the most controversial limit in calculus 1

[@pnoceyyecnon3204]

Come on its clearly approaching 1. Using the fundamental theorem of Engineering, sin theta is simply equal to theta. Therefore, theta / theta = 1.

[@TTFMjock]

I see what you did there

[@vano__]

but its a fact thats comes from taking first terms from taylor series for sin, which requires differentiating sin, otherwise I dont think theres a reason for thinking sinx is approx x for small values

[@vano__]

by the way, he say it in the video..

[@sebastiansalazar9735]

Te ves muy fresco 😎👌

[@Tomaplen]

yeah, and limit just means evaluate at the point

[@beatoriche7301]

It depends on your definition of the sine function; this argument only works if you define the sine function through angles and the unit circle. However, in order to even rigorously define what an angle is, you have to use either the complex exponential, measure and integration theory, or another tool in this lane; from a mathematical standpoint, it is not at all clear what it means for a circular segment to have some length. For the sake of simplicity, most analysis textbooks define sine and cosine via the exponential function or via their Taylor series expansions; in both cases, the derivative of sine being cosine follows very easily from differentiating the power series in question term by term, which is alright because it has an infinite radius of convergence (and power series can always be differentiated term by term within their open disk of convergence).

[@liamhopson2913]

Exactly. Couldn't have put it better myself 👍🏻

[@youtubeviolatedme7123]

This guy probably has more students cheat off of him than girls cheat on him... and that's okay bc I'm both a virgin and an idiot edit: or girl. or other.

[@ginosuinoilporcoinvasivo8216]

Could you provide additional information about this topic? I'm interested

[@beatoriche7301]

@@ginosuinoilporcoinvasivo8216 If I may ask, what exactly would you like to know more about? The complex exponential? Measure theory? Defining sine and cosine via their power series expansions?

[@liamhopson2913]

@@ginosuinoilporcoinvasivo8216 It all depends on your definitions. There is lots of areas (particularly in pure mathematics such as analysis) where you need to be incredibly precise with your definitions. For example we all know that sin(theta) can be considered as "opposite/hypotenuse" or as a power series, and we'd hope that these two notions are equivalent (but starting only from the power series, it isn't obvious that it has anything to do with triangles and equally starting from the triangle it's not obvious it would even have a power series!). This is often the case, there are different definitions of integration that each apply to a certain class of functions, and we hope that when we consider two different definitions of integration considered on the same function for which the definitions make sense, we hope that they would agree (which they infact do). This is the idea behind Lebesgue integration (involving measures) as a more general form of Riemann integration (the typical "area under the curve" approach). That got very long winded and philosophical, but I hope it made sense! Loads more examples come up in Complex analysis when you can start from different definitions and propositions and still unlock the same theorems. Hope this helps

[@__skillz]

I just used this as an example of when to use L'Hopital's Rule when tutoring someone yesterday 😳

[@anmoljhamb9190]

F

[@Jordan-zk2wd]

@@bugraaydn3602 Think about it this way: 64/16 = 4/1. Just cancel the 6 on the top with the 6 on the bottom and you gwt your answer. Is this correct? Well, no. You get thw right answer, the method is easy to describe, but that doesn't mean applying the method here is correct just because it gives a correct answer. Similarly it is with lim as x->0 of sin(x)/x, if you use L'Hopital's rule to get it is the wrong method because you can't understand how to get the derivative of sin(x) without understanding what the answer to limit as x->0 of sin(x)/x (try and derive it using the limit definition of a derivative and prove it yourself). This means when you use L'Hopital's rule here, you are assuming the answer to the limit before you make it in a subtle way, and if you assume the answer before starting you haven't really worked it out. For other limits you apply L'Hopital's rule to, you don't tend to run into similar issues and it essentially is just due to the properties of the sine function in particular.

[@michaelmulla6178]

@@bugraaydn3602 In order to find that its cos you need to know what he is trying to find out

[@teachforjoyresearchphysics]

M a physics guy but love this fella

[@realbignoob1886]

*oh frick*

[@yourmathtutorvids]

I will never not read l’hopital as “le hospital 🚑🏥🥼” 😂

[@yourmathtutorvids]

@hawkturkey 🙈🙈🙈

[@blackpenredpen]

You will never “not” read lollll

[@muhammadhussainsarhandi9928]

@@blackpenredpen hahaha never not, (two negatives together make one positive) Btw, excellent video respected Sir,

[@ianmoseley9910]

Muhammad Hussain Sarhandi Several languages commonly use the double negative as an emphasis, rather than the apparent literal meaning

[@PersonalMiner123]

this

[@osmeridium]

Yeah my calculus teacher couldn't explain why so thank you

[@derdualeraum356]

I learned in the Analysis I class that the trigonometric functions are defined by: sinx:= x/1! - x^3/3! + x^5/5! - ... cosx:= 1 - x^2/2! + x^4/4! - .... So d/dx(sinx)=cosx. So I can use LH rule...

[@aram9167]

These definitions assume that the derivatives of sinx and cosx are cosx and -sinx respectively.

[@Kerlyos_]

@@aram9167 No. We are not calculating the Taylor expensions here. We're only taking these two series as definitions for sin and cos Just as the exponential function is often defined by it's series as well. It then follows that the derivative of sin is cos It's only the power rule -- Note that you can also define sin and cos with the exponential It also follows that the derivative of sin is cos

[@ronaldjensen2948]

These functions can be shown to come from the Taylor series, can they be derived another way? We can show lim n->inf (1-1/n)^2 gives the Taylor series for e....

[@zapzya]

The problem is that if you define them this way, you can use LH rule, but you don't know if they match up with circle angles and such. You need to prove a lot of the inherent properties exist as well, like periodicity and convergence (perhaps not too hard, but necessary). Even if they share some of the properties, you still need to show that they are equivalent to the sin and cos functions defined by using circle angles, if you want to use them for that type of analysis. It's a bit of a trade off. You have an easier time proving the limit sin(h)/h, but you need to show it's a well behaved function and that it matches up with circles afterwards.

