

That won’t work right?

3/0?

[fallow igloo]

For sure

[neon iron]

yes

[jolly agate]

0?

[fallow igloo]

It won’t work

[neon iron]

the denominator has x

[fallow igloo]

Limit is a concept of finding the neighbors of the value

[jolly agate]

What is the difference if it would be from the left instead ?

0^-

Any difference at all?

[ruby tree]

Yes, opposite sign

[jolly agate]

So +4?

[ruby tree]

Oh for (a)?

I thought you were asking about (e)

[fallow igloo]

Same

[jolly agate]

Well it was just a theoretical question of a)

[ruby tree]

For (a) it doesn't matter

[jolly agate]

How come ?

If it came from the left instead

Of right

[ruby tree]

The limits from both sides are the same and equal to the value at that exact point

[fallow igloo]

Left lim = right lim = lim

[ruby tree]

The function is continuous at that point

[jolly agate]

And for example e)

3/3-3

[fallow igloo]

If lim exists, it is equal to both left lim and right lim

[jolly agate]

What does it tell me ?

3/(3-3)

They meet at 0?

[ruby tree]

Here you can't just substitute

I guess you can think of it like (3+)/((3+)-3)

[jolly agate]

Okay, but won’t it be +3/0?

[ruby tree]

No because that's just not a number

[jolly agate]

Soooo for this function there is no defined values ?

Limit ****

[ruby tree]

Well, a limit can also be +inf or -inf

[jolly agate]

Ohhhh ye !!!

And how to determine if it’s + or - infinity ?

[ruby tree]

More formally it would be (3+ε)/((3+ε)-3), with ε positive and arbitrarily close to 0

[jolly agate]

Sooo the denominator gets as close to -inf and the top gets as close to +inf?

[ruby tree]

No

The numerator is just 3

[jolly agate]

Oh okay

[ruby tree]

The denominator goes to 0 but what matters is from which side

Is it positive or negative?

[jolly agate]

Posetive

[ruby tree]

So you get +inf

[jolly agate]

Yes, we get close to 0 from the right ause

Side

Or?..

[ruby tree]

Yes that's what the + in 3^+ means

Draw the graph, you'll see why it's +inf

[jolly agate]

Didn’t understand that, but can’t you look at the 3^+

And see that we close in from the right, which tells us that it is + inf

Is this to any help?

[ruby tree]

This tells you the limit of 1/x at 0+ for example

[jolly agate]

Where ???

[ruby tree]

1/x = x^{-1}

Fourth row, q < 0

[jolly agate]

Ohh

But then it’s negative?

[ruby tree]

No it says +inf

[jolly agate]

Oh ye wrong line

One last question, so the sum in the denominator is = 0, or?

How come it is not 0 then

And not infinity

[ruby tree]

Precisely because it's the denominator

This is a graph of 1/x

You can clearly see that when x approches 0, it goes to either infinities

[jolly agate]

Yes

But the x/(x-3)

Is that the graph?

[ruby tree]

No but it's similar

[jolly agate]

Ok

[ruby tree]

The numerator doesn't matter, it's 3, it doesn't contribute because you get either +inf or -inf anyway

The denominator is 0, but that just means the function is undefined there

The limit from the right is +inf, similar to 1/x at 0+

[jolly agate]

And how does 0!!!!! Become + infinity

[ruby tree]

The point is that when you evaluate the limit, you do not get 0 in the denominator

You approach 0

[jolly agate]

We approach 0 but never gets to it

[ruby tree]

Just compute the values of x/(x-3) for x=4, then x=3.1, then x=3.01, then x=3.001, and so on

You'll just get higher and higher values

At the limit, it becomes +inf

[jolly agate]

Okay i think I might understand it somewhat

For B then

It’s similar ?

Well wait that must be 0

[ruby tree]

You can also look at the graph

This is x/(x-3) along with the line x=3

[jolly agate]

Ohhh okay

[ruby tree]

For (b) it's different

[jolly agate]

We get close to 3 but never 3

[ruby tree]

Yes

[jolly agate]

Since it’s a rational fraction?

Or whats it’s called

1/x

[ruby tree]

Not sure what you mean

[jolly agate]

Me neither I’m trying to translate from Swedish..

It’s division

Is it because of that

[ruby tree]

About x/(x-3)?

[jolly agate]

Yes

[ruby tree]

Well yes it's a fraction

You divide by a smaller and smaller number

[jolly agate]

Ye and it gets bigger and bigger

[ruby tree]

Yes

[jolly agate]

For B then, what are the play rules ?

What is the first step I should do in order to try and find the limit

If I can not graph it

[jolly agate]

Oh I’m sorry, should have posted it again

[ruby tree]

No worries

What's |x| if x < 0?

