Limit, trig1 maybe 2 | Mathematics | Page 1

brave wigeon Dec 31, 2024, 5:54 PM

#

.

rn_image_picker_lib_temp_233fc5f0-86e9-42fa-80d3-063ea40def18.jpg

silver vortexBOT Dec 31, 2024, 5:54 PM

#

brave wigeon .

#

Do not ping the Moderators, unless someone is breaking the rules.
Do not ping the Helper Moderators, unless there is a conflict between helpers.
Do not ping other members randomly for help.
Ask your question and show the work you've done so far. If you've posted a screenshot of a question, specify which part you need help with.
Wait patiently for a helper to come along.
If the Helper has answered your question, remember to thank them with the Mathematics Ranks bot and close the thread with:

+close
Feel free to nominate the person for helper of the week in #helper-nominations
If you're happy with the help you got here, and the server overall, you can contribute financially as well:

brave wigeon Dec 31, 2024, 6:03 PM

#

provided :
[AB] is tangent to the quarter unit circle at C
m(COA)=x
Area(BOA)=P(x)
Area(COB)=S(x)
asks for,
lim p(x)/S(x)
x->0

I first did,
P(x)/S(x)=|BO|.|OA|/2/|BO|.|OC|/2=|BO|.|OA|/|BC|.|OC|
Then to find the distances
|BO|=t (some value that i gave between the missing area that completes BO after being added by sinx) + sinx
using euclids theorem to find t,
cos^2x=t.sinx
t=cos^2x/sinx
|BO|=cos^2/sinx + sinx = (cos^2x+sin^2x)/sinx = 1/sinx
did the same for |OA|
cosx+v=sinx^2x
v=sin^2x/cosx
|OA|=(sin^2x+cos^2x)/cosx= 1/cosx
then wrote for P(x)/S(X)

(1/sinx)(1/cosx)/|BO|.1 (because |OC| is radius)

#

now im stuck on how to find |BC|

#

i tried to use pythagoras theorem to try to find it but

#

that doesnt really work to give me an answer i can work with

worthy sapphire Dec 31, 2024, 9:06 PM

#

is the question asking where the limit is pointing?

#

i mean

#

to what P(x)/S(x) is pointing when x is pointing to 0?

river stirrup Dec 31, 2024, 9:09 PM

#

answer is 6

worthy sapphire Dec 31, 2024, 9:10 PM

#

you're not meant to give the answer, and 6 isn't even a choice

river stirrup Dec 31, 2024, 9:10 PM

#

worthy sapphire you're not meant to give the answer, and 6 isn't even a choice

thats the point...

worthy sapphire Dec 31, 2024, 9:12 PM

#

#

Anyway if we do use the pythagorean theorem to find BC, we would start with OC^2 + BC^2 = BO^2

abstract rain Dec 31, 2024, 9:14 PM

#

river stirrup answer is 6

?tag helping

wraith badgeBOT Dec 31, 2024, 9:14 PM

#

**When helping people, please do not dump the full solution. **
Remember that these people are trying to learn concepts, and our goal should be to reinforce them. Giving them a solution will not help.

**Instead, give them hints and guide them toward the correct solution. **
You can ask things like "what part did you get stuck on?" or "what have you tried so far?" or give them a small starting hint. This ensures better learning. For more tips, reach out to Schlaumau (@schlaumau).

worthy sapphire Dec 31, 2024, 9:14 PM

#

abstract rain ?tag helping

its fine 6 isn't even an answer

river stirrup Dec 31, 2024, 9:15 PM

#

i was tryna be ironic no_pain

worthy sapphire Dec 31, 2024, 9:15 PM

#

BO^2 = (1/sinx)^2

abstract rain Dec 31, 2024, 9:16 PM

#

im solving this myself, but ||the two red triangles outlined are congruent right||?

worthy sapphire Dec 31, 2024, 9:17 PM

#

1/sin^2(x) is equal to 1 + cot^2(x)

#

OC is assumed to be equal to 1

#

so 1^2 + BC^2 = 1+cot^2(x)