[@selfification]

@@zapzya See the top comment above. It's hard to rigorously define angles and lengths for curves. In particular, here's a fun question. What's pi? The ratio of the radius of the circumference of a circle and its diameter? What's a circle? You basically took a broom and swept a transcendental series definition underpinning sine and cosine into a number called pi and just decided to symbolically manipulate it. You don't get periodicity, convergence or anything else automatic. You just a priori assumed it would work just because you could draw a circle. Does that work on any underlying field other than reals? That's why most analysis classes start with a series definition of exp(x). It's not the Taylor series for e^x. It's the definition of the exp operator. It works for matrices. It even works for the exponentiation of operators (when you start fiddling with Lie groups and algebras). You later show that the Taylor series is unique and immediately conclude that this definition of exp must coincide with its Taylor series.

[@YellowBunny]

At my university sin(x), alongside exp(x) and cos(x), was defined as the (Taylor) series before the concept of Taylor series was introduced for more general functions.

[@areyousrsrightneowbro]

just use the fundamental theorem of engineering: sin(theta) = theta for small angles of theta = theta/theta = 1

[@Prxwler]

6:55 I'm glad you noticed that after "don't" you can not have "neither" :D Always improving :)

[@blackpenredpen]

Thanks 😊

[@mustafamalik4211]

Lol when I saw cosh, I was confused for a solid 5 seconds trying to figure out where the hyperbolic functions came from.

[@p.abhijitsuhas26]

Me too lmao

[@smashingstuff2454]

Cosh x means hyperbolic cos. You also can have hyperbolic Cosh Sinh Tanh Cach Sech Coth These are hyperbolic trig functions

[@Tyler-cm6vk]

As a high school student, I only know so little, but I really love using the squeeze theorem to prove it approaches 1. Even though it takes more time, it's much more satisfying for me.

[@shivangsrivastava6024]

Exactly that level of satisfaction after solving a complex limit 🤩

[@dariog3738]

real physicist like me just uses the small angle approximation and writes θ/θ guess i'm just different 🤷🏻‍♂️

[@idjles]

That is exactly Using Taylor’sexpansion . I think this whole video is wrong.

[@tomasbeltran04050]

@@idjles according to this: ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/unit-5-exploring-the-infinite/part-b-taylor-series/session-99-taylors-series-continued/MIT18_01SCF10_Ses99c.pdf you have to know how to derivate sin(x) before going Taylor. I think your whole comment is wrong

[@nasekiller]

not sure if you are joking, but this approximation only works because we know that the limit of sin t / t is 1

[@tomasbeltran04050]

@@nasekiller demonstrate it without l'Hôpital's rule

[@nasekiller]

@@tomasbeltran04050 how you show it, is completely irrelevant. But you can only use the Approximation sin t = t for small t, because the limit of sin t / t is 1

[@MaxxTosh]

I feel like the table of values method might be more egregious, because it’s not an actual proof and you don’t know the true limit

[@ronaldjensen2948]

and you're left with the question of how the table values for sin(theta) were created in the first place? Taylor expansion or its cousins?

[@roberthrusecky9439]

I think this is only circular reasoning if you use those particular proofs of dsinx/dx=cosx/taylor series of sinx as "the proof". For example, I could derive the taylor series by using the series for e^(ix) and Euler's formula from complex analysis. This does not require previous knowledge of the limit.

[@Apollorion]

But how has Euler's formula been proven?

[@Apollorion]

@ゴゴ Joji Joestar ゴゴ That's an ambitious claim which would be nice to be true. Can you quote such a proof or provide us a link so that we can verify?

[@christianalbertjahns2577]

Imagine finding more complicated proof of d(sin x)/dx = cos x that doesn't rely on knowing the value of lim x -> 0 for (sin x)/x just to make sure using L'Hopital rule for lim x -> 0 for (sin x)/x won't cause the trouble of circular reasoning

[@manik1477]

you can’t stop me!! MUA HUAHA

[@blackpenredpen]

😂

[@blackpenredpen]

😂 😂

[@blackpenredpen]

😂 😂 😂

[@jhanzaibhumayun5782]

Judging by his replies, I'm pretty sure he is mad.

[@djvalentedochp]

@@jhanzaibhumayun5782 pretty weird replies

[@iansamir18]

If we are working with circles we can find that the derivative of a circle e^ix = cos x + i sin x is just i*e^ix = i cos x - sin x, therefore the imaginary parts show us that (sin x)’ = cos x without limits.

[@moshadj]

It depends on how you define sin(x). The definition given in Rudin's Principles of Mathematical Analysis of sin is its power series expansion. Then you can define pi as the smallest positive value for which sin(x) = 0. And you can describe sin's geometry directly from the Fundamental theorem of Trigonometry sin^2 + cos^2 = 1

[@neinicke9086]

i love how this video turns up the day we go over l’hopital’s rule. and our first example was sin(x)/x 😂

[@TheCrashbandicoot36]

so cool, didn't know that. And now to find the geometry demonstartion of the limit (thanks for adding it on the description)

[@zdrastvutye]

there is a powersum of sin(x) =x-x^3/6+x^5/120... divide it by x will result in 1-x^2/6+x^4/120...

[@lagrangiankid378]

It's not necessarily circular reasoning since you can prove d/dx(sin(x))=cos(x) with methods different than the limit definition of derivative.

[@justacutepotato2945]

What other method we can use?

[@thomasrl194]

Geometric

[@martinepstein9826]

"methods different from the limit definition" To be clear, the limit definition is the only definition. Every single method for evaluating derivatives relies on what a "derivative" is.

