The Limit (do not use L'Hospital rule)

[@alexatg1820]

How physicists do the limit: In small θ, sinθ≈θ And 0 is very small So sinθ/θ=θ/θ=1

[@JoQeZzZ]

The small angle approximation actually comes from this exact reasoning!

[@miso-ge1gz]

unironically good enough

[@86400SecondsToLive]

@@miso-ge1gz It's not a solution for the same reasons pointed out in the video. He used the Taylor approximation and you can only do that when you know the derivative.

[@miso-ge1gz]

@@86400SecondsToLive But he got the same answer? Or do you have to prove everything on the math tests

[@86400SecondsToLive]

@@miso-ge1gz As explained in the video: You need to know the limit to know the derivative of sin(x) (hint: the limit is the derivative at 0). If you don't know the limit, you don't know the derivative. If you don't know the derivative, you have no approximation via Taylor. Without the approximation, his solution is not a valid solution. You do not need to prove everything on math tests, but you also can not assume the claim to be true in order to prove it true.

[@emperorpingusmathchannel5365]

Do not use circular reasoning! *Proceeds to use the unit circle to define the inequality*

[@blackpenredpen]

GLaDOS lollll I love this!!!!

[@JensenPlaysMC]

@@dp-zn8bd "Circular"

[@bensolomons4299]

ahah ahaha i love it!

[@DianaHatter]

of course your name is GLaDOS

[@icew0lf98]

@@dp-zn8bd ahahahah sarcasm ahahaha

[@restcure]

"of course, we're all adults now, this angle is in radians" - I love it!

[@capsey_]

Everybody knows that real men use clock time for angles

[@Abdullah-pu2dr]

@@capsey_ can't wait till I have to calculate cos(9:28)

[@tolgaerdonmezorigamitutori1605]

Damn, I knew this was legit, but no one believed

[@Ketsur0n]

@@capsey_ ah yes

[@GRVTY3]

@@Abdullah-pu2dr remember, it's periodical. the only angles you need are from 00:00 to 01:05

[@zsoltvastag416]

Engineer solution: Book says it's 1. So it's 1. qed.

[@sharpfang]

No! Saying it is 1 is insufficient! It is 1, but with 0.0001% accuracy!

[@blightedcrowmain8236]

@@sharpfang brr

[@mokaakasia4636]

@@sharpfang ammm, no. It's 1, but with any Eps. accuracy , limit definition ref.

[@sharpfang]

@@mokaakasia4636 Get back to your cave, you mathematician.

[@GarGlingT]

You cannot graduate with engineering stuff, you need to pass a proper calculus.

[@Selicre]

"circular reasoning" I don't know if this was a trigonometry pun, but it is now.

[@banderfargoyl]

Yup. Trigonometry is invalid.

[@marcushendriksen8415]

Hehehe I've used the phrase multiple times in my own maths journal when exploring trig; it's a quality pun

[@reelgangstazskip]

@@marcushendriksen8415 >"maths"

[@marcushendriksen8415]

@@reelgangstazskip what's wrong with that? Last I checked, the subject was called "mathematicS", not "mathematic"

[@joryjones6808]

When my Calc 2 Teacher was doing a similar problem he said "by no means is this a pun, the reasoning ends up to be circular."

[@ahmedalfuwaires]

I love how enthusiastic you are about math. Need more educators like this

[@Notnessi]

that also why i like his channel. You know he loves maths

[@TIO540S1]

You need to prove that the arc length is greater than sine and less than tangent. “You can see from the picture” isn’t good enough.

[@blackpenredpen]

Rob Ryan then we can compare areas!

[@samporterbridges577]

Yeah comparing areas is how i learned it

[@BigDBrian]

how would you compare the areas? what areas are you comparing?

[@AlexandreRibeiroXRV7]

@@BigDBrian the small right triangle with the green side, the circular sector with angle theta, and the big triangle with the blue side.

[@mrbatweed]

Does area correspond to length? Koch snowflake?

[@tomaszlosinski875]

I often have no idea what he's on about but his enthusiasm makes me keep coming back each time. I'm addicted now.

[@tollboothjason]

It may be easier to tackle theta's plus/minus issue before taking limits. f(x) = sin(x) and g(x) = x are odd functions, and (odd function)/(odd function) = even function. Thus h(x) = (sin(x))/x is even, so its limit going to zero from either direction is the same. Thus we can say without loss of generality that theta>0 and proceed from there.

[@n0ame1u1]

Well, since we all know sin(θ)=θ, lim θ→0 sin(θ)/θ = 1

[@trace8617]

good one lol

[@gregoriousmaths266]

Dang it I just commented that but now I see this Well hello fellow engineer

[@walrusninja3581]

Gregorious Maths I’m only in high school calculus right now, but how is this an engineer thing? Isn’t it true for all fields of math that during near zero situations, sinx=x

[@alvachan88]

@@walrusninja3581 This is just a guess. I think this is especially close to home for engineers cause of the timing of 20 oscillations to find the value of g that we did in secondary school. Our physics teachers would always yell "small oscillations" without telling us why. Only in high school after learning the content of this video did we realise smaller oscillations isn't to reduce the effect of air resistance.

[@bradyzimmerman8380]

Lol a physics major I see

[@craftbuzzwonky4752]

When I saw the thumbnail, I was like: The man has gone mad obiviously. Then I noticed that it was the factorial sign and not an exclamation mark for a new discovery in math!