[jolly agate]

Posetive ?..

[ruby tree]

No

Do you know what |x| does?

[jolly agate]

Absolute values

[fallow heart]

Yeah so... Do you know the definition of abs value?

[jolly agate]

Kinda, may need to repeat it tho

[fallow heart]

Apart from making everything non negative (which is a actually a property, not the definition)

[ruby tree]

Basically you can define it like this:
|x| = x, if x >= 0
|x| = -x, if x < 0

[fallow heart]

For sure, you need to know that (not only for limits, but also for all the future math you'll encounter)

[jolly agate]

This i rembeber !

[fallow heart]

Awesome, that's what you need here

[ruby tree]

So do you agree that

$\lim_{x \to 0^-} \frac{x + |x|}{x} = \lim_{x \to 0^-} \frac{x - x}{x}$

[jolly agate]

No

[thorny flame]

Nel

[ruby tree]

Sorry, 0- not 0+

[jolly agate]

Yes !

No worries 🤣

[ruby tree]

So what's the numerator here?

[jolly agate]

-x?..

[fallow heart]

[jolly agate]

Hahahaha I’m sorry

[fallow heart]

What's 6 - 6?

[jolly agate]

0

[fallow heart]

Yeah better

[ruby tree]

The numerator is x-x, the denominator is x

Considering x -> 0+, can you describe with words what each of them equal?

[jolly agate]

Each of the x’s?

[ruby tree]

No, the numerator and the denominator

[jolly agate]

X+X

We approach X from the right

[ruby tree]

I don't know why I keep writing 0+

[jolly agate]

I’m a lost cause ain’t I

[ruby tree]

Considering x -> 0-, can you describe with words what x-x equals, and then what x equals?

[jolly agate]

No I can not

What ever we put in, will be 0/ something

Always 0?

[ruby tree]

Yes

Exactly

x-x is always exactly 0

x however is not quite 0, it's something less than 0 but arbitrarily close

[jolly agate]

The /x you thinking of !

?

Alone x ?

[ruby tree]

Yes

The denominator

[jolly agate]

Why does it not become 0😢

[ruby tree]

Because it's a limit, that's kind of the point...

The function is not defined at 0

[jolly agate]

Okey

[ruby tree]

Try (c) and then look at the graph

[jolly agate]

That Will be x+x / x

Since we come from the tight to 0

Right

[ruby tree]

You need parentheses but yes

[jolly agate]

And the x is not defined at 0

HOW does the answer become 2????????

[ruby tree]

You had (x+x)/x, you just have to do some basic algebra on it

What's x+x?

[jolly agate]

2 x?

[ruby tree]

Yes, and what's 2x / x?

[jolly agate]

2

[ruby tree]

Right

[jolly agate]

Ohhh

I Will try the rest on my own.. wish me luck and thank you so so so much.

[ruby tree]

[jolly agate]

.close

[topaz sinew]

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[hollow pumice]

I have no idea how to graph a function, I missed the lecture for this as I was out

[gentle night]

You draw the x and y axis

Then scale them with numbers

After that you place the definite points of the graph

[hollow pumice]

What're those?

[gentle night]

The points that are clearly defined in the properties

Meaning doesn't have limit

Like 3 and 5

[hollow pumice]

so the definite points on 3 and 5 are 1 and 4?

[gentle night]

Yeah these are the y values of the points

[hollow pumice]

oh so 1 and -2 are the x?

[gentle night]

Yeah

[hollow pumice]

so does that mean 3 and 5 when sketched are are just points on a graph?

[gentle night]

Yeah

[hollow pumice]

okay, got that

[gentle night]

Then you slowly get to use the other information

Like for example to the right of the one there is only 2 properties
1 and 8

From 8 you know that as you are going to infinity the function approaches -1

And from 1 you know that the graph on the right touches the point 1

So you start drawing from point 1 a line to the right that approaches -1 as you keep going to the right.

But never reaches -1

[hollow pumice]

like this?

[gentle night]

And draw it like a curve that starts to get down quickly to -1 at the start then it starts to flatten out with time

[gentle night]

No

There are many mistakes there

Ok 2 mistakes

Not that many

Also I recommend drawing the graph with a pencil

To erase things Easley

[hollow pumice]

It is

[gentle night]

Ok

The first one did you check property 2

It says that the graph on the left of the one starts from 0

And what does the property 7 say

[hollow pumice]

Infinity?

[gentle night]

It doesn't just say infinity

You know how to read limits right?