[@lagrangiankid378]

@@justacutepotato2945 you can define the sine function as a taylor series and then calculate the derivative using the power rule, or you can also prove that the rate of change of the sine function is equal to the cosine using geometry, or you can calculate the derivative numerically substituting values, or you can also define the sine as an inverse function of an integral, or well, you can also define sine as the function whose derivative is cosine. And those are only a few examples: no need to know the limit, no circular reasoning

[@justacutepotato2945]

@@lagrangiankid378 won't there be like only one conventional single for the definition of sine?

[@AbouTaim-Lille]

Instead of using the Hospital rule u can use the Taylor series to calculate the limit. Sin (x )/x = (1/1! .x¹ - 1/3℅.x³ + 1/5! x^5 - ....) /x = 1/1! - 1/3!x²+ 1/5! x⁴ - .... And here all terms except the first term tend to zero as they have x. Abd the first one is 1.

[@landsgevaer]

The fact you can find a circular argument for a conclusion does not make the conclusion invalid, it makes only this particular argument invalid.

[@Jordan-zk2wd]

Yes, and that is exactly why it is incorrect to use L'Hopital's rule here. Maths questions are not about simply giving an answer, but demonstrating an understanding, show your work etc.

[@landsgevaer]

@@Jordan-zk2wd So where do you end? You would prefer to start with Hilbert's Principia to arrive at 1+1=2 on page 160 or so, and take it from there? You would hardly get anywhere in your entire lifetime. Here, the validity of l'Hopital can be shown by applying the definitions of a derivative (in terms of a limit), but the point is that that would work with any ratio of two continuous and differentiable functions that approach zero (whether sin(x)/x or x^2/tanh(x) or anything similar), and that IS De l'Hopital's rule. Not allowing De l'Hopital here is akin not allowing functions of real numbers until you show that you understand those properly too. It is perfectly fine to build on proved work of others. If we can see further it is because we stand on the shoulders of giants, and such.

[@numbers93]

@@landsgevaer the reason for not using l’hopitals here is that it involves computing the derivative, whose computation in turn relies on the limit being well define. This is why the logic is circular. It does not mean we have to go all the way back to the axioms to justify anything; we just can’t have something being true because it is true because it is true because it is true because it is true ad infinitum

[@landsgevaer]

@@numbers93 But wait, these are different limits!? If it had said sin(x)/x^2 then the derivatives of the numerator and denominator do exist, but the limit of the ratio doesn't; contrariwise, if it had said |x|/(|x|+1), then the derivatives don't exist but the limit of the ratio does. These are different things. Both are limits, but entirely different limits. So how can it be circular!?

[@numbers93]

@@landsgevaer what are you on about? what do the existence of the other two limits you are introducing here have to do with the circular logic illustrated by the video regarding the proof of the limit of sin(x)/x?

[@Utesfan100]

It is only circular if we use l'hopital's rule or Taylor series to find d/dx sin x itself. Once we have found that d/dx sin x = cos x by some other means we can freely use this in future l'hopital's rule or Taylor series calculations.

[@nasekiller]

i dont know how it was for you, but when i had calc 1, we actually defined sin and cos via a power series. if you do that, you can easily see that sin' = cos by differentiating each summond individually.

[@Taterzz]

i'll just use the small angle approximation- "No Taylor Series!"

[@vivic2939]

proof by graph: you can see its close by seeing the graph Q.E.D

[@johnnath4137]

The small angle approximation for sine comes from the Taylor series, which explains how we know it is a first order approximation. That for the cosine derives from the same source, which shows it is a second order approximation, which in turn explains why the cosine approximation for small angles is so good.

[@GAPIntoTheGame]

@@johnnath4137 that was the joke. He said “I’m gonna use small angle approximation but not a Taylor series” as a joke because some people would probably say that unironically, without realizing the small angle approx is literally an application of the Taylor series.

[@Beny123]

@@johnnath4137 how do you know that . What if Taylor series started from the understanding of the small angle concept in the first place ?

[@johnnath4137]

@@Beny123 I see where you are coming from. Historically, the trig functions emerged from geometry, and the small angle approximation was known well before the advent of calculus and the Taylor series. But the geometrical ‘proof’ of the result is not rigorous and is not good enough for a modern proof. The root of the trouble is that the simple geometrical definitions of the trig functions are themselves wanting in the requisite rigour. Possible ways of supplying this rigour is to define these functions in terms of integrals (as has been done by G H Hardy in his seminal A Course of Pure Mathematics) or in terms of power series (after the manner of Karl Weierstrass). Unfortunately, this means that a rigorous treatment of the trig functions has to be delayed until one is well into a first course of real analysis. In this approach, the limit of (sinx)/x as x → 0 is trivially 1: simply divide the power series for sinx (which you will note constitutes the definition of sinx if we use the Weistrassian approach) by x and substitute x = 0. Of course this all presupposes that the whole theory of the convergence of power series has been previously set up. The integral approach of Hardy is even worse: here the theory of Riemann integrability needs first to be developed. But it is right that school age children should be introduced to the trig functions via geometry and that they should be shown the intuitive plausibility of the small angle approximation. But when it comes to strict proof of results, the trig functions need first to be re-defined analytically.

[@OtherTheDave]

2:22 This is why I write the parentheses. No need to worry about confusion because there is none.