[@george.6157]

You're so cool :) I just thought of this a few hours ago because of prooving the derivate of sinx and this limit made me confused and now you explain it, best timing ever

[@blackpenredpen]

G. E. Ahhh nice!!!! : )

[@Kdd160]

U can do this by Maclaurin-Taylor series expansion of the sine function

[@zeffar99]

@@Kdd160 no you can't it's still circular reasoning

[@Kdd160]

@@zeffar99 yup i actually realized months after i had already written my comment 😆😆😆

[@MrCigarro50]

Thank you, we saw this proof in class, but my students enjoy this video to clarify some questions.

[@martinepstein9826]

9:14 Technical point: once you've written 1 >= lim(...) >= 1 you've already used the squeeze theorem. The fact that c >= d >= c implies c = d is just a basic property of ordered sets, not a calculus theorem.

[@dakkoshite]

This is among my favourite proofs to demonstrate in my high school calculus class. It paves the way to have students think using a different mindset.

[@lackdejuranez7084]

It's really beautiful, isn't it?

[@discretelycontinuous2059]

I'd be inclined to use the taylor series

[@Peter_1986]

I just learned the Taylor Series today, so I am very eager to practise it at the moment. I will totally use it on this expression as well.

[@gerardmonsen1267]

But the Taylor Series uses derivatives to create the terms, so you end up running into the circular argument problem again.

[@valeriobertoncello1809]

But that still uses derivatives

[@Pete-Prolly]

1st) I know he'll use Squeeze Theorem 2nd) the video title didn't say don't use derivatives it said "don't use L'Hospital" 3rd) DiscretelyContinuous didn't say use "L'Hospital" or "Derivatives" and 2 of you got on his f***ing case over it; he said "use Taylor Series" and one COULD use the McLaurin or the Taylor series without differentiating "using rules," but by using "the limit definition" of the derivative for the 1st few terms and then use "proof by induction" to show the series converges such that: lim [ (ƒ(θ+Δθ)-ƒ(θ))/Δθ ] Δθ→0 such that ƒ(θ) = sinθ/θ = ∞ (∑ (θ²ⁿ⁻¹)(-1)ⁿ⁺¹/(2n-1)!)•(1/θ) ⁿ⁼¹ & I'd convert θ to radians 1st & create a parameter t=θπ/180 & say that ƒ(t) = sin(t)/t = ∞ (∑ (t²ⁿ⁻¹)(-1)ⁿ⁺¹/(2n-1)!)•(1/t) ⁿ⁼¹ and evaluate lim [ (ƒ(t+Δt)-ƒ(t))/Δt ] Δt→0 and solve after expanding this McLaurin series into a Taylor series. You can then see whether or not the series converges to 1, or some other value, or not, but I leave that as an exercise for the reader.

[@gerardmonsen1267]

Lοɢɪco Λόгος ̙ Again, you run into circular reasoning. How do you find the Taylor Series of sin? Take its derivative and evaluate it at a point. To obtain the Taylor Series of sin, you have to already know the derivative of sin. Thus, you end up with the same circular reasoning problem that you’d run into with L’Hospital’s Rule - that you can’t find the derivative of sin using a method that uses the derivative of sin. The squeeze theorem avoids this problem.

[@Alfred-pr1qf]

you are a great teacher! In the past I was afraid of math, but with you everything is much clearer and simple! thanks a lot

[@Matt-tw3oz]

my calc professors were very cool. they'd give partial credit even if they say "solve for x using [certain theorem]", but we did it another way. i was always of the mind to just give students all the theorems/rules on the front page of the test. IMO, calc isn't about memorizing facts like history class but is more about working a complicated problem with a properly filled toolbox. just give me the toolbox, and i'll take it from there.

[@user-en5vj6vr2u]

Professors literally tell you to use a certain theorem so that you learn how to use each “tool”

[@ninoporcino5790]

The squeeze theorem, here in Italy we call it "the two cops theorem" because it's like two cops taking a thief to prison, they don't let him escape from the sides

[@MysteryHendrik]

In Germany we call it the ‘sandwich criterion.’

[@qhw9]

MysteryHendrik in kuwait we say sandwich theorem

[@Crazmuss]

That boring, lets call it chikan theorem!

[@longsteinpufferbatch4949]

@@Crazmuss is that Japanese?

[@draaagoo7799]

we call it bocadio said theorem

[@aliexpress.official]

6:10 You can't just say on an exercise/test that tanx>x>sinx because it can be seen from drawing. you'll be required to prove that as well.

[@JiminatorPV]

Can you not proof that with f(x)=tan(x) and g(x)=x functions, since they both are continuous in the interval [0 , pi/2), and f(0)=0 and g(0)=0, for the tangent to take a value inferior to the arc lenght, the derivative of f(x) has to be minor than 1, and since secant squared is always >= 1 in said interval, tangent must always be equal or greater then the arc lenght.

[@stewartzayat7526]

Unfortunately, it's not a straight line, so I think you would need to do more work

[@lautyx18]

Isn't tan(x) ~ sin(x) when x->0 ?

[@stewartzayat7526]

@@lautyx18 yes, basically you are saying lim as x->0 (sin(x)/tan(x)) = 1, you can show that as follows: lim as x->0 (sin(x)/tan(x)) = lim as x->0 (sin(x)/(sin(x)/cos(x))) = lim as x->0 (cos(x)) = 1

[@CarTLA]

You can proof that by using the triangles you draw in the unit circle. I don't know how to explain it properly without an image, though.

[@sugarfrosted2005]

I would put a warning (DO NOT USE LHOPITAL'S RULE!!) on the exam.

[@hamanahamana3799]

You could also use L’Hopital after proving d/dx(sin x) = cos x through Euler’s formula: e^ix = cos x + i sin x Derivative of both sides... ie^ix = d/dx (cos x) + i d/dx (sin x) = - sin x + i cos x By multiplying Euler’s formula by i and matching up the real and imaginary parts of the derivative of Euler’s formula, we can see that d/dx (sin x) = cos x and d/dx (cos x) = -sin x Now I’m wondering how one could prove Euler’s formula without using Taylor series, or by proving the Taylor series without using the derivative

[@ptrexy]

Yeah, me too. Have you found any proof?

