"And if you put zero in here..." *stands in shame as he realizes he has committed an atrocity*
@bprpfast
2 жыл бұрын
😂
@AZunon
2 жыл бұрын
a mathematical atrocity indeed, so much so i didn’t understand a single thing as everything went out of my teenager brain
@aabahdjfisosososos
2 жыл бұрын
@@AZunon how old are you
@aabahdjfisosososos
2 жыл бұрын
@@AZunon I’m 15.
@MikeB3542
2 жыл бұрын
I'm sure he can squeeze out a limit out if he tries hard enough
@lunaas1029
2 жыл бұрын
Guys, I think he finally reached his lim
@Ktm2191
11 ай бұрын
His limx→ ∞
@onyemasuccess3411
11 ай бұрын
😂😂😂
@calculus988
8 ай бұрын
LOL
@purplrshadowyay
8 ай бұрын
That was good
@bradathebread
6 ай бұрын
Limit of bprp approaching insanity
@matheusferesturcheti288
2 жыл бұрын
For those who don't know yet, this is the Fundamental Trigonometric Limit, which evaluates to 1
@matheusferesturcheti288
2 жыл бұрын
In fact I think he has at least one video on it, about not using L'Hopital to solve this limit
@solo4554
2 жыл бұрын
@@matheusferesturcheti288 in this vid did he use l'Hopital or not ?
@AminePvPDZ
2 жыл бұрын
@@solo4554 no , he did not.
@rengarmain212
2 жыл бұрын
Its not even that hard, u can sandwich it with cos x and 1 or alternatively 1/cos x both works fine
@chayanaggarwal3431
2 жыл бұрын
@@matheusferesturcheti288 you could evaulte by Taylor expansion or sandwich theorem and trigonometry
@Wutthehel0
2 жыл бұрын
I love how he says nervously at the end - "and, and...and have a look."
@hartree.y
2 жыл бұрын
The dude forgot the answer
@livelydodo
2 жыл бұрын
it's small angle approximation right? sinø = tanø = ø when ø is very small. so as ø tends to very small => ø/ø = 1
@sanatandharma6009
2 жыл бұрын
@@livelydodo yes but sin 0 degree =0 This is not angle approximation
@AK-vb9ks
2 жыл бұрын
@@livelydodo the actual proof comes from taylor
@Daniel31216
9 ай бұрын
@@AK-vb9ksNo it doesn't. You need to know the derivative of sinx to do that.
@Predaking4ever
2 жыл бұрын
This what it is really like teaching math because you know what to do next, but it would involve future topics.
@leonhardeuler675
2 жыл бұрын
This is what happens when you upload multiple videos everyday. He's finally snapped.
@bprpfast
2 жыл бұрын
😂 😂
@agooddoctorfan651
Жыл бұрын
😂😂
@benshapiro8506
Жыл бұрын
mahn, no 1 is perfect.
@Kidnapper656
4 ай бұрын
@@benshapiro8506 except me ofc
@Fintogy
4 ай бұрын
As an engineer, sin(1/x) = 1/x so x*1/x =1 lim 1 = 1
@gdtargetvn2418
2 ай бұрын
☠ I hate how valid it is
@babarb9234
2 ай бұрын
😂😂@@gdtargetvn2418
@TeamGCS
2 ай бұрын
To formalize this and make it legal you would have to say that sin is asymptotically equivalent to x as the value inside sin approches 0, because it is the case for 1/x, then you would have: x*sin(1/x) ~ x*(1/x) = 1
@PowerUpStudio_
Ай бұрын
sin(1/x) approaches 1/x as x approaches infinity because of its taylor series x-x^3/6+x^5/120-... as x approaches zero the other terms become negligible so sin(1/x)≈1/x as x->inf so lim(x->inf)(xsin(1/x))=1
@GutReconIkaros
Ай бұрын
@@PowerUpStudio_ to prove it is its Taylor series, you need to use the derivative of sin at 0, which is the limit you want to evaluate.. But it is still a good idea for limits involving a product of sin with a polynomial, or if you suppose alrady given the equality dsin/dx (0) = cos(0) = 1.