[@blackpenredpen]

I should have done that! It reminds me of the time doing series Sin(n)/ln(n) And you can imagine how bad it would be with out ( )

[@mikebenson8045]

@@blackpenredpen sinn/lnn well obviously the two n's cancel out so its si/l but as we known from Roman numbers L is fifty and i is the imaginary unit so sinn/lnn is i/50s

[@OtherTheDave]

@@blackpenredpen Oh it’s not just you. All my math profs would omit parentheses. It’s not so bad here because “sinh(+)” doesn’t mean anything, but it drove me nuts in classes where we had stuff that should’ve been written “sin(h)*x”, and because people also like to omit the “*”, it’d get written sinhx and you had to just magically know if the prof meant sinh(x), sin(h*x), or sin(h)*x. Heaven help you if there was also a variable “hx”, because there was at _least_ a 50/50 chance that the prof would be in such a habit of omitting the “*” that they’d write “h*x” as “hx” even though they’d already said “hx = h+x” or whatever earlier in the problem.

[@compphysgeek]

and we usually write function arguments in parentheses, e.g f(x). Wonder why people decided they don't have to do it for trigonometric functions. Would it be acceptable to write fx = x² for example?

[@passecompose7484]

I read a paper about education. Conventional method to prove the limit of sin θ / θ, using area properties on a circle, is also a logic circulation, because the area's strict definition needs an integral of sqrt(1-x^2). The problem is, it requires calculus of trigonometric functions. So maybe the most rigorous way is to define sin and cos as their taylor series.

[@SummerFrost23]

Apparent not that rigourous, they avoid proving the series at first. More like setting their own rule. Limit of sinθ /θ has no circular reasoning if you consider the concept of convexity to set up the straight/arc length inequality and obtain the limit as 1 using squeeze theorem.

[@davannaleah]

One simple geometric proof is to show, on a circle, as the angle (in radians) gets smaller and smaller, the sin of theta approaches theta itself, so the limit expression is just theta/theta, cancelling out becomes 1.

[@The1RandomFool]

In complex analysis I actually use L'Hospital's rule for lim (z-k)*cot(pi*z) as z approaches an integer k so that the cosine in the numerator cancels with the cosine (derivative of sine) in the denominator. This limit is often used in the calculation of difficult series. Another reason to use L'Hospital's rule is for something like lim (z - exp(pi/4*i)) / (z^4 + 1) as z approaches exp(pi/4*i). The reason is that it's far easier to enter a single value into 4*z^3; factoring the denominator and cancelling makes it nasty after taking the limit.

[@philipschloesser]

Joke's on you. I'm actually not using the Taylor series of the sine. Just the power series we used to define it

[@kay-lideneef28]

For the derivative of sin with the lim. U can use also the formula of simpson to do the top of the fraction.

[@cardflopper3307]

This is a very eye opening video... I always knew Lim x-->0, sinx/x was a tricky limit but now it's clear why

[@dlz5709]

For some reason, l’Hopital’s Rule isn’t taught in France. We only use the limit of the rate of change (not sure that’s the right name).

[@blackpenredpen]

Wait why?! Isn’t Guillaume de l’Hôpital French?

[@Ben-wv7ht]

Hmmm , I can tell it’s told ! But as far as I know, our teachers want us to use limited development (wich is actually the same)

[@dlz5709]

@@Ben-wv7ht Actually I spoke without knowing. I’m only in first year of prépa, and I believe we’ll only learn about limited development at the end of the first semester (outside of physics/engineering). Still, I skimmed through the limited development chapter and it seems like they never present it as L’Hopital Rule.

[@yannld9524]

@@blackpenredpen Yes he is, and indeed L'hospital was publied by him but it was Johann Bernouilli (a Swiss mathematician) who probably invented this method. In fact in France we prefer use a Taylor approximation to compute such limits, in french we call that a "développement limité". For example we will write sin(x)=x+o(x) with the "little-o" notation, so sin(x)/x=1+o(1) and conclude that the limit is 1. Another example : 2sin(x)-sin(2x)= 2(x-x^3/6 + o(x^3))-(2x-(2x)^3/6+o(x^3)) = x^3+o(x^3)=x^3(1+o(1)) and x-sin(x) = x-(x-x^3/6+o(x^3)) =x^3/6 + o(x^3)=x^3/6 (1+o(1)) so [2sin(x)-sin(2x)]/[x-sin(x)] = 6+ o(1) and thus the limit as x approches 0 is 6. I think we don't learn l'Hospital rule because it is just a particular case of Taylor approximation, and with this tool we don't have to be worry about the conditions over the limits or about g'(a) is not 0.

[@Ben-wv7ht]

@@dlz5709 j’espère que ça ne te dérange pas que je répondes en français Le développement limité se fais quasi toujours lorsque x -> 0 et fais apparaître les premières dérivées Si tu utilises l’hôpital 1 fois tu utilises la première dérivée donc un DL d’ordre 1 2 fois 2 eme dérivée donc DL d’ordre 2 Etc ..

[@edooox8916]

You can dimostrate the limit with the definition of the taylor serie of sin(x). And working with that by multiplying with 1/x which is the denominator

[@parasgovind6271]

I just find it so funny that you begin releasing videos on topics that I begin to learn that very week! Thank you so much!!!

[@Tomaplen]

2:30 cosh(+) (hyperbolic cosine of plus)

[@chinter]

Oh my cosh

[@jackkalver4644]

I’ve come up with 2 proofs that d/dθ sin θ = cos θ that don’t use this limit OR an angle sum identity. Euler probably used one of them.

[@calculus988]

Can you share it with me. I'm curious

[@tttkitu]

So to derive sin t / t you need to assume d sin t = cos t; to derive d sin t you need to assume d sin t / t = 1 That's the "circular" part, so we use what seems to be inductive or "geometric" proof. Not rigorous, but then again inductive reasoning commonly crops up in proofs we consider rigorous.

[@Evilope]

I never considered the circular logic in applying L'Hopitals in situations like this; thanks for sharing!

[@jackkalver4644]

I took the challenge and found TWO ways of proving that d/dx sin x = cos x without using lim x->0 (sin x /x), lim x->0 (cos x /x-1/x), or sin(x+h). One involves a helix, the inverse rule, and the definition of arc length. The other involves the derivative of a parametric function. One day, I’ll be teaching students these proofs, giving them the chance to prove the limit and the angle-sum identities.