[@RandomPerson-lp2tl]

But you just used d/dx (cos x) = - sin x and d/dx (sin x) = cos x without proving it. So its still circular reasoning.

[@spaghettiking653]

Like Random Person pointed out, this is still assuming the values of d/dx(sin) and d/dx(cos), but I think it can work if you use the Maclaurin series expansions of sine and cosine and differentiating them to legitimately get their values.

[@hamanahamana3799]

​@@spaghettiking653 My original comment was showing how Euler's formula implied d/dx(cos) = -sin and d/dx(sin) = cos; since e^ix = cos x + i sin x, d/dx(e^ix) = ie^ix = i(cos x + i sin x) = -sin x + i cos x = d/dx(cos x + i sin x), which gives d/dx(sin) = cos and d/dx(cos) = -sin after equating real and imaginary parts. That's not a full proof that d/dx(sin) = cos, though, since all the proofs of Euler's formula I know use d/dx(sin) = cos implicitly (this is what Taylor or Maclaurin series do). If we can prove Euler's formula without that dependence, though, that fully proves d/dx(sin) = cos and d/dx(cos) = -sin. Sorry if that wasn't clear originally.

[@spaghettiking653]

@@hamanahamana3799 Oh, sorry, please forgive me as I wasn't reading your comment clearly. I see what you were trying to say now, my bad :)

[@rachitgoyal8457]

This is the dream guy who can do all your maths homework with a smile 😃.

[@tydude]

nice to see the proof. my calc prof just told us to memorize this limit without explaining how it works

[@HagenvonEitzen]

6:13 "From the picture you can see" is not a proof that the curvy arc is shorter than the straight tan segment. Things are a lot easier when you compare areas of small triangle < pie < big triangle

[@Emilia333g]

I think proof by steve is my new favourite way of proving limits

[@neutronenstern.]

Me: The thumbnail isnt correct Also Me: He lied to me Now Me: Oh 0! is indeed 1 Me :(

[@matthewhartley3147]

6:12 It's clear from the graph why sin theta is less than theta. The line connecting the two points of the arc is longer than sin theta, and the arc is longer than the line. However, it is not clear from the graph why theta is less than tan theta. Sure, the line to (1,tan theta) goes higher than than the arc does, but the arc curves. If the arc curved enough, then the arc would be longer.

[@Gold161803]

I had the same thought. To avoid that, I compare areas instead of lengths: The area of the triangle bounded by two radii and the chord (not drawn) is 1/2*1*1*sin(theta). That's less than or equal to the area of the sector, which is 1/2*(theta)*1*1. That's less than or equal to the area of the big triangle with a base of 1 and height of tan(theta), whose area is of course 1/2*tan(theta). You can do all your algebra from there :)

[@wesleydeng71]

@@Gold161803 Area of the sector = theta/(2*pi)*pi*1^2 = theta/2 < area of the big triangle = tan(theta)/2 => theta < tan(theta)

[@valeriobertoncello1809]

The implicit assumption is that a circle is nice and smooth, and any small enough arc on its circumference looks like a chord.

[@jaimeduncan6167]

Yep it's a Gap. The proper way to do it is using the areas. Do it yourself I am sure you are going to enjoy it and it's not difficult.

[@Samuel-kq4ip]

kzbin.info/www/bejne/jouQnWtmn9aMa7c&ab_channel=PrimeNewtons this video explains it well if anyone else is still confused

[@dmytro_shum]

Can you make a video of example (with the uncertainty of 0/0) where l'hopital's rule gives the wrong answer?

[@credoma1]

yes, please

[@duckymomo7935]

It’s not L’Hospital failing but the conditions don’t apply Lim x-> ∞ x/(x + sinx) = 1 Lim x-> ∞ 1/(1 + cosx) = DNE Problems: 1) the denominator, derivative version, goes through 0 over the entire interval (0, ∞) 2) the “L’Hospital limit” doesn’t exist therefore L’Hospital rule doesn’t apply (it doesn’t say what happens to the original limit)

[@RunstarHomer]

But L'Hopital's rule doesn't give the wrong answer

[@dashyz3293]

the worst l'hopital's can do is just make you fall into a really long line of repeating l'hopital's over and over again. I am not sure if there is such a limit that makes you do it forever, but l'hopital's cannot fail if you use it right.

[@kohwenxu]

@@dashyz3293 You can also get stuck if you use L’Hopital’s rule blindly, because you can get stuff that goes to denominator equaling 0, but the numerator is non 0 and therefore you’re stuck

[@iabervon]

With a more geometric definition of the derivative (or of differentials), you can prove the derivative geometrically, and the limit with L'Hopital's Rule. There's a nice proof based on tiny arcs of a circle being effectively straight.

[@stephenbeck7222]

For the 0- limit at the end, perhaps a little more formal way of saying it is that y=sin x is an odd function and y=x is an odd function, both of which we know without calculus. An odd function divided by another odd function is even, thus the 0+ limit must be equal to the 0- limit.

[@grigoriichrrnyshov5621]

man that is the beautiful exsplanation, you show it very clear

[@RachidOUSALEM]

Sinθ < θ < Tanθ needs to be prouved...

