He’s trying so hard to not use L’Hospital’s rule😄

@bprpfast

2 жыл бұрын

😂

@baghirovali2361

2 жыл бұрын

Yeey, exactly🤣🤣

@danielandresariaslopez

2 жыл бұрын

Yes ❤

@spacerocks9740

2 жыл бұрын

This is easy without l hospital

@user-lk1fw1lp8b

2 жыл бұрын

@@spacerocks9740 easiER, I would say

"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

Guys, I think he finally reached his lim

@Ktm2191

9 ай бұрын

His limx→ ∞

@onyemasuccess3411

9 ай бұрын

😂😂😂

@calculus988

7 ай бұрын

LOL

@purplrshadowyay

7 ай бұрын

That was good

@bradathebread

5 ай бұрын

Limit of bprp approaching insanity

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

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

8 ай бұрын

​@@AK-vb9ksNo it doesn't. You need to know the derivative of sinx to do that.

This what it is really like teaching math because you know what to do next, but it would involve future topics.

This is what happens when you upload multiple videos everyday. He's finally snapped.

@bprpfast

2 жыл бұрын

😂 😂

@agooddoctorfan651

Жыл бұрын

😂😂

@benshapiro8506

10 ай бұрын

mahn, no 1 is perfect.

@Kidnapper656

3 ай бұрын

​@@benshapiro8506 except me ofc

As an engineer, sin(1/x) = 1/x so x*1/x =1 lim 1 = 1

@gdtargetvn2418

Ай бұрын

☠ I hate how valid it is

@babarb9234

Ай бұрын

😂😂​@@gdtargetvn2418

@TeamGCS

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_

8 күн 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

For those who want a trigonometic solution: 1 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

4 ай бұрын

Technically, it's ≤ and ≥. Or =, if you're a programmer.

@srr9281

Ай бұрын

Wondering why or how your first line of math is derived? The rest is just math grammar after that.

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.

Pay attention to the sum being solved - ❌ pay attention to his had when he writes with two markers at a time - ✔️

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

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"

Wait. He’s still trying to approach 0 without putting 0. 😂😂😂

Me using Desmos and scrolling really far to the right: **I am 4 parallel universes ahead of you**

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

9 ай бұрын

Prove the diferrentation of sin x is cos x without using this result.

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.

The way he switches pen is so amazing 🥰

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.

I’m really proud of myself right now, that’s how I do math! Finally a common factor with a genius

How i evaluate it: Sin(x)=x for small values of x and this "x" is 1/x which approaches 0, so this is true. Because of this, we can write this as x times 1/x, which is x/x, which is 1 when approaching infinity, or anything for that matter. I havent taken a calculus class and dont know if this is right.

through "sandwich criterion", this evaluates to 1

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

Finally isn’t sin theta over theta = 1? I’m new to calculus why did he seem to not like to do that😅

@user-jd6kc4of3s

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

@Princesstherine

2 жыл бұрын

kzread.info/dron/Qd9lRbOu1eSpWNHWkVu7FA.html

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

11 ай бұрын

Love the way you answered it

Perhaps you can use the Squeeze Theorem identity, namely (sinx)/x as x approaches 0 equals 1.

We still love you, Chow.

@victorvictory8470

2 жыл бұрын

Way to go!

Alternatively we can use pinch theorem to find the value of limit as 1

Your facial expressions when you realized something was wrong made me laugh 😂😂

L'Hospital rule be like: *Am I a joke here?*

@anonimousweb4833

11 ай бұрын

He is an MIT teacher who discourages people to use L'Hospital rule while proving this limit

@sohamacharya171

9 ай бұрын

Lhopital here would give you a circular proof. This limit is best evaluated by the sandwich theorem

@MathProdigy-qg5gx

2 ай бұрын

@@sohamacharya171the squeeze theorem?

@namansabhagani

Ай бұрын

@@MathProdigy-qg5gx yep, both are the same thing

Man the marker switching is effortless, goddamn

U gotta do the OG proof using geometry 😅

😂😂yoo what happened why are you suddenly chocking 😂😂😂 damn you good really good 😂😂😂

Did anyone notice how well he changes the pen between his fingers, damnnn

You can just divide both numerator and denominator by 1/x

I’m more impressed by his dual use of the markers!

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.