[@Engy_Wuck]

how do we know that sin(a+b)=sin(a)cos(b)+cos(a)sin(b)? Doesn't this already require geometric proof (or Eulers Identity)?

[@jotave49]

Can you make the geometric proof for this? That would be awesome to see and understand.

[@davidbrisbane7206]

It's interesting that people say you can't use L'Hopital's rule in such cases, but everytime you do, you get the right answer.

[@compphysgeek]

I agree with the reason you can't use L'Hôpital, but if I define the sine function as a series just like I can define the exponential function as a series, then using Taylor Series should not be a problem since I don't need derivatives of sin. Although having said that I use the series definition of sin, L'Hôpital also becomes a viable option again.

[@h4c_18]

I used the sum to product formula and ended at cos(theta)*(lim h->0 sin(h/2)/(h/2)), similar technique can be used for the cosine function

[@davidbrisbane7206]

Actually, if you _define_ the sine and cosine functions analyically, then their definitions are _completely_ independent of their definition using a unit circle. Then you derive the usual properties of the sine and cosine function based on the above analytical definition of the sine and cosine function.

[@davidbrisbane7206]

I agree. That's how real analysis books define it and then show that all the usual properties of sine and cosine hold.

[@ketomousketo3345]

But if you derivate the top you get cosø, and if you derivate the bottom you get 1 because its the variable. And the limit when ø goes to 0, the cosø=1, and 1/1=1 so it basicly works.

[@bryantwiltrout5492]

As soon as I saw theta as the denominator, I knew what was about to happen. My calculus teacher in college told me if I ever saw that, to do what you said to do at the end. He actually did a great job going through the problem too cause I was confused as hell at first.

[@eminjoy5541]

This is basic differentiation where instead of dy/dx we use dy/d(theta).....so that results in cos(theta)/1 = cos (theta)

[@nanamacapagal8342]

Geometric proof because he asked for it (not a professional at this. I tried): There is a ball attached to a rope on one end. When I spin the rope holding the other end, the velocity of the ball is tangent/perpendicular to the rope. The distance is just x because if i rotate x radians the ball moves x radians as well. But for really small values of x the curviness of the path is negligible and the path of the ball looks practically straight. So sin(x) is really close to x for small values of x. Since our limit is focusing on when θ gets close to 0, we can replace our sin(θ) with θ. θ/θ is 1 so our limit is equal to 1 and we are done.

[@monopton5212]

Use complex exponential to find the derivative

[@iambic-kilometer]

Here's a fun 2nd/3rd semester calculus way, a la parametrization. Suppose we define two functions x(t) and y(t) with the following properties: x(t)^2 + y(t)^2 = 1 (we're constrained to the unit circle); x'(t)^2 + y'(t)^2 = 1 (we're parametrizing by arc length); x(0)=1, y(0)=0, and y'(0) > 0 (we start at the point (1,0) and head in the positive y-direction.) We want to show that y'(t) = x(t) for t near zero. Differentiate x(t)^2 + y(t)^2 = 1 implicitly with respect to t: 2x(t)x'(t) + 2y(t)y'(t) = 0, leading to x(t) x'(t) = - y(t) y'(t). Multiply x'(t)^2 +y'(t)^2 = 1 on both sides by x(t)^2: x(t)^2 x'(t)^2 +x(t)^2 y'(t)^2 = x(t)^2. Replace x(t)^2 x'(t)^2 with y(t)^2 y'(t)^2: y(t)^2 y'(t)^2 + x(t)^2 y'(t)^2 = x(t)^2. This leads to (x(t)^2 + y(t)^2) y'(t)^2 = x(t)^2; use x(t)^2 + y(t)^2 = 1 to arrive at y'(t)^2 =x(t)^2. The initial conditions x(0)=1, y(0)=0, and y'(0) > 0 lead to y'(t) = x(t) for t near zero. There are continuity arguments that preclude transitions from y'(t) = x(t) to y'(t) = -x(t).

[@VivBrodock]

>:3 y=sin(x) ≡ y=x for x values close to 0 -> lim(x->0) sin(x)/x ->lim(x->0) x/x -> 1 anyways you don't have to do a geometrical proof, you can use squeeze theorem on the interval (-pi/2, pi/2) sin(x)≤x≤tan(x) -> 1≤x/sin(x)≤1/cos(x) ->cos(x)≤sin(x)/x≤1 lim(x->0) cos(x) =1 lim(x->0) 1 =1 lim(x->0) sin(x)/x = 1

[@Devp1006]

you can derive the Taylor series for 0 by using multiple angle formulae - no differentiation required.

[@eugeneimbangyorteza]

How about defining sintheta in terms of (e^(i theta) - e^(i theta))/2i ? That will also do it shortly.

[@aram9167]

e^(ix)=cos(x)+isin(x) is derived either from Taylor series or differential equations, so no.

[@eugeneimbangyorteza]

@@aram9167 I thought they were just derived from hyperbolic trigs. I guess i was wrong

[@Kerlyos_]

@@aram9167 Well yes. And no. You can use the series as DEFINITIONS of sin and cos The same goes for the exponential function. Then you can prove that e(ix)=cos(x)+i.sin(x) Without knowing any derivative -- In fact, if you define sin and cos by their series, it immediately follows that the derivative of sin is cos 🤷

[@aram9167]

@@Kerlyos_ The issue here is that bprp defines sine as being the y-coordinate of the unit circle, no? You can define it with power series, sure, but that definition just seems completely arbitrary and nonsensical compared to the geometric definition.

