I was wondering that, too.

So was I

If you work out the proof of this, you need the taylor series expansion of sin and cos. To derive those, you need the derivative of sin and cos.

@@beardedboulderer2609 Not if you _define_ sine and cosine to be the real and imaginary parts of the complex exponential. Most analysis courses don’t define sine and cosine via the unit circle as high school math courses do, but either via their Taylor series expansion or the complex exponential. And since it’s pretty easy to show that the power series defining sine and cosine have an infinite radius of convergence, they can be differentiated term by term (that’s a famous theorem in complex analysis; complex power series *always* converge within a circular subset of the complex numbers - though this circle may have infinite or zero radius -, and for all complex numbers within this circle, the function given by the power series is differentiable with its derivative being equal to the series obtained by differentiating term by term), which immediately yields their derivatives without requiring the limit in question at all. A lot of the time, there is no one fundamental way to define things in math; Euclid’s lemma in number theory takes considerable effort to prove going by the usual definition of primes, but sometimes in abstract algebra, it is more convenient to define prime elements directly via Euclid’s lemma, and then it’s just the definition of a prime. And when you get to higher levels of mathematics, you enter fields where, more often than not, people have spent decades fleshing out the definitions easiest to work with. If you go by the naive definition of sine and cosine via the unit circle, the limit in question is certainly not trivial. However, while there is a good reason high schoolers aren’t immediately confronted with the power series definitions of sine and cosine (despite it being much more powerful, generalizing seamlessly to complex numbers and even matrices), I haven’t seen a single analysis textbook that defines trig functions via the unit circle; it just makes things much more complicated than they need to be, and you’d have to flesh out a full theoretical apparatus to even make the definition of these functions rigorous, since, if you start doing analysis, it’s not at all clear what an angle or the length of a curved line mean (that already requires quite a bit of measure theory).

I would say you are mot being very clear. Because your sine is NOT the sine of the problem. At least, we don't really know that immediately, do we? By your definition, I understand that lim_{x→0} Yoursin(x)/x = 1 is pretty much automatic. But Yoursin(x) just uses the same name as that sin(x). You would have to show they are the same.

I substituted x = tan θ

@@drpeyam No, the upper bound is also valid since x+1 is bigger then √(x^2+1) for positive x ( because if you square both of them you get x^2+2x+1 in comparison to x^2+1, there's a 2x difference.)

That's exactly what I thought

x doesn't equal square root of x squared, that's the absolute value of x, but since the limit goes to positive infinity the absolute value of x is just x so you kind of have a point.

How can you actually calculate this limit then?

@@skylardeslypere9909 I believe you need to "squeeze" sin(x) between 0 and 1, meaning that towards infinity the x will dominate the value of sin(x), eventually becoming x/x which cancels out to 1.

I work on plowing the fields

Except Guillaume de l'Hôpital also spelled his name l'Hospital (the circumflex usually shows a missing s - the french for forest is forêt, for example, and the words are (unsurprisingly) related).

@@davidgould9431 amazing, such Interesting stuff

if you defined them that way, have fun proving all their other properties from there as well

They follow from the standard arguments since it's easy to verify that |cos x + isin x| = 1, |d\dx cos x isin x)| =|-sin x icos x| = 1, cos 0 + isin 0 = 1. So, (cos x, sin x) is the expected point on the unit circle; since C(+) over R is isomorphic to R^2, we're done (modulo relatively easy, if tedious, analytic proofs of Euclid's axioms).

Hello sir, How do you prove that sin’=cos, from the very barebones basic definition of derivative?

That’s basically the same as what he did, though.

Already done, I think it’s called don’t use l’hopital

@@drpeyam OK but can I make a video that I put on KZfaq about my way to find the second limit (limit as x goes to infinity sqrt(x^2+1/x) please.

If it had already known before doing that limit that the derivative of sin is cos, that it's redundant , but first of all it was proven that the limit of sinx/x is 1 by other methods then to find out that actually the derivative of sin is cos.

@@IoT_ I agree. It's only circular if the limit is being used to show that d/dx(sin x)=cos x.

not all derivations of (sinx)’ = cos(x) involve sinx/x. eg let y= e^ix =cosx+isinx, sinx=Im(y), y’ = iy= icosx -sinx , (sinx)’ = Im(y’) = cosx. Euler's identity depends on the series definitions of sin, cos and exp. sinx/x is not used

No, that’s assuming the limit exists

There’s a video on that :)

0/infinity = 0 and infinity/0 = infinity (or undefined)

@@drpeyam so 0/infinity is not indeterminate... huh... omg of course it is xD it always decreases

cos-1/x is also circular reasoning

@@drpeyam would (cos^2(pi/4*((x-2)/x))-sin^2(pi/4*((x-2)/x)))/x be circular reasoning? Sry for super specific example, im just unsure how to figure out if is circular

The last part was epic👌

Wait, doesn't this contradict L'Hopital 's rule?

@@anshumanagrawal346 No, the rule requires that the derivative limit is actually equal to something.

@@zlatanibrahimovic8329 Yeah, I know that now but didn't at the time of posting this reply

Yeah

😱😱😱

No! If that were true you could replace sin(x) with x^2

Aaah....okay dr.peyam thank you so much 😄

@@drpeyam maybe you can make a video about this?just wondering..

@@mohammadfahrurrozy8082 there is that video

???

Dr Peyam is everything ok with you and fm?

I remember that, haha

Ok

Someone didn’t watch the whole video 😉

😂

it is so simple

None taken, but that’s not how the education system works here. Usually here one proves the limit of sin(x)/x here using the proof with triangles, but then the students don’t care about it because they can just use l’Hopital’s rule, and they think L’Hopital’s rule is a substitute of the proof, which it is not. Of course if you can independently prove the sin(x)/x formula somehow then sure you can use l’Hopital’s, then sure, go with it, but that’s not what blackpenredpen and I are saying. We are talking about the 99% of students who do that, not about the 1% like you or me. We’re talking from experience here

@@drpeyam Sure... but in that case you should make a distinction between deriving theory, and applying results (as is usually the case on a test). When theorems are built, they have to be built in a certain order, and we can't assume what we want to prove it. And definitely, within that context of building the theory, the triangle proof of the limit of sinx/x is fundamental and using l'Hopital's rule to prove it in that context would be ridiculous since we haven't proven l"hopital yet. I think students can and should be taught this. That's all good. My issue is when you say it is "mathematically invalid" or saying it is "wrong" when simply calculating a limit like this. As I mentioned all this can really be avoided by just telling the student on the test, "solve this without l'Hopital's rule or solve this only using the squeeze theorem". The student will learn the triangle proof and all is good. And it doesn't give the misunderstanding that it is "mathematically wrong" to apply l'Hopital's rule in this situation. It just emphasizes that the student should know and understand the triangle proof as it is a fundamental result. I know you guys only want to improve understanding, but there's some confusion among some viewers asking... "ok, l'Hopital's rule is mathematically wrong to use here, where else can't it be used?" I'm just going by the comments I see.

But did you use l’Hopital?

@@drpeyam no

Very good