the most controversial limit in calculus 1 

I see what you did there

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

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

Te ves muy fresco 😎👌

yeah, and limit just means evaluate at the point

@hawkturkey 🙈🙈🙈

You will never “not” read lollll

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

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

this

F

@@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.

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

M a physics guy but love this fella

*oh frick*

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

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.

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

@@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?

@@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

Me too lmao

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

@@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

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....

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.

@@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.

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

@@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

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

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

@@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

Thanks 😊

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?

Exactly that level of satisfaction after solving a complex limit 🤩

😂

😂 😂

😂 😂 😂

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

@@jhanzaibhumayun5782 pretty weird replies

But how has Euler's formula been proven?

@@Apollorion there are some awesome proofs that don't use the taylor series of sinx,cosx,e^x

@@hybmnzz2658 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?

Here are a bunch math.stackexchange.com/questions/3510/how-to-prove-eulers-formula-ei-varphi-cos-varphi-i-sin-varphi Some of the proofs rely on the derivative of sinx though so be careful. Some more noteworthy mentions: -arguing that e^ix represents an object whose velocity is always normal to it (hard to explain it has a physicists intuition) -mathologer has a video on e^ipi for dummies where he shows e^ipi=-1 using only the properties of complex multiplication and the limit definition of e. I think when generalizing and rigourizing this it becomes a problem though as you may need sinx/x limit anyway.

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

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 ( )

@@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

@@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.

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?

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

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.

@@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.

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

@@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.

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.

@@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.

@@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

@@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!?

@@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?

My Teespring link is in the description. It’s called the “derivative on your wall”. I am also doing a 20 off discount with the code “superfan20” from now to thanksgiving. Thanks.

because you were using degree instead of radius. your calculator read it as “sin(x^deg) / x “

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

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

@@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 🤷

@@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.

@@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

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.

Oh my cosh

What other method we can use?

Geometric

"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.

@@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

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

A Maclaurin Polynomial, is a special case of the Taylor Polynomial, that uses zero as our single point

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

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)

@@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.

@@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.

@@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 ..

Technically this is not an exception. L'Hopitals rule does confirm limx->0 (sinx/x) = limx->0 (d/dx sinx). The theorem works, you just can not evaluate d/dx sinx. Anyways I know what you mean though. Check out wikipedias page on Lhopitals rule and counterexamples when not following the hypothese correctly.

You could derive the angle addition formula by geometry 😃

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

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

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.

Both

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!

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

Yes, he has a video about this already

Yea, I called that “the limit” but I realized I never wrote down the part that derivative of sin(x) requires this limit thus it still caused lots of questions. This video is to show the circular reason part

The geometric intuition is not so far from the intuition looking at the limit which you understand. Let me walk you through: Lets say we have a function f(x,y) and a direction vector u. As you know, the directional derivative would be defined as limh->0 f(X+hu) - f(X) / h , where the capital X is a vector that stores the point we are evaluating the derivative at. To be clear, X=(x_0,y_0) a vector. Now geometrically we imagine a surface f, and that f(X+hu) - f(X) / h would be like a secant line analogous to single variable calculus. Where your confusion comes is that you think since h approaches zero the magnitude of u does not matter. But think again because the h on the bottom of the denominator is the same as the h on top. They are approaching zero but are the same number along the way! So since we are dividing by h, we imagine the secant line picture and that the h corresponds to the distance between the xy coordinates of X+hu and X. Now it should be clear why u has to be a unitary vector; X+hu needs to be a distance h from X where we evaluate the function.

Additional comment: if the geometric interpretation does not click for you think about how directional derivatives are computed. Directional derivatives can be computed by taking the dot product between the gradient of the function and the direction vector. What would happen if the direction vector was only in the x-direction (u=(a,0) perhaps)? Well the dot product would only give you the partial derivative with respect to x times a. We impose the condition a=1 so that the partial derivative with respect to x is the same as the directional derivative in the x direction. We do this because it is convenient for directional derivatives to be a generalization of partial derivatives. And so we impose the condition that u is a unit vector.

@@hybmnzz2658 Thanks a lot for both comments

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

@@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".

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.

The problem is not getting the right answer. Circular reasoning is a fallacy for a different reason.

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.

Can you share it with me. I'm curious