[@brunobautista6316]

@@Kerlyos_ but to define the series expansion you must know the derivative in first place (and probe that cos(x) is actually the derivative) so... no, you can't define sin(x) and cos(x) with its series *if you want to probe the derivative of sin(x)*. Either way, yeah, of course you could do that

[@Bodyknock]

A slightly shorter way to say this is lim x->0 sinx/x = lim (sin(0 - x) - sin(0) )/ x = d/dx sin(0). In other words the limit you want is exactly the derivative of sin at 0. So you can’t just say “I know the derivative at 0 is 1” to calculate that the derivative is 1.

[@amarmerabti8338]

The lim (x>0) of sin(x)/x is the same thing to the lim (x>0) of (sin(x) - sin(0))/(x - 0) In other meaning it's the definition of the derivative of sin(x) when x=0 So we can just derive sin(x) which is cos(x) and calculate cos(0) which is 1 . And that's it.

[@oenrn]

But d/dx sin x = cos x is something you get FROM the limit, so you can't use it to prove the limit itself, that's circular reasoning.

[@alexandersanchez9138]

Of course, you can always use the definition sin(z) := [e^(iz)-e^(-iz)]/2i. Then, you can use L'Hopital's rule without circular reasoning because the derivative of the complex exponential function (defined as a power series) is known.

[@oenrn]

But you need the Taylor series to prove Euler's formula.

[@alexandersanchez9138]

@@oenrn Not necessarily. You can actually use just the theory of differential equations. It's not to hard to show that the function satisfying the IVP f'(x)=if(x), f(0)=1 is e^(ix) = f(x) = cos(x)+isin(x) using two elementary methods (integrating factors, and laplace transform). This proves Euler's theorem without any Taylor series.

[@antonyqueen6512]

It was obvious from the start that that the limit is equal to the derivative of the sine function at teta equal 0, which is equal per definition to: Lim[(sin(x)-sin(0))/(x-0)], x->0. Now if you want to assume you don’t know the derivative of the sine function, the problem can be solved as follows: When x 0 is x And Lim[sin(x)/x], x->0 is equal Lim[x/x], x->0 which is equal 1.

[@tusharkaushalrajput]

sin 18° = −1±√5/4= -φ/2 Exactly why sin18° is related to golden ratio. I don't get why it has relation or it just natural symmetry.

[@georgeb8893]

I think it may have something to do with being able to divide a circle into 5 equal sectors using only compass and straightedge. ( cos 72 = sin 18 if constructed - and you do have there since it only involves a square root, would give it to you 5*72=360). There must be something about the symmetry of the situation that will tell us that the golden mean (I almost wrote golden meme LOL) is happening somewhere there: there is some a and b and c there with c=a+b and b/a=c/b=phi, the golden mean. Now that I think more, the dividing the circle by 5 and the 5 under the radical may be just coincidence. Okay: if you draw a regular Pentagon and then connect with segments those vertices which are not already connected, you will have inscribed in the Pentagon a perfect 5 pointed star. And the ratio of the star segment lengths to the Pentagon segment lengths will be the golden mean!

[@georgeb8893]

@Tushar Rajput I started daydreaming about your question - see the above post.

[@MonkoGames]

If you forget what the derivative of a sin wave is (or cos, -sin, -cos) just draw it, and draw the derivative(the slope) of each point

[@VietTran-du4ip]

If we can show lim (cosx - 1)/x = 0 as x -----> 0 and lim sinx/x = 1 as x ----> 0 independently (without using L' Hopital Rule) in proving d/dx (sinx) = cosx then we have d/dx(sinx) = cos x . So we can still apply the L' Hopital Rule to lim sinx/x to confirm that limit sinx/x = 1 as x --------> 0

[@thekid8784]

this goes under the conditional IF no one knows what the d/dx(sinx) is. of course, we all know what it is. therefore, we still can use l’hopital’s rule. he literally says it: what IF we do not know d/dx(sinx) is cosx? lmaoaoaoaoaoa cmon yo

[@EhsanBasiri0]

I think we can do it with hopital The derivative of sin(∆) is ∆'×cos(∆) we know the ∆ is the number we give , so derivative of it is 1 so the derivative of ∆ is the ∆' = 1 In the end we have cos(∆) when ∆ -> 0 so the cos(0) = 1 Am I wrong pls let me know if I ❤

[@andypan4936]

I think 3Blue1Brown did a geometric proof for the derivative of sin(x) with respect to x, without using the limit definition. With the geometric proof for the x-derivative of sin(x), I am not sure if using L’Hopital’s Rule in this case would still be circular reasoning. kzbin.info/www/bejne/iWHCootqi6-bg7M

[@r3v1v3r4]

We were doing exercises in class today and the teacher just told us about this, 2 hours later i watch a random youtube video from my reccomanded: do not use l'hopital on this limit

[@antoine5571]

Is it possible to say that sin(x)/x = (sin(x)-sin(0))/(x-0) = sin'(0) = cos(0)=1 (when x goes to 0) ?

[@elihowitt4107]

You can prove the derivative is cos in many ways(geometricaly for example), you don't have to evaluate the limit in question

[@idjles]

That doesn’t help, because that formula is how limits are defined!

[@younesabid5481]

At 2:12 you used the sum of two angles formula right!? Well wasn't that derived from the complex definition of sine that we got from Euler's formula? And to get there we would have known the derivative of sine so that we can work out the Taylor series.

[@blackpenredpen]

You could derive the angle addition formula by geometry 😃

[@artsmith1347]

It could be derived that way, but proofs for most high school students do not rely on Euler's formula.

[@younesabid5481]

@@blackpenredpen thanks for the info! I'll check that out

[@stephenbeck7222]

Basically every precalculus/trig book will show the sum angle rule for sine and cosine using basic geometry. The formulas are quicker found using Euler but it’s a good exercise to do it the long way. “The Limit” has more ‘funny business’ than the sum angle formulas.