[hollow pumice]

Ik how to solve like

The equations

This stuff

[gentle night]

I mean about reading them
Like if I asked $\lim_{x\to1^+} f(x)= 2$
Do you actually understand that

[thorny flame]

Sherif Player

[hollow pumice]

I don't

[gentle night]

That's the problem

Okay what about
$$\lim_{x\to1}f(x) = 2$$ would you understand that

[thorny flame]

Sherif Player

[gentle night]

Without the +

[hollow pumice]

the y point would be 2?

[gentle night]

No

It means that as you get more and more closely to point 1 with function f(x) you gets more and more closely to the value of 2

It doesn't indicate that
f(1) = 2 though

[hollow pumice]

okay then

[gentle night]

Like this

Here the limit as you get closer and closer to 1, f(x) gets closer and closer to 2
But actually f(1) = 0 here

[hollow pumice]

the + and - mean which side of the graph its coming from, right?

[gentle night]

Yeah
The positive means to/from the right of the value

The negative means to/from the left of the value

[hollow pumice]

and if theres neither, it comes from both sides?

[gentle night]

Yes

[hollow pumice]

Okay, got that

what about with infinity?

so as x approaches +/- infinity

[gentle night]

As x approaches + infinity means as x keeps going to the right
The value of the function gets closer and closer to L
But it actually doesn't get equal to L as it goes

[hollow pumice]

okay

[gentle night]

Negative is the same but to the left

[hollow pumice]

so each of these criteria, so i'm understanding this right

2 point, and 6 different lines?

[gentle night]

Not exactly because there are properties that refers to the same line

Like 1 and 8

[hollow pumice]

how do I know which properties refer to which?

[gentle night]

You understand what the property wants you to draw
You should not try to figure how many lines you should draw
Instead you like try and connect the properties together

If you looked on the properties you will find that to the right of the 1

There is only 2 properties

Which are 1 and 8

The first one explains the limit of f(x) as x goes to 1 from the right

And 8 explains how the line would approach as x keeps going to the right

If you think about it, it is the same line, which starts from 1 and keeps going to the right

[hollow pumice]

So like that?

[gentle night]

Read property 8 again

[hollow pumice]

-1?

[gentle night]

What does f(x) approach as x approaches infinity

[gentle night]

Yup -1

[gentle night]

It doesn't start from 1 though

Also don't make it reach -1

[hollow pumice]

leave some space?

[gentle night]

Make it like it is approaching -1 without getting to be -1

[gentle night]

Yeah

[hollow pumice]

so where does it start from?

I dont understand

[gentle night]

I want to make sure you do understand

So first you put the dots

Which is done

Then we start looking for applying other properties

What property 1 says?

[hollow pumice]

(1,1)
(-2,4)

[gentle night]

Yeah we finished that part

[hollow pumice]

as x approaches 1, it gets closer to the y value 1?

[gentle night]

Yeah but approaches one from right or left?

[hollow pumice]

from the right because it says +?

[gentle night]

Yes

So if you looked at any other property there is nothing to be done to the right of the point (1,1) except that property (property 1) and property 8

[hollow pumice]

ohhh because +infinity ok

so I do what I did before, but starting at the point 1,1?

[gentle night]

Yeah if you looked at the other properties you will find them happening at the left

[gentle night]

Yeah but do you understand why we start from that point

[hollow pumice]

because its on the right of the graph?

[gentle night]

You didn't get my point
We start from this point because of property 1 that says as we approach 1 from the right we get closer and closer to y value 1

Which is that exact point

[hollow pumice]

okay, I think I got that then

[gentle night]

Okay

Now let's think of where to start the graph at the left of point 1,1

Do you see the property that will tell us where

[hollow pumice]

is it 6 or 7?

[gentle night]

These talks about x = -1

Not x = 1

[hollow pumice]

so it's property 1?

or 2

[gentle night]

Why 1

We are now talking about the left of the point x = 1

[hollow pumice]

well it says as x approaches 1 on right side of the graph

so 2 because its talking about the left side

it has a -

[gentle night]

Yes property 2 is the one

So what does it say

[hollow pumice]

as x approaches 1 on the left side of the graph, it gets closer to y value 0

[gentle night]

It doesn't mean on the left side of the graph
It means on the left side of the value

Like on the left of x= 1

[hollow pumice]

got it

[gentle night]

As we approach x = 1 from the left y approaches 0

So what does that say about the point we are starting at the left of x = 1

[hollow pumice]

This?

[gentle night]

No

You have a problem with graphing limits by the way, I would recommend watching a video about it

[hollow pumice]

suggestions would be much appreciated

[gentle night]

https://youtu.be/7Q2HwTHcxA0?si=dAGtVEsHb6HNB3ib

[hollow pumice]

alright so sshould I watch this video and come back?