@shahenazmalek5882
8 ай бұрын
Pay attention to the sum being solved - ❌ pay attention to his had when he writes with two markers at a time - ✔️
@liamryden558
2 жыл бұрын
this is extremely epic
@hellhusk
2 жыл бұрын
and this is how to comment properly
@DadicekCz
Жыл бұрын
@@hellhusk And this is how to answer properly
@Charismatic_philosopher
Жыл бұрын
@@DadicekCz and this is how to look properly
@kingbeauregard
2 жыл бұрын
It's like when you bump into a celebrity on the street but you're so star-struck you can't remember what they appeared in.
@RamsLiff
2 жыл бұрын
For those who want a trigonometic solution: 1 < x/sinx < 1/cosx 1 > sinx/x > cosx use limits as x -> 0 , and 1 > sinx/x > 1 so because of the squeeze theorem, sinx/x as x->0 is 1 too
@isavenewspapers8890
5 ай бұрын
Technically, it's ≤ and ≥. Or =, if you're a programmer.
@srr9281
3 ай бұрын
Wondering why or how your first line of math is derived? The rest is just math grammar after that.
@Rando2101
Ай бұрын
@@srr9281 I'm guessing geometry, not really sure how tho
@nicolastorres147
2 жыл бұрын
Wait. He’s still trying to approach 0 without putting 0. 😂😂😂
@symiangearhead8983
2 жыл бұрын
After trying so hard in limits to not get a 0/0 form and ending up getting a 0/0 form in the end , "YOU HAVE BECOME THE VERY THING YOU SWORE TO DESTROY"
@ChandanYadav-dq3pj
Ай бұрын
But isn't lim X approaching to 0 sin(x)/x equal to 1
@Rando2101
Ай бұрын
@@ChandanYadav-dq3pj it's kinda hard to explain why
@vloggerdeb9017
2 жыл бұрын
The way he switches pen is so amazing 🥰
@micah2455
Жыл бұрын
Me using Desmos and scrolling really far to the right: **I am 4 parallel universes ahead of you**
@vimuthabeysinghe6
2 жыл бұрын
Finally isn’t sin theta over theta = 1? I’m new to calculus why did he seem to not like to do that😅
@ChickenPowder
2 жыл бұрын
Yeah it’s 1
@butterchickenproductions5141
2 жыл бұрын
he has a video on why you can't use L'Hôpital's rule for it
@juanitome1327
2 жыл бұрын
@@butterchickenproductions5141 right but it wasnt l’hopital it was equivalent infinitesimals right?
@ignaciodemiguel3683
2 жыл бұрын
@@juanitome1327 exactly
@Professional-Hater
9 ай бұрын
only when theta -> 0
@Harsh-ll9wp
2 жыл бұрын
And it is 1, pretty simple ah ha
@karanbhura1378
9 ай бұрын
Did anyone notice how well he changes the pen between his fingers, damnnn
@kevinning8027
2 жыл бұрын
U gotta do the OG proof using geometry 😅
@gustavoespinoza7940
2 жыл бұрын
My favorite trick for limits is turning them into differentiation like with this problem as x approached 0 sin(x)/x = (sin(x) - sin(0))/(x-0) = d/dx sin(x) evaluated at x = 0 which is just cos(0) = 1
@sohamacharya171
11 ай бұрын
Prove the diferrentation of sin x is cos x without using this result.
@eternal2686
4 ай бұрын
Can’t you just use substitution here? Lim x->inf xsin(1/x) Infsin(1/inf) As x -> inf of sin(1/x), the value of the function becomes smaller and smaller Therefore, we have infsin(1/inf) = inf(1/inf) = 1 Does this work or no?
@nvapisces7011
11 ай бұрын
Should be 0^+, because it is coming from positive direction. Then evaluate the famous limit with squeeze theorem or you can use McLaurin series where if the angle is small, you can ignore powers of x³ and above
@Rando2101
Ай бұрын
The series uses the derivative of sin(x). In order to prove the derivative of sin(x), you prove lim(x->0) sin(x)/x = 1 Don't do that. That's circular reasoning.
@panos21sonic
5 ай бұрын
Man the marker switching is effortless, goddamn
@themandel2017
2 жыл бұрын
Sin x over x goes to 1 as x approaches 0 because rotating a circle nears a stable line velocity as you are on a horizontal angle and that parallel is vertical
@RishaadKhan
2 жыл бұрын
yes, he's already shown a whole video and how to not use lhopital's rule
@johnr.aucoin1990
2 жыл бұрын
@@RishaadKhan Where's it !!!!