[@skilz8098]

It's simple we have sin(t)/t = 0/0 = indeterminate by basic mathematics or basic arithmetic. Through various proofs and theorems that are all related to intersecting lines and angles they generate by vector notation and linear equations to the properties of right triangles and circles along with the use of tangents both lines and trig functions... by induction it is proven. This property holds for all circles and the same two triangles that can be generated by the radius and it's extended line from the linear equation that generates that line to the tangent line of the circle that is also a perpendicular bisector to the x-axis. It is inferred through induction. It has already been proven. This is why he stated in the beginning we can do this through the use of Geometry! Another proof of induction is to state that the polynomial equation f(x) = x^2 where all of x > 0 is a 1 to 1 mapping of all of the possible pairings of the side of a square to its area. for example consider these sets of points where the x is the length and the y is the area (0.5, 0.25), (1,1), (2,4), (3,9), (4,16) ... (n, n^2) where n is +. We only do this because we don't think of length of a side of an object nor its area to be negative... So if you have Area by induction we can find the length of one of its sides as long as we know the polynomial function that gives us the area. So for example if we have the function f(x) = 3x^2 + 4x - 2. This is full quadratic. The first term gives us an area of a square. then added to that area is some length and a constant. If we take the value of 3 and evaluate this function f(3) = 3(3)^2 + 4(3) - 2 = 27 + 12 - 2 = 37 units^2. Now let's see what the "pseudo length of this would be" (meaning taking its derivative and going down to a lower dimension of space). f'(x) = 6x + 4. Now evaluate it with the same input f'(3) = 6(3) + 4 = 18 + 4 = 22 units. Here by induction we can see that the highest order of degree within a given polynomial will determine the spacial dimension we are in. So by example x^0 = 0 dimension, x^1 = 1 dimension, x^2 = 2 dimension, x^3 = 3 dimension and x^n = n dimension. There are somethings that you don't have to prove because they are understood. Just as the fact that the derivative of sin(t) = cos(t) and the derivative of cos(t) = -sin(t). We all know that they are understood to be just that and don't require proofs. Go ahead and try to write up a proof for those two derivatives and let's see how long it takes... I'll gladly accept his proof as a form of induction just by knowing and understanding Geometry. Here's one for you prove that 0 = 0 and that 1 = 1... I dare you; good luck with that because they are abstract ideas...

[@achyuthramachandran2189]

Yes, comparing the lengths of those lines and arc isn't enough. The way I was taught was to compare the areas of the figures formed, namely the triangle with height sin x, the sector with included angle x, and the triangle with height tan x.

[@marceloescalantemarrugo6391]

With the mean value theorem.

[@MrCoffeypaul]

It can be seen on the diagram to be true!

[@seroujghazarian6343]

Easy. The adjacent is always smaller than the hypotenuse

[@stxavierskotbaggi413]

For a very small value of x, sin(x) ≈ x As x is approaching 0 meaning a very small number *lim(x->0)* [sin(x)/x] = *lim(x->0)* [ x/x ] (bec x is infinite small) = *lim(x->0)* [1] = 1

[@Galileosays]

L'hopital is permitted, since we can define sin (theta)= {exp(i.theta)-exp(-i.theta)}/2i, and the derivative is {i.exp(i.theta)-(-i)exp(-i.theta)}/2i=cos(theta). In other words there is no geometric proof needed. If one would argue that complex theory has to be proven too, then it would imply that every mathematical problem requires first the proof of the whole history of mathematics starting with Euclides 1+1=2.

[@mychaelsmith6874]

You would have to prove that your definition is the right definition of sin.

[@hamiltonianpathondodecahed5236]

And also the euler's identity exp(ix)=cos(x) +i sin(x) Has came from the taylor series of sine and cosine crntered at zero Which in turn came from differentiating sine and cosine. *CIRCULAR REASONING*

[@nuklearboysymbiote]

Actually you do have to prove the complex stuff to be true first, but it doesn't imply you have to start from 1+1=2. and if you've ever looked at the proof you'll know that representing a complex number as a power of e is proven USING the fact that the derivative of sine is cosine… CiRcULar ReAsONinG

[@Sonny_McMacsson]

@@mychaelsmith6874 If we follow that logic to its conclusion, we have to prove all of mathematics from counting sheep on up.

[@thiantromp6607]

embustero71 which we do have to

[@MadScientyst]

Really simple from Geometry (or PreCalc), when angle θ is small: Area of triangle < Area of Sector < Area of Tangent Thus, 1 < sinθ/θ < 1 So by the Squeeze Theorem sinθ/θ=1 when θ is small...QED

[@jimallysonnevado3973]

how did you conclude that the tangent line is longer than the curve there might be a possibility that the curve is longer because its a curve

[@mrfreezy7457]

In this case, it is sufficient to prove that tan(x) >= x for all [0 < x < pi/4] since we're in the first quadrant and we're going to x=0 anyways. If you plot y=tan(x) against y=x, you'll find that tan(x) is always greater than or equal to x on our interval, so we're safe here. Edit: Another solution would be to use areas instead of lengths. i.e. 0.5(cosx)(sinx)

[@mariomario-ih6mn]

(edited)

[@chessandmathguy]

It all depends on whether we're asked to calculate the limit or to prove that the limit is 1.

[@wurttmapper2200]

Defining sin and cos in terms of complex exponentials, you can prove the derivative of sin is cos; then you can use L'H

[@carultch]

You need Taylor series to do that, and that requires proofs that depend on this rule. There's always going to be a hidden detail behind the scenes that depends on the limit of sin(theta)/theta, no matter what method other than this you do, to prove this rule.

[@MusicalRaichu]

I think you can use the geometric derivation to prove that d/dx sin x = cos x. Using loppy's rule for sin x / x would then be valid.