[gentle night]

This for general understanding @hollow pumice

But for the specific problem you can watch this
https://youtu.be/LJCLiCOr4ek?si=2PHshTr2aw316toM

I recommend watching the first one before getting to this to be able to understand the solution better

[hollow pumice]

alright, so i'll watch these then and try it

Should I ping you when done?

[gentle night]

I have no problem with that

[hollow pumice]

Alright, i'll be back then

thank you

[topaz sinew]

@hollow pumice Has your question been resolved?

[hollow pumice]

@gentle night alright so

I sorta understand a bit better

what the video didn't explain was how to deal with properties like 4 and 8

[topaz sinew]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[hollow pumice]

.reopen

[topaz sinew]

✅

[gentle night]

ok so you applied propertyies 2,3, and 5 correctly but you did property 7 wrong

can you see how you did it wrong

@hollow pumice ?

[hollow pumice]

hm

[gentle night]

read property 7

[hollow pumice]

as x approaches infinity, it gets closer to y value -1

[gentle night]

7 not 8

[hollow pumice]

oh whoops

as x approaches -1 from the right side, it gets closer to infinity

[gentle night]

so in you graph
as x approaches -1 from the right side, why does it get closer to negative infinity

[hollow pumice]

ohhhh its supposed to be going up?

[gentle night]

yeah that is what property 7 means

it needs to go up for y to approach infinity as x goes to -1 from the right

[gentle night]

ok that is almost correct

[hollow pumice]

regarding property 6?

[gentle night]

but read property 4

[hollow pumice]

as x approaches negative infinity, it gets closer to y value of 2

[gentle night]

ok in your graph as x approaches negative infinity, y gets closer to 0

[gentle night]

that is y gets closer to 1

[hollow pumice]

I can't even my own graph whoops

[gentle night]

the other thing is that why did you make that line that goes to -2,4

[gentle night]

almost correct

[hollow pumice]

What'd i do wrong

[gentle night]

i will explain later but answer the previous question

[hollow pumice]

The dotted lines?

[gentle night]

no

the arrow

that points to -2,4

[hollow pumice]

I'm sorry I don't know which one you're talking about

[hollow pumice]

I honestly forgot why I even drew that

[gentle night]

yeah, it is not stated in any of the properties so just delete it

now about the minor mistakes.
for a function to be called function it needs to have the property that for any 1 input only 1 output is given from the function
so that is why any function needs to pass something called the vertical line test, which is a test where if you draw a vertical line any where on the x - axis it needs to intersect f(x) only at 1 point.
Now what is that have to do with the graph
the property 5 says that f(-2) = 4
which means that at x = -2, y = 4 and there should be no other point under or above it. that is why you need to put a hole on any line you draw that passes below or above it

so you will need to put a hole under that point on that line

the other minor mistake is that hole with the dot in it, which does not make since mathimatcly, like the hole means that that the line does not contain the number it started at but because you put the dot it actually means that the line contain the number it started with so just erase that hole and you will be ok

that is it

@hollow pumice send the picture here to make sure of it

[hollow pumice]

I drew a new one to keep it legible

[gentle night]

very cool
just make the right line curvy and smooth not bent like that

[hollow pumice]

Like that?

[gentle night]

like that

[gentle night]

and that hole should be on the line, it doesn't make since to be in the air

[gentle night]

okay that is it

correct

[hollow pumice]

Alright, I'm gonna try to do all this without looking at the past steps now

Thank you so much for the help

[gentle night]

you're welcome

[topaz sinew]

If you are done with this channel, please mark your problem as solved by typing .close

[hollow pumice]

.close

[topaz sinew]

Channel closed

Closed by @hollow pumice

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[topaz sinew]

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• Show your work, and if possible, explain where you are stuck.
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[abstract jay]

Sam and Tom each choose a positive integer, which they share with a third
person (but not with each other!). They are then told that 280 is either the sum
or the product of the two chosen numbers, and are asked whether they know the
number chosen by the other person. Sam says, ”I do not know Tom’s number”.
Tom then says, ”I do not know Sam’s number”, to which Sam replies, ”Now I
know Tom’s number!”. What can you say about the number chosen by Sam?

[mortal steeple]

is this from mind your decisions?

https://youtu.be/Ka-Ge3PdVOM?si=EM8Bu5zKP6bWqdmI

a similar problem

@abstract jay

[abstract jay]

thanks for sharing the video

i think i can solve it now 👍

[mortal steeple]

wlcm

[abstract jay]

thank you have a nice weekend!!

[mortal steeple]

weekend? whats a weekend?

[topaz sinew]

@abstract jay Has your question been resolved?

[topaz sinew]

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[neon iron]

how am i supposed to find the taylor series of this

can i just find the taylor series of cos(6x^2)

then i subtract by 1

and divide by x^2?