@user-mg3xb4ds4b
2 жыл бұрын
@@johnr.aucoin1990 its in his main channel blackpenredpen
@jameszhang9326
2 жыл бұрын
"Da Hospital wants 1!" - Cheerios
@KingGisInDaHouse
2 жыл бұрын
The engineering trick works here. Sin(1/X)≈1/x
@platosbeard3476
2 жыл бұрын
The engineering trick uses a Taylor series to approximate sin, which is problematic because the use of derivatives of sin lead to circular reasoning - i.e., the limit of sin(x)/x is required for the derivative of sin (lim h->0 sin(x+h)/h). Having said that, and I'm a little rusty with this stuff so don't take it on faith, the power series for sin can be done w/o calculating that particular limit. I'm thinking Newton calculating Pi here. If that's the case, then circular reasoning can be avoided by using that particular construction. I may well be misremembering this, though
@Rzko
2 жыл бұрын
that's an equivalent when x goes to 0, theres nothing wrong here, sin(x) ~ x when x -> 0
@Rzko
2 жыл бұрын
@@arishalpin5883 I don't get what you mean by Taylor expansion, Taylor serie? This symbol "~" means "is equivalent", and it is a powerful thing to find limits. f(x) ~ g(x) when x-> 0 f(x) = g(x) + ε(x) with ε(x) goes to 0 when x -> 0. Maybe you are american and don't use ~ and DL's in analysis, but it is the more efficient, no need to use Taylor series or things that complicated to just find a limit
@lanzji1345
2 жыл бұрын
@@arishalpin5883 As if engineers would bother with Taylor when doing this kind of approximation... Taylor is used by mathematicians to find out that they can't blame engineers for being stupid when doing this!
@thaddeusal-britani1099
2 жыл бұрын
In engineering class: ≈== Well, approximately.
@4dotaonly
2 жыл бұрын
through "sandwich criterion", this evaluates to 1
@jeremylim51
Ай бұрын
Normally my physics lesson when it gets so difficult like this during my 3rd year uni… i just answer either 0 or 1
@sohamkale262
Ай бұрын
bro messed up so hard, idk how he managed to not freak out
@Barc112
10 ай бұрын
BPRP: "... and have a look... and this right here... and.. and have a look...😢😢" Me: Oh no, he's melting 😭😭😭
@laughingtime7618
8 ай бұрын
Here answer would be 1 Explanation when theta is very small, perpendicular and hypotenuse are equal and sin theta could be written as theta,so theta divided by theta is 1.
@alphabeta2589
2 жыл бұрын
L'Hospital rule be like: *Am I a joke here?*
@anonimousweb4833
Жыл бұрын
He is an MIT teacher who discourages people to use L'Hospital rule while proving this limit
@sohamacharya171
11 ай бұрын
Lhopital here would give you a circular proof. This limit is best evaluated by the sandwich theorem
@MathProdigy-qg5gx
3 ай бұрын
@@sohamacharya171the squeeze theorem?
@namansabhagani
3 ай бұрын
@@MathProdigy-qg5gx yep, both are the same thing
@bhubaneswarsingha6607
9 ай бұрын
Man you should leave some link so that i can watch part 2 immediately without wasting my time searching for part 2
@marczxp8088
Жыл бұрын
that slight of hand tho🔥
@bijipeter1471
6 ай бұрын
Thank you, sir
@Subham-Kun
9 ай бұрын
Bro be like : I know that I have to use the L'Hôpital rule, but I cannot
@marsahere2621
2 жыл бұрын
I’m really proud of myself right now, that’s how I do math! Finally a common factor with a genius
@austinmuse6029
9 ай бұрын
His brain did the thing Desmos does when you ask it to do something near impossible, just "Um, yeah, this looks right, oh you're still typing? Um, lag."
@Ra-Zorant
Жыл бұрын
We gonna talk about how smoothly he switched between the two markers😂
@husklyman
2 жыл бұрын
I'm still waiting for the ONE moment when this limit is solved
@twens11
2 жыл бұрын
It's actually solved
@husklyman
2 жыл бұрын
@@twens11 if you know the solution, take r/woooosh
@actionoverloaded887
Жыл бұрын
@@husklyman 1
@louisrobitaille5810
Жыл бұрын
[Sin(x)]/x = 1.