[@nikoszervo]

Finally, I understand! I have a really good book about Calculus, but I never quite understood this limit.

[@SummerFrost23]

Those give series definition for e^x , sin x or cos x by throwing away geometric proof are like claiming it is "naturally true" but out of nowhere.

[@lutherlessor4029]

Since the limit is equivalent whether the variables in the expression and limiting variable, isn't this just a self referential equation where we can isolate the terms with "limit of sin(var)/var," factor it out, then solve for the desired limit?

[@proguyz78]

Hey guys, does this work? lim x->0 (sinx) = 0, lim x->0 (x) = 0, lim x->0 (sinx) = lim x->0 (x) Divide both sides by lim x-> 0 (x) and get: lim x->0 (sinx/x) = 1

[@finmat95]

Following this logic: lim x->0 (sinx) = 0, lim x->0 (x^2) = 0, lim x->0 (sinx) = lim x->0 (x^2) Divide both sides by lim x-> 0 (x^2) and get: lim x->0 (sinx/(x^2)) = 1

[@amirparsi4165]

It’s odd, since going the wrong way still gets you to the right answer. (I mean cos(0)/1 equals 1)

[@Tomohiko_JPN_1868]

Yes, you will get the right answer via L'hopital. Because in fact it is NOT a wrong way. But it is Not Logical way at all, as the video says.

[@erasmorecami2320]

It Is enough to think of the geometrical meaning of the numerator and denominator ti get the immediate result! Math is the art of getting results without unnecessary calculations...

[@nanaagyasijnr8615]

Who else noticed the Pokemon tank he's holding

[@CrownedFalcon00]

Maybe this is just my mind as a physicist (with a love of math) but, I see no problems with using L'H rule or taylor series to DEMONSTRATE (not prove) this limit. It is a little circular but the derivative of cosine is known for all physics and engineering. Also we know L'H rule works on really messy limits that we do in physics many with ratios of trig functions (flippantly I Know). It is just from a practical standpoint easier to use L'H rule and more efficient. I appreciate the thoroughness of your statement and I would never prove this using LH rule, but to say it can't be used on expressions like that would render my work incredibly inefficient. It might be useful to clarify practical use of a rule vs definative proof of a math statement. As a side note Demonstrating something is true should always be backed up by formal proof. I would never call using LH rule on a function like that a proof somewhere in literature. However, I'll leave that to math guys to do formal proofs of this stuff. I'm too busy doing physics work to rigorously show the limits of those kinds of functions all the time. Perhaps amend your statement as dont us LH rule to prove sin(x)/x=1? For demonstration purposes with the asterisk to find the proof somewhere else it should be fine.

[@Cbon-xh3ry]

Agreed we know this limit or Taylor development and just use it. If I understand correctly he says we should not assume we know it because it involves the limit we’re looking for. Who cares ? I don’t Lol

[@CrownedFalcon00]

@@Cbon-xh3ry I think this attitude mathematicians have comes from a genuine place of trying to make sure our statements are not logically inconsistant. Something I appreciate as a Theoretical Physicist. Too many physicists and engineers use the tools of math in ways they weren't intended. This would be okay if they knew what they were doing but many of them don't and that can lead to some results that don't make sense. Heck I run into this all the time when trying to define properties of fluids vs solids because definitions that seemed to be the same in one regime are different in a different regime and without careful work you end up with condricitions. What erks me about this video it doesn't teach nuance about the mathematical toolkit one may develop. Being able to push forward even if the exact method isn't rigorous has lead me to some pretty critical insights in my research. Going through the exact mathematics is something done afterwards once I get a result and I need to prepare for a paper. My PI has been a big advocate of making sure you are precise once you know you have a result but don't let precision weigh you down when you are trying to just make progress on a problem.

[@Cbon-xh3ry]

@@CrownedFalcon00 totally understand I’m just saying in this specific case it could confuse more than teach, but I could be wrong. My son’s teacher was trying to solve this and used the hôpital’s rule cause she couldn’t find the answer. They haven’t even started to work on derivatives 🤣🤣. I showed him the geometric solution but he’s not ready yet ha ha

[@thedalek8884]

Doesn't using Taylor series yield the right answer though? If you divide the Taylor series for sin(x) by x you get 1 - some functions of x. When you take the limit of that as x approaches 0 you just get 1. Unlike using l'hospitals rule it seems to yield the right answer.

[@hewhogoesbymanynames]

L'ohpital also yields the right answer. The point is that you can't know the derivative of sinx until you know the limit in question. You also need the derivative for the Taylor series.

[@ThePeterDislikeShow]

Is there any relationship between the summed angle sin formula with the product rule? They look so similar to one another!

[@nmmm2000]

Best explanation ever. I looked for this from my university years... :)

[@littlenarwhal3914]

I mean this is both correct and incorrect. It cimpletely depends on your definition of sine and cosine and fortunately the most common and useful definition is to define them directly from their power series (probably as the real and imaginary parts of the complex exponential and its power series expansion). Once that's done you dont get any issues with the derivative since you can differentiate term by term. The properties of these functions that make them good for dealing with angles come as a by product. My personal favorite definition of an angle is to go through the dot product and the cosine function.

[@gpetty91]

Yay, now I have a video I can share instead of going through the motions of the argument for the billionth time.

[@a.k.s7613]

I have a black pen. I have a red pen. Ahhhh. Blackpenredpen

[@xevira]

Simple if anyone knows derivatives or have seen your videos: that limit is the fundamental definition of the derivative of sin Θ at 0, meaning you can't use the derivative via L'Hôspital's rule since that is akin to circular logic.