[@mainlakshayhoon]

It could simply done by using that When Q≈0 then, sinQ≈Q≈tanQ So, Lim (Q→0) SinQ/Q = Lim (Q→0) Q/Q = 1

[@willstray4267]

Love how the thumnail gives away the answer as 0!

[@ThePharphis]

I see a lot of similar criticism in the comments, just wanted to voice my own similar criticism: Once you have validly proved a conclusion via logical argument, it can be used as a TRUE premise in any other logical argument (this is how the entirety of math works, after the axioms are in place). Thus, if we PROVED the derivative using valid logical arguments and true premises, the conclusion is TRUE and can be used as a TRUE premise in our argument to prove this limit = 1.

[@MagicGonads]

Especially if you take the definition of sin and cos to come from exponentials and differential equations rather than principly geometrical

[@otakurocklee]

Yes. Thank you.

[@YourAverageLink]

Did not think I'd finally find out why the tangent function is named as such in this video!

[@ActionJaxonH]

I’ve seen the full proof using the Squeeze Theorem before, but I never knew about L’Hospital rule (I’m learning Calculus for fun on my own time). I had HEARD of the rule, and that it was important, but this is the first time I actually saw what it was from myself. Wow, that’s a powerful rule. Really really powerful

[@奥特曼-b8e]

Jaxon Holden some limits counter l’Hospital though

[@soumyadipbanerjee2074]

Well, to evaluate the left-hand limit we can just use the symmetry of sin(x)/x. As sin(x)/x is an even function it has to be symmetric about the y axis. Thus, as sin(x)/x approaches 1 as x approaches 0 from the right-hand side, by symmetry, it also has to approach 1 as x approaches 0 from the left-hand side. Thus the limit as sin(x)/x approaches 0 exists and is equal to 1.

[@psionl0]

If d/dx sin x = cos x was proven using Euler's identity then L'Hospitals rule would be perfectly valid.

[@Super1337357]

Lim(x->0)sin(x) is not sin(0) though.

[@psionl0]

@@Super1337357 It is.

[@thexoxob9448]

The thing is euler's identity was proven by derviatives

[@psionl0]

@@thexoxob9448 The thing is that since e^ix has a real part and an imaginary part, we can write e^ix = f(x) + i g(x). Without using any trigonometry at all, we can show that |e^ix| = 1 (ie e^ix lies on a unit circle) and that f(x) has the properties of sin(x) and g(x) has the properties of cos(x). The only sketchy thing is establishing that x is an angle measured in radians.

[@dang-x3n0t1ct]

What if we prove d/dx sin(x) = cos(x) by the limit definition of the derivative

[@thexoxob9448]

For 8:22, you can't take the limit on all 3 sides, as that would be circular reasoning of what the squeeze theorem states. (It states if g(x)c h(x) = L, then limx->c f(x) = L. This implies that you can take the limit of all 3 sides, not the other way around.)

[@aurithrabarua4698]

I don't use L' Hospital rule.. I use the 'Sandwich Theorem' to state that lim sinx = 1 x→0 x

[@SuperV2g2to]

Who is the limit on the left side and on the right side? Pls

[@DiViNe_-jv6we]

Definitely -1 +1 coz sin angle intake minimum and maximum value -1 and +1 respectively.

[@tomkerruish2982]

That's how it's done in Apostol.

[@sajjadmazhar8390]

I used to watch your videos during my school and college days... I'm 25 today and still I'm interested to watch your videos ❤

[@doctormaddyson]

In Russian school it called “The first perfect limit” (Первый замечательный предел).

[@Kitulous]

мы быстренько пробежали пределы в 10-м классе, перешли на производные, и никаких Лопиталей и замечательных пределов мы не проходили( щас на логарифмах сидим (11 класс)

[@doctormaddyson]

Kitulous, это прискорбно, мне в школе с этим повезло. А так, второй замечательный предел-то вы должны были обязательно пройти, так как второй замечательный предел - это определение числа e, которое в показательных и логарифмических функциях постоянно встречается.

[@imadhamaidi]

Russians are the best mathematicians.

[@GemaPratamaAditya]

gordon freeman

[@anarmehraliyev1286]

In english it is called the first remarkable limit. Similar names.

[@wediadi6788]

3:16 The circumference of a full circle is given by the formula 2πr, 2π being the angle size of a full circle in radians. Hence it follows that the arc length that corressponds to any angle θ in radians is given by θr. When it is a unit circle, the arc length is equal to θ.

[@reetjaiswal3950]

I often battle with the question, " Is this true just in my case, or in all cases?" when I'm writing objective based tests. And most often it turns out the latter was true. Should I lose out on time to confirm, or move on with the best choice?

[@PartyMain]

I actually learnt a different way to solve it from physics. So sin(¥) of very small angles is actually equal the ¥ itself. So if we limit sin(¥) to 0, it means sin(¥) = ¥. Therefore we have ¥/¥ = 1. Same goes for the tangent

[@rudycummings4671]

That solution looks too easy. Something looks fishy. Please check with your math prof

[@PartyMain]

@@rudycummings4671 it's in our formula cheat sheet, handed out by our prof themselves. Limiting sin(x)/x & tan(x)/x is 1

[@andrewweirny]

So that's why sin / cos is called "the tangent." I'd like to say "I always wondered" but the truth is "I never wondered."