@ANASzGAMEOVER
Жыл бұрын
Love the way you answered it
@Rookie1706
Ай бұрын
How do you do it 🤔 I had a similar question on my test just a few days ago and being new to limits having not even learned limits approaching infinity until yesterday, I was confused how to do it, everyone guessed including me, but I did attempt the question. The question I believe was limx->0(sin(x)/x).
@rushdhammalikck
9 ай бұрын
Bro sandwich theorem 😂
@dorian4387
2 жыл бұрын
MFW you try your best to avoid circular reasoning
@gamingzeraora443
9 ай бұрын
Calculus is intuitive. Ramanujan sir made number theory more unintuitive than light at a subatomic level 🔥
@kkpb5916
Жыл бұрын
why isnt it just zero using the formula lim x->infinity sinx/x ? If you write lim x->infinity x sin(1/x) as lim x->infinity (sin1/x)/1/x as he wroted it, you will get the formula . Like you will have the same thing up and above, with x->infinity, like the formula. Someone pls help
@urnightmare1199
11 ай бұрын
Take derivative of both numerator and dominator , you will end up with cos(0) = 1
@brianding8542
2 жыл бұрын
I saw this limit immediately and my brain immediately turned it into 0/0
@DebarghyaKundu-d8u
Жыл бұрын
x is approaching to ♾️ and sin function gives a dancing value in between [-1,1] . so infinite multiplied with finite value lying in between [-1,1] will be infinite
@davep8221
11 ай бұрын
The rest is left as an exercise for the student.
@blank4502
11 ай бұрын
Bro hit his limit on RAM
@elirol4628
11 ай бұрын
It becomes a standard limit which approaches 1
@Paiadakine
6 ай бұрын
I will never use math like this.
@jormdeworm
4 күн бұрын
I literally got this exact question a few days ago
@similosihlegqalisisa8211
11 ай бұрын
Who else is like that when explaining to others? Me I relate😂
@sunsetflory
2 жыл бұрын
as soon as i've seen the last line i screamed "OOONE". so glad that i still remember fundamental limits after 4 years when i didn't use them
@hardiksharma9498
Жыл бұрын
1st ques of limits class😂😂
@akshaysriram8559
2 жыл бұрын
Unexpected ending😆
@KingGisInDaHouse
Жыл бұрын
When you use this though you technically have to claim it’s zero from the right hand side.
@7ymke
3 ай бұрын
Fun fact due to the Taylor expansion of sin x*sin(n/x) as x goes to infinity is always equal to n
@thwinaung2107
2 жыл бұрын
I'm not an expert but sin(1/x) where x tends to infinity already approaches sin(0) which approaches 0. So 'x * sin(1/x)' is inf*0 which approaches 0. Isn't it?
It’s 1. Every first semester calculus student knows the limit as x->0 (sin x / x) =1
@mrincredible4381
Жыл бұрын
Math is so difficult even he got nervous😢
@ranaabdullahranaabdullah3296
2 жыл бұрын
Sir I do same fault with same question in precalculus course I also said to ma'am is that if we take x to in division then we apply squeeze theorem. Them ma'am make me correct
@Sai-su4yf
5 ай бұрын
His lim--> L'hospitals rule
@Samir-zb3xk
11 ай бұрын
You can do a geometric proof to solve lim x-->0 of sin(x)/x in case anyone is wondering. (You can also do L'Hôpital's rule of course but we're going to ignore that for the sake of the proof)
@Muertee00
6 ай бұрын
BRO GLITCHED 😂😂😂
@Purvesxh
Жыл бұрын
Real solution is here: Sin 1/x oscillates between -1 and 1 So, lim x→0 (x sin1/x) = 0 x [-1,1] =0
@randomuser1010
Жыл бұрын
This was exactly my thought, but according to this video it’s 1, im gonna have to ask my professor i guess
@Rando2101
Ай бұрын
Yes, but the limit isn't x->0
@mianumer2712
Жыл бұрын
I don't know what is math even at the age of 18.5😅😅😅
@MrPlaiedes
2 жыл бұрын
He panicked at the end when he tried to use the Lambert W fn...