#

we subtract 1 from both sides and get to BC^2 = cot^2(x)

#

thus BC = cot(x)

#

cot(x) = cos(x) / sin(x)

#

i will leave it to you here: cot(x), the answer you should normally get for BC, is useful
extra hints: ||tan(0) = 0 ; 1/cos^2(x) = 1+tan^2(x) ; remember that x is pointing to 0, so P(x)/S(x) is pointing to something too||

#

if i did anything wrong please correct me, it is entirely possible that i made a mistake here

#

however i had a lot of fun doing it, thank you for posting

#

the answer i got is coherent from a geometrical point of view, so that is reassuring'

worthy sapphire Dec 31, 2024, 9:33 PM

#

abstract rain im solving this myself, but ||the two red triangles outlined are congruent right...

not explicitly stated, so i just assume that no

brave wigeon Dec 31, 2024, 10:42 PM

#

worthy sapphire is the question asking where the limit is pointing?

No just the value of the limit

#

Which would be

#

P(0)/s(0)

#

So i guess not that much of a limit question and more trigonometric

worthy sapphire Dec 31, 2024, 10:43 PM

#

Ok yes

#

That was the answer i was looking for

#

So

#

Did you plug cotx into the equation for BC?

#

In the form of cosx/sinx

brave wigeon Dec 31, 2024, 10:45 PM

#

No I just tried to use pythagoras for |BC| which did not give me a result that looked like i could do anything with

#

So I just posted it here

#

Cuz I spent 15+ mins on this atp

worthy sapphire Dec 31, 2024, 10:45 PM

#

brave wigeon No I just tried to use pythagoras for |BC| which did not give me a result that l...

What result did you get?

brave wigeon Dec 31, 2024, 10:46 PM

#

Well I'm not on my desk right now but

#

It'd be

#

cos^4x/sin^2x+sin^2x= m^2+1

#

So to solve for m which is just a value i gave to |BC| right now

worthy sapphire Dec 31, 2024, 10:48 PM

#

Oh that's not what i did

worthy sapphire Dec 31, 2024, 10:48 PM

#

worthy sapphire Anyway if we do use the pythagorean theorem to find BC, we would start with OC^...

I did this

brave wigeon Dec 31, 2024, 10:48 PM

#

sqrt((cos^4x/sin^2x) +sin^2x -1)=m

worthy sapphire Dec 31, 2024, 10:48 PM

#

Building on you previous answers

brave wigeon Dec 31, 2024, 10:49 PM

#

worthy sapphire I did this

Thats what I did here though

worthy sapphire Dec 31, 2024, 10:49 PM

#

Which were 1/sinx and 1

brave wigeon Dec 31, 2024, 10:49 PM

#

Wait im stupid I already found

#

|bo| as 1/sin

#

Ur right

#

Wtf

#

Wait no but i did do that

#

Then thats just

#

Sqrt(1-sin^2/sin^2) =|bc|

#

But how does that help

#

At this point plugging this in would still give me a 0/0

#

No?

worthy sapphire Dec 31, 2024, 10:53 PM

#

yes but 1-sin^2 = cos^2

brave wigeon Dec 31, 2024, 10:53 PM

#

Wait no then its just the sinx simplify and im just left with 1-sin^2

worthy sapphire Dec 31, 2024, 10:53 PM

#

and we are back to cotan()

brave wigeon Dec 31, 2024, 10:54 PM

#

Ok yeah

worthy sapphire Dec 31, 2024, 10:54 PM

#

with cos^2 and sin^2

#

so we have cos/sin

brave wigeon Dec 31, 2024, 10:54 PM

#

Does the sin not simplify?