[@peterdecupis8296]

this video highlights the lacks of the traditional formulation of goniometric functions; the graphic "evidence" can not be directly employed in a theoretical deduction but can only "inspire" the statement of axiomatic definitions. In the case of point, the circular arc lenght may be rigorously defined only by an integration theory that is grounded on the continuity and completness properties of real numbers (e.g. Riemann); for instance, in this context we can as a first step define the arcsin function as an integral function, that is strictly crescent in the range [-1,+1]; therefore the sin function (restricted in the range [-pi/2, pi/2]) is defined through functional inversion. The limit sin(theta)/thets as theta tends to 0 would be replaced by the limit t/arcsin(t) as t tends to 0 and this would be directly evaluated by applying integration properties to the integral function arcsin(t)=integral{[(dy/dt')^2+(dx/dt')^2]^0.5} dt'} from t'=0 to t'=t, with y(t')=t'; x(t')=[1-t'^2]^0.5; since the integrand function, i.e. [1-t'^2]^-0.5 is continous in finite intervals around 0 we can use the intermediate value theorem for computing the definite integral from 0 to t and with easy calculations we obtain arcsin(t)=t*(1+o(t)) as t tends to 0; then our limit is immediately evaluated as 1.

[@peterdecupis8296]

As an alternative, we can also work with areas, with an analogous method

[@JohnRandomness105]

The problem with using l'Hospital's rule is that it's circular reasoning. You use sin(x)/x --> 1 to derive the derivative formulas. One geometric way is to squeeze the areas. BTW: I hope that I misread the thumbnail. Sin(x)/x most certainly does not approach zero as x approaches zero. EDIT: It's not really clear to me that the arc length satisfies the inequality. The areas do, because they are inside each other. Okay, I went back to the thumbnail. It works if you read "0!" as zero-factorial, which is one.

[@manuelferrer6501]

Its clear that sin(x) < x < tan(x)on bigger angles, for example in 45°= π/4 , tan( π/4 ) = 1, bigger than x , cuz pi

[@epicbird08]

The very reason why we use radians. Wonderful!

[@milhousekid]

I remember a teacher was showing me the derivative of sin, using l'hopitals rule and I had this exact question. Unfortunately he didn't explain it as well as you

[@leviszhou711]

曹老师讲的挺好的 帮我把大一的夹逼定理都复习了一遍。👍

[@Koisheep]

I thought you would talk about non-Hausdorff spaces and I was like "WHAT ok that's wild"

[@just_another_guy6915]

lol same

[@mjones207]

One of the students I tutor recently asked me *why* lim (θ→0) [(sin θ) / θ] = 1. I showed him a graph of y = sin θ superimposed over a graph of y = θ. The more we zoomed in toward θ = 0, the closer the graphs were, meaning their quotient became closer and closer to 1.

[@奥特曼-b8e]

mjones207 some functions are not that well-behaved though

[@seankelly8906]

Haven't watched it yet but I'm guessing he's going to mention how that limit of sinx/x, which is the derivative of sine at 0, changes when you use different radian units. In fact for any real r, f(x) = e^(rxi) gives a parametrization of the circle but with angular velocity, or derivative of sine at 0, equal to r in absolute value. The real way this is resolved is by how you define sine: if you don't define it such that it has unit derivative at 0, or r=1 above, you're working with a different function. One nice way to define sine to make things work as expected is to specify a certain unit speed parametrization of the circle in the plane, then make sine the y-coord projection. Or you can use the differential equation definition, where of course one of the conditions is that this limit, the sine derivative at 1, needs to be 0.

[@lightyagami6647]

Nyc method btw😅😂

[@nomic655]

Very nice method to prove this, slightly different than how we have it in the books. But for us, the limit sinx/x is a formula, so we don't have to prove it when we use it, unless there's a specific question that wants you to prove it.

[@rex17ly62]

Why can't we conclude that sinx/x=1 at x=0? Because the limit from 0+ and 0- both go to 1, so doesn't that mean that the function converges to 1?

[@isaacaguilar5642]

Nova :3 thats what i was thinking too

[@user_2793]

If plugging in a specific value gives you an indeterminate form, then the function is said to be undefined at that point. That just makes more sense.

[@BabyBugBug]

I like this! I think this goes to show for the 1000th time why math isn’t hard, it’s just shifting around terms and manipulating equations to equal something you are looking for. When terms in an equation don’t look like this, fiddle with them and see how or if you can make them so. It took me a long time to learn this but it’s very valuable.

[@x4wi483]

ig if you define sin(theta) and cos(theta) with taylor series you know that d/dx sin(theta) = cos(theta) from the power rule sooooo in this scenario teacher shouldn't technically give 0 points if they didn't specify witch definitions we use ¯\_(ツ)_/¯

[@jonpress6773]

You can't do that because determining the taylor series of a function requires you to know all its derivatives anyway, including the first derivative. So you're back to the same circular reasoning.

[@pete4416]

I've heard it called the Etau theorem, the sandwich theorem. And today, the squeeze theorem. It's so unserious I love it

[@kmarasin]

I think it's unfair to disallow the use of a basic definition like (sin x)'=cos x without explicitly starting it. It might be true that the means of showing this definition using the 1st fundamental theorem of calculus relies on knowing the limit in the problem, but that's not the ONLY way to show (sin x)'= cos x. We could use similar "geometrical" methods there, too. Also, I think calling it circular logic is a little TOO accurate. On math exams, questions like this are posed in such a way that the student can show knowledge of definitions by cranking through an operation using those definitions. Completing a logic circle is exactly what they're supposed to be doing!

[@jelmerterburg3588]

Totally agree on this one. Given that it has already been proven (and flat out stated as given in early math) that the derivative of sin x is cos x _and_ that l'Hôpital's rule is not given with any exceptions about which derivatives you can use, this should be fair game. If anything, it shows that the derivative of sin x and l'Hôpital's rule are perfectly consistent with each other.

[@TheCrashbandicoot36]

7:50 i kept thinking about sin(x) being positive..., until i realized it is a geometric demonstration, and you were using the first cuadrant, were sin(theta) is positive. Great demonstation.