@r1ain.602
2 жыл бұрын
Alternatively we can use pinch theorem to find the value of limit as 1
@AlekVen
2 жыл бұрын
I'm a physicist, this limit is 1 cuz sin(x) = x
@visiontravels4569
Жыл бұрын
For a very small angle , bro . Don't apply that every time
@cybertar
10 ай бұрын
For a small angle yes
@Rando2101
5 ай бұрын
Meanwhile engineers:
@Rando2101
Ай бұрын
The final limit is 1 Proof using geometry.
@LukeDurham03
18 күн бұрын
Some say he’s still trying to make it work some how
@vrajdesai120
8 ай бұрын
1 is answer bcoz limø->0 sinø/ø=1 where ø=x
@lapicethelilsusboy491
2 жыл бұрын
I can solve this one without the L'Hôpital's Rule
@np-gu9kg
2 жыл бұрын
Sinx = x -(x^3)/3!+(x^5)/5!.….....(Taylor series) Similarly, substitute 1/x in place of x Then take the 1/x common from numerator and denominator as well And you will be left with higher powers of (1/x) which will be zero..And the final answer will be 1
@JiminatorPV
2 жыл бұрын
I think the point is that he does not want to use calculus, if he wanted to he could have usted l'hopital rule.
@Rando2101
Ай бұрын
The point is not to take the derivative of sin(x) to find the limit.
@sairajmhatre5097
2 жыл бұрын
* CUTS THETA IN ANXIETY*
@vegan__girl
2 жыл бұрын
I think the limit would be equal to 1, after the substitution.
@psych0-shorts
11 ай бұрын
SQUEEZE THEOREMMMM
@Alness5
9 ай бұрын
Just use the maclaurin expansion lmao
@Rando2101
Ай бұрын
Maclaurin expansion uses the derivative of sin(x), which is just as bad as using l'Hôpital
@zandertaljaard5522
Жыл бұрын
Perhaps you can use the Squeeze Theorem identity, namely (sinx)/x as x approaches 0 equals 1.
@abubakarumar4713
Жыл бұрын
O my God, tutor has seized
@josephstalin4202
2 жыл бұрын
How do you guys figure out what is what and what to do..
@shahenazmalek5882
8 ай бұрын
That's why it is famous 😂
@sanjaysownworld2744
Жыл бұрын
It's 1 from sandwich theorem
@rushilpatel7418
2 жыл бұрын
this makes sense if you use maclaurin series for sin x and take reciprocal of everything (you get sinx = x and sin1/x is 1/x) and x(1/x) is 1. pretty sure this is very hand wavey but it makes intuitive sense. it’s funny how maclaurin is x approaches zero but this is for x approaches infinity
@krish-502
2 жыл бұрын
Use expansion of sine.
@Bluewaters701
2 жыл бұрын
Heyyyyyyy something similar came today in my engg exams today , wow
@sayantanmazumdar9371
2 жыл бұрын
Uh sin(x) = x for small x 1/x is small for big x thus sin(1/x) as x approaches infinity is 1/x so x*1/x x cancels out leaving 1
@arifyesehehehehhewahahahah3445
9 ай бұрын
when x is close to 0, sin x ≈ x ≈ tan x; cosx ≈ (2-x²)/2 ≈ 1.
@Riv70077
Жыл бұрын
Man didn’t want to differentiate the numerator and denominator to use the l hospitals rule
@darrenhepperle4854
14 күн бұрын
He confused himself. Never try going fast when doing advanced math. Slow and steady. Maintain focus.
@MariyappanJ-v5s
8 ай бұрын
And answer is 1 cos sin(x)/x=1
@MexicanDre
2 жыл бұрын
Oh my God dude it's been so long since i seen your videos you have an epic goatee/beard!!!
@Pain544
2 жыл бұрын
He's the type of guy who's like- ok what if you put this here? WAIT GOOD QUESTION THO...
@amritraj7640
4 ай бұрын
I mean, this a case of standard Trigonometric limits. Lt x-->0 sinx/x =1. Or if you use L Hospital's rule, diffrentiate numerator and denominator wrt theta, We will get cos(theta)= cos(0)=1. So both ways the answer is "1". There is nothing so hard about this problem.
@DaMeowster
2 жыл бұрын
Isn't it zero from the right? I mean obviously because the limit exists both halves are equal but it deserves explanation
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