[@emanueleloschi6363]

I really like how for you all seems so simple :)

[@peterdecupis8296]

In modern analysis goniometric functions can be rigorously defined in terms of the complex function theory and/,or through a consistent measure theory of smooth curves (as the circle is); in this way all the "familiar" analitic properties are rigorously proved without any "petition of principle". Anyhow, the "pupil-oriented" proof in terms of arc and segment comparison of the limit sinh/h as a h tends to 0 can be evolved to a rigorous theoretical standard, because it represents the base concept of the Archimedean measurement of a circle: in the case of point you can axiomatically define the ciircle lenght as the coincident limit of two sequences, i.e. a minorant, crescent sequence of the perimeters of inscribed polygons and a majorant, decrescent sequence of the perimeters of circumscribed polygons, as the number n of sides tends to infinity. In this way, if we give this limit the traditional symbol 2pi, sin(2pi/n) and tan(2pi/n) would be defined as the side of the inscribed and circumscribed n-agon, respectively;; the equality of limits of the relevant perimeters, i.e. n sin(2pi/n)=n tan(2pi/n), both converging to 2pi, shall imply all the analitic properties. In fact, from the axiomatic definition of the circle lenghth in terms of a couple of minorant and majorant convergent sequences we get: sin (2pi/n)

[@SummerFrost23]

Geometric proof can be rigorous as well. But people just copy whatever they learnt in analysis and repeatedly claim that series definition is the only rigorous solution.

[@TTFMjock]

Are there any other exceptions to l’Hospital?

[@holyshit922]

From geometry get necessary inequalities and then use squeeze theorem

[@yoavboaz1078]

Can’t you mark lim y=h->0(sin h/h) and then y=bla-bla-bla +cos(theta)*y?

[@vivic2939]

my brain has now double volume after this video

[@ethanbeck5535]

Hey BPRP, would love some help with this calc 3 problem! ∭(𝑥−2)𝑑𝑉 V where 𝑉={(𝑥,𝑦,𝑧):((𝑥−2)^2)/9+((𝑦+3)^2)/25+((𝑧+1)^2)/16

[@jamiewalker329]

Not convinced. Most mathematicians define sin(t) as (e^(it) - e^(-it) )/2i (see mathworld.wolfram.com/Sine.html) and it's derivative can be determined from that of e^z - which is analytic everywhere and is it's own derivative. (Also you can immediately prove the Taylor series from this definition, as e^z is defined as an infinite power series!) Thus from the chain rule, we can very quickly see that the derivative of sine is (e^(it) + e^(-it) )/2 which is the definition of cos(t). At no stage in this derivation would we make use of the limit of sint/t as t tends towards zero, so using l'Hopital's rule is absolutely not a circular argument. Indeed, I would be interested in seeing your geometric proof that sinz/z tends to zero for as z tends to zero but when z is complex!

[@blackpenredpen]

You are talking about complex analysis. This is a calc 1 topic.

[@jamiewalker329]

@@blackpenredpen Hmm - I'm talking about maths. You also roundly reject the power series method that calc 2 students use, and their method is completely valid as a power series definition of sine is entirely equivalent to the complex exponential above (and much more versatile to generalisation than the unit circle definition - it allows us to do sine of a complex number or a matrix etc...)! I have a problem with how adamant you are about "NO!". There is no caveat, no nuance, no aside, no ambiguity in definition, no "this isn't the whole story...". Just a resounding "wrong".

[@joaco287]

I'm sorry if this question is kind of stupid but, in multivariable calculus, the directional derivative is always defined with a unitary vector, I can understand why this is necessary when doing the limit and end up multiplying by a constant, but I don't get the geometrical interpretation of getting different values of the derivative depending on the length of the vector you choose

[@joaco287]

@ゴゴ Joji Joestar ゴゴ Thanks a lot for both comments

[@htikeaungwai5513]

Really like your videos ! 💙 From Myanmar 🇲🇲 (Burma)

[@alessandrocasagrande3056]

It's pretty simple if you use the Taylor series for sine in 0.

[@Ben-wv7ht]

Correct me if I’m wrong but doesn’t it depends of the definition of the sine function ? I mean , if it’s defined as its taylor series then we can use l’hopital, excuse me if it’s not clear

[@ashtonsmith1730]

could i use a different reason that dsin(x)/dx=cos(x) like 3b1b? edit:time to remove the comment heart

[@blackpenredpen]

Could you remind me on how he did it?

[@medsaifeddinekamoun2921]

@@blackpenredpen Here's the video : kzbin.info/www/bejne/iWHCootqi6-bg7M , the proof starts at exactly 12:38, it's pretty intuitive and very easy to comprehend to the point of asking yourself : how come nobody saw this before , even me 😂😂😂😂😅 ?? But despite it feeling trivial after watching 3b1b's explanation, you kinda feel that it needs a bit more rigor, at least if you want to actually take it as a final proof and be able to use l'hopital's rule in computing the limit in your video while still being able to sleep at night😂😅. This might be a good idea for another vid by you !! You can maybe title it : "BPRP debunked"😂😂 or "L' hopital's rule shall not be violated" 😂😂😂😂 .... and with that you'll be the saviour of millions of calc students all over the world 🦸‍♂️🦸‍♂️🙏 ..... Btw I reaaaaaaallly love this channel and your vids ❤️❤️❤️❤️❤️❤️❤️ and keep them coming ❤️❤️.

[@ashtonsmith1730]

@@blackpenredpen i used that way of derivative of sinx to show sinx/x is 1 as x->0

[@kazamaki6586]

Where is the different between lim sin(h)/h and sin(theta)/theta.... ? When h->0 and theta ->0

[@FrogworfKnight]

Thats the point of this video. That in order to apply L'hopital's rule to solve this problem via proof, you would need to do the derivative of sin(x), which in the process gives us the same thing we are trying to solve via proof.

[@ςγτε]

Can you explain about d/dx (e^x) same way and how we end up with the e^x series