[@ib9rt]

How can you assert that the arc length is less than tan θ? When θ is small the arc length approaches the diagonal of a rectangle and the diagonal of a rectangle is longer than any of the sides. Therefore the assertion is not obvious and must be proven.

[@NotBroihon]

It indeed must be proven. My approach is to use the derivatives of both tan and the arc. The derivative of arc is simply 1 and the derivative of tan is sec² (quotient rule of sin/cos). At theta = 0 both derivatives are 1, but for all 0

[@assalane]

I don't see how it approaches the diagonal of a rectangle. Which rectangle, the one with side sin(θ) and tan(θ)? But this does not become a rectangle because you will always have the hyp of the second triangle > 1. it will always be a square trapezoid.

[@angelmendez-rivera351]

assalane Exactly my point.

[@nuklearboysymbiote]

Compare areas: 1. right triangle of base 1, height sine 2. circle sector made by the angle 3. right triangle of base 1, height tangent You will see the areas go 1

[@uzair.mhaf.s]

I like how i watched the wholevideo and i aint even doing A level maths yet! Saving this for future use

[@JAzzWoods-ik4vv]

me: learns summation rules for sine and cosine. What you gonna do now, eh?

[@davidseed2939]

The simplest approach is a series expansion of sin θ = θ-θ^3/6 +Higher order terms. so sin θ/θ =1 - θ^2/6 + otherr terms in θ which tend to zero.

[@benjaminbrady2385]

Absolutely amazing video as always! And it actually made me come up with a new question! Couldn’t you “vertically approach the limit” by using the complex world? Like, sin(i+0)/(i+0) to sin(0.00001i+0)/(0.00001i+0) Or also negative i?

[@tejarex]

Yes. As with the real version, one may ask if (lim z->(0,0) sin(z)/z) exists and whether it is unique. With an infinity of directions, complex limits 'more ofter' depend on the direction of approach. I would not be surprised if this is such a case.

[@SuperRousku]

Since sin(θ) is an analytic function, it is in C^\infty and thus has a unique derivative irrespective of direction. As the derivative of sin(θ) is equal to the limit by definition, the limit must also have a unique value irrespective of direction of approach.

[@AngeloLaCruz]

100% After all these years I finally grasp this Proof completely.

[@98danielray]

to be fair nothing says you cant use a tool that came from a limit to prove that same limit.we do it all the time with other theorems. would be circular reasoning if it was the first time proving them. whichever the case, great video. thank you

[@dlevi67]

Especially since it is possible to determine the analytic derivative of sin(x) without using the lim x->0 sin(x)/x ...

[@ihave3heads]

The problem is that you never prove L'hopitals in first year calc.

[@98danielray]

@@ihave3heads indeed. it is proven in analysis

[@dlevi67]

@@98danielray Can y'all please stop assuming that academic programs are the same the world over?

[@angelmendez-rivera351]

dlevi67 Can you please stop being an arrogant prick and pretending that the derivative of sin is going to ever be proven analytically in a Calculus 1 course without using the derivative definition? Jesus Christ the hypocrisy. This is the problem with math in KZbin.

[@calebfoxschool]

I'm in an online class for calc right now, which requires zero proofs or anything at all. So I've gone the whole semester without realizing where the derivatives of any of these trig derivatives come from, and it felt so random until I watched this and another proof video of yours. Thanks so much

[@Selicre]

I'm not quite sure if I'd consider the comparison of the three values rigorous. To me, it'd seem like theta would be the largest, since it's fairly close to being a diagonal in a rectangle; is there a more direct proof of this comparison somewhere?

[@blackpenredpen]

Selicre [Hyper] this is good since theta is small

[@d1o2c3t4o5r]

blackpenredpen the variation of this proof where the areas of the triangles are compared addresses this criticism.

[@MarkVsharK]

I was also having a tough time it, but the inequality is valid as theta and tan(theta) are equal at theta=0 and tangent increases faster than the angle. An easy was to see this is the graph y=tan(x) and y=x together, but you can also see this algebraically by comparing the derivatives. d(x)/dx=1 everywhere and d(tan(x))dx=(sec(x))^2 has a minimum at x=0 therefore tan(theta) must always be greater than or equal to theta. Technically this only proves this from 0 to pi/2 but a similar argument can be made for the limit approaching zero from below, just that the negatives would cancel. I hope this helps :)

[@linkmariofan8921]

@@MarkVsharK But you can't use the derivative of tan as proving what its derivative is requires the limit

[@MarkVsharK]

Linkmario Fan I wasn’t so much trying to prove the final result of the limit in the video, but rather show that inequality is valid for anyone who didn’t want to accept it without proof

[@haakoflo]

Define: exp(i*phi) = exp(i*theta*n) = (exp(i*theta))^n => lim(n->infinity=>theta->0) [ (1+i*theta*n/n)^n = (exp(i*theta))^n ] | By one of several common definitions for exp(x) => lim(theta->0) [ (1+i*theta) = (exp(i*theta)) ] | using principal roots(assumes phi < pi): => lim(theta->0) [Im(1+i*theta) = sin(theta)] | using Euler's formula: => lim(theta->0) [sin(theta)/theta = 1]

[@תומרסבן]

Or we can just prove that (sinx)'=cosx in another way...

[@ihave3heads]

The difference quotient is the DEFINITION of the derivative.

[@Gold161803]

@@ihave3heads You can still choose to represent that difference quotient algebraically or graphically, which can result in different proofs

[@kimothefungenuis]

3b1b actually did it

[@gregoriousmaths266]

Noooooo my worst nightmare haha Why don’t we just compromise and use the Taylor series expansion instead

[@David-km2ie]

True, or we choose another definition of sin like the series def

[@L开门见山]

After years of my calcus lessons now I wonder why x

[@RalphDratman]

That is great! But the answer was the same when you did it the wrong way! Circular reasoning gave the right answer. Is there an example where circular reasoning gives the wrong answer?