And it is 1, pretty simple ah ha

Excellent video wonderful 😊😊😊😊😊❤❤❤❤

Ptdrrrr 🤣🤣🤣 « and…and…and have a look 👀 »

I appreciate his transparency. He uploaded the short anyway Everybody fumbles sometimes!

@mrjnutube

7 ай бұрын

Nothing like that. Believe me, he manipulated the function XSin(X) into Sin (θ)/θ deliberately in order to introduce the L'Hôpital's rule.

Since theta tends to 0, can't we use the small angle approximation? Sin theta~theta, so sin theta/theta tends to 1, which is the limit of the function.

Unexpected ending😆

My engineering brain just go: sin(1/x) becomes 1/x so the limit goes to 1

his brain exploded on that one

You can use the theorem where the limit of the area of a certain triangle is smaller than sin(x) /x and the area of an arc is bigger than sin(x) /x but they both tend to 1. Hence so does sin(x) /x

It's an inderminant case if we Directly put limit then it gives us 0 * (a number between -1,1) which is domain of sin theta hence 0

German here. I have learned that you guys pronounce "lim" like "limit" while we in GER do not use "limit" but "limes" instead. "Limes" is latin and comes from the border fence/wall between the roman empire and the germanic region in ancient times. Is there a special reason that you say "limit"? What can I learn here?

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

Replace 1/x by logx so final expression is Xsin(logx)

Love the way you just change the markers

His marker switch was ultimate 🤯🤯

It should actually be the limit as theta→0 from the right side only. If you have it as theta→0 you're actually evaluating the original limit for both x→-∞ and x→∞ and hoping they're equal.

Fun fact : plug in a huge value as x(in degrees) in the limit and you'll get pi

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

10 ай бұрын

This was exactly my thought, but according to this video it’s 1, im gonna have to ask my professor i guess

Famous but very easiest limit question

Finally he explaining 😂

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."

BPRP: "... and have a look... and this right here... and.. and have a look...😢😢" Me: Oh no, he's melting 😭😭😭

Just use the maclaurin expansion lmao

All the engineers laughing their ass off.

this joke is like 100 levels up there

Oh my God dude it's been so long since i seen your videos you have an epic goatee/beard!!!

In this question sin1/x is approaching to zero and 1/x is also approaching to zero , if take ratio of sinx and it's tangent passes through origin Y=x (where x is tending to zero) then the ration should be one .

Man you should leave some link so that i can watch part 2 immediately without wasting my time searching for part 2

We gonna talk about how smoothly he switched between the two markers😂

Bro got kirito's dual wielding skill

I will never use math like this.

Calculus is intuitive. Ramanujan sir made number theory more unintuitive than light at a subatomic level 🔥

"Da Hospital wants 1!" - Cheerios

😂😂😂😂😂😂😂😂😂😂 I thoroughly enjoyed this!

Take derivative of both numerator and dominator , you will end up with cos(0) = 1

man got pre calc flashbacks when he saw sintheta/theta

Salute that you didn't edited

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

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

He just got stuck at last 😂😂😂😂

He is talking!! He is talking!! HE IS TAAAAALKING!

when x is close to 0, sin x ≈ x ≈ tan x; cosx ≈ (2-x²)/2 ≈ 1.

Bro be like : I know that I have to use the L'Hôpital rule, but I cannot

When theta approaches zero, sin theta also approaches zero, so that limits will give 1.

Bro sandwich theorem 😂

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?

I'm confused Squeeze theorem tells me this limit should be between -∞ and ∞. That ain't very useful. Any way for me to selve this using squeeze theorem?

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)

I am a firm believer in the the squeeze theorem

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

9 ай бұрын

For a small angle yes

@Rando2101

4 ай бұрын

Meanwhile engineers:

WHY WHY WHY AM I SEEING THIS AFTER MY CALC EXAM

@atasarac_

8 күн бұрын

I crossed it out

It approaches to 1 pretty easy to visualise from graph of this function

it's simple but my eye tell me I can't do it

Can I use squeeze theorem for evaluation of this limit

Thank you, sir

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

This is not wrong but very clever. Only real mathematicians know this trick😊

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?

The skill how he changes marker pen;)

It is the definition of derivative of sin evaluated at 0, which is cos(0)=1.

Man didn’t want to differentiate the numerator and denominator to use the l hospitals rule

Fun fact due to the Taylor expansion of sin x*sin(n/x) as x goes to infinity is always equal to n

Isn't this the sinc(x) function which has a limit of 1 as x approaches 0?