#

Not at my desk rn so maybe im missing smth

worthy sapphire Dec 31, 2024, 10:54 PM

#

BC = sqrt (cos^2/sin^2)

#

=cos/sin

#

[1/cos(x) * sin(x)] / [cos(x) / sin(x)]

#

because you were stuck at [1/cos(x) * sin(x)] / BC

#

we arrive at sin(x) / cos(x) * cos(x) * sin(x)

#

we can remove sin(x) from the top and bottom

brave wigeon Dec 31, 2024, 10:58 PM

#

Yep

#

So its just

#

Cosx

worthy sapphire Dec 31, 2024, 10:58 PM

#

we arrive at 1 / cos^2(x)

brave wigeon Dec 31, 2024, 10:58 PM

#

Which cos0 is 1

worthy sapphire Dec 31, 2024, 10:58 PM

#

that is equal to 1 + tan(x)

brave wigeon Dec 31, 2024, 10:59 PM

#

worthy sapphire we arrive at 1 / cos^2(x)

Isnt it 1/cosx?

#

Bcs the sqrt

#

From bc

worthy sapphire Dec 31, 2024, 10:59 PM

#

yes

brave wigeon Dec 31, 2024, 10:59 PM

#

Oh no

#

Yeah

worthy sapphire Dec 31, 2024, 10:59 PM

#

but there is already cos x in 1/cos x * sin x

#

and we multiply it by the cos x from the BC

brave wigeon Dec 31, 2024, 10:59 PM

#

Right my bad I forgot ab that

worthy sapphire Dec 31, 2024, 10:59 PM

#

because we are dividing so we cross multiply

#

so 1/cos^2(x) = 1 + tan(x)

brave wigeon Dec 31, 2024, 11:00 PM

#

Yah I was thinking of p(x) as 1/sin not 1/sincos

worthy sapphire Dec 31, 2024, 11:00 PM

#

if x -> 0

#

then what does tan point to?

brave wigeon Dec 31, 2024, 11:01 PM

#

0

worthy sapphire Dec 31, 2024, 11:01 PM

#

ok

#

so 1 + tan(x) points to what, if x points to 0

brave wigeon Dec 31, 2024, 11:01 PM

#

1+0=1

#

Yah

worthy sapphire Dec 31, 2024, 11:02 PM

#

so P(x)/S(x) -> 1 as x -> 0

#

now think of it this way

#

as the angle portrayed by x gets smaller

#

the area of the triangle COB, which represents the area S, gets bigger

#

if x reaches 0

#

then the angle BOC will equal the angle BOA

brave wigeon Dec 31, 2024, 11:03 PM

#

Oh wait I think I get what you're getting a

#

They're going to be the same triangle so

#

Wow

worthy sapphire Dec 31, 2024, 11:04 PM

#

so the area of the bigger triangle/the area of the smaller triangle will be 1

brave wigeon Dec 31, 2024, 11:04 PM

#

Thats kind of cool and seeing that could've made this a 2 second question

worthy sapphire Dec 31, 2024, 11:04 PM

#

yes but yk what its very good that you approached it the first way, i often want to go fast and do it through "logic" but most of the time teachers want the pure algebraic stuff lol

brave wigeon Dec 31, 2024, 11:05 PM

#

Well I guess seeing both of the approaches is thebest outcome

#

Next time I see smth like this

#

I can just think of it the fast way

#

While knowing the logic on algebra

#

Okay, thank you for the help

worthy sapphire Dec 31, 2024, 11:06 PM

#

you're welcome! best of luck in upcoming tests

brave wigeon Dec 31, 2024, 11:06 PM

#

$close

#

Um wrong command

#

Okay well I don't remember the command so

worthy sapphire Dec 31, 2024, 11:07 PM

#

+close

brave wigeon Dec 31, 2024, 11:07 PM

#

+close

glass templeBOT Dec 31, 2024, 11:07 PM

#

brave wigeon +close

Please thank your Helpers before closing!

Please thank the helpers who assisted you by clicking the buttons below. You can thank each helper only once. Once you're done, click "Close Post" to close this thread.

glass templeBOT Dec 31, 2024, 11:07 PM

#

glass temple

Thank you for your feedback! dark matter has been awarded 1 helper_points . They now have 925 . They have 0 daily left for today.

#Limit, trig1 maybe 2