[@blackpenredpen]

Hmmmm that's interesting and I will think about it!

[@jacksainthill8974]

The simplest form is probably this, I think: 0 = 1, therefore 0 = 1.

[@alekseikhalin5808]

I doubt circular reasoning can produce wrong answers, the problem with it is different. By using circular reasoning, you try to prove a statement by using something, that requires that exact statement to be true; one can say you try to prove a statement true by saying it is true. It doesn't produce anything useful to mathematics, which is a goal overall. Actually, if you start with the wrong statement, by using circular reasoning, you might prove the wrong statement to be true; if that's what you were wondering, then yes.

[@Tetration]

If your premise is wrong, then your conclusion will also be wrong (assuming the rest of the argument works). In circular reasoning, the conclusion is a premise used to prove the conclusion. So the premise is wrong if and only if the conclusion is wrong. If for example, the limit as x->0 of sinx / x = 1/2, then the derivative of sinx would be: lim h->0 of cos(x + h/2)(sin(h/2)/(h/2)) = 1/2*cosx. Then we would have: lim x->0 of (d/dx sinx)/(d/dx x) = lim x->0 of 1/2*cosx/1=1/2*cos(0) = 1/2 This would “prove” that the limit is 1/2 but relies on the knowledge that the limit is 1/2. In short, the truth value of the premise is equivalent to the truth value of the conclusion for circular reasoning.

[@iwersonsch5131]

I am infallible because this statement says so. This statement is correct because I am infallible.

[@aelialaelia477]

man this is brilliant I'm bingewatching your whole channel

[@tigtig1873]

if we use definition of differential like this lim(x→0) sin(θ) - sin(0) / θ - 0 it gives the answer!!

[@patrykp.3731]

can you give proof that the derivative of sin is cos without using a limit of sin(x)/x at 0?

[@OrekunChen]

@@patrykp.3731 We just use e^iθ = cos(θ) + i sin(θ) then, (e^iθ)' = ie^iθ = -sin(θ) + i cos(θ) = cos'(θ)+ i sin'(θ). (I'm derivating over dθ) By identification of real and imaginary part, we have the conclusion. Then you can ask me why we have the first equality. For me, that's just a definition. And i'm pretty sure that we dont need any derivative to demonstrate the existence of that thing.

[@JustMaiyak]

I was desperately looking for this comment

[@Slecker95]

How do you know or prove the inequality sin < theta < tan besides looking at the picture. My point is that even looking at the picture it doesn't seem like that's necessarily the case (sin looks to be the same length as the arc length) and I'm wondering if there's a rigorous or analytical of showing that relationship.

[@akshataggarwal4002]

Hey bprp thats you proving sandwich theorem

[@איתןאיילשטיין]

5 years later, i have simpler solution, When the angle approach 0 so do the redian. When the angle is 0 sin(0) is equal to the redian. So when the angle approaches 0 the sin(angle) aproches the redian. So its 1

[@coloboquito]

In russian squeeze theorem is called "the theorem about two policemen" (теорема о двух милиционерах). Those two policemen are capturing the criminal between them, forcing him to come where they want

[@soubhagyarajkhandual]

In India its called "the Sandwich theorem". I didn't know that a single theorem can have different names in different countries.

[@arthur-vi1ub]

In France it is called the "Policemen's theorem"

[@astreroid118]

Thank you so much for this! I missed a day of school and was stumped until I saw you video.

[@kalu7655]

This has never actually happened to me. In fact, we pretty much established this limit as a fact on one of our classes and just continued on with much harder problems. I think I prefer the way we did that because it lets us focus on less trivial stuff.

[@okiharyonugroho2750]

In 30 years of my life, I never felt this clear

[@semiawesomatic6064]

Sinθ≈θ for small θ. So the limit becomes θ/θ is 1

[@JensenPlaysMC]

circular reasoning again

[@KelfranGt]

Is this trick acceptable?

[@Sam_on_YouTube]

I was wondering the same thing. It looks like you are more or less proving the small angle approximation. On a test, why wouldn't you presume that is already known?

[@bleppss2769]

Sam When you are first introduced to these concepts you are tested on knowledge on the subject area so you need to show an understanding of it. In classes where you need to know things like that going into the class you might be able to get away with it

[@Sam_on_YouTube]

@@bleppss2769 I did this stuff in school 20 years ago now, so it is hard to remember what you would already know as of the point you get any particular question. Thanks.

[@joshuamendez425]

i watched this last year when I was in calc 1 and idk why it didn't click but now that I'm calc 3 i was brushing up on limits and this was crystal clear, thanks blackpenredpen

[@mafazabrar4349]

oh, I think I know how to solve this one, I just have to- BPRP: HAS THIS EVER HAPPENED TO YOU

[@sebastianaspheim]

There was a question like this on a old norwegian exam i was practicing on that used the the areas to end up with the inequality actually! Was a great question to learn from that really tests ones understanding of limits, triginometry, and math in general.

[@aleladebiri]

I feel it's time for an epsilon - delta proof of this.

[@cgandcats]

looking for that

[@kee1zhang769]

the arc length θ is less than tanθ requires justification. Compare the area of the sector and the outer triangle formed by tanθ

[@YourPhysicsSimulator]

"This one right here: one, one is always one, so this is one". 8:42 -Blackpenredpen, 2018.

[@孙林可]

The part of tanθ > θ is real hand-waving. You should say "leave to viewers as an exercise".