Hope you all enjoyed! Don't forget to check out the sponsor of the video: ►Get my favorite, free calculator app for your phone or tablet: MAPLE CALCULATOR: www.maplesoft.com/products/maplecalculator/download.aspx?p=TC-9857 ►Check out MAPLE LEARN for your browser to make beautiful graphs and much more: www.maplesoft.com/products/learn/?p=TC-9857
I am really curious about what you have to say regarding this limit, but can't bring myself to watch the video, because I disagree per se with the premise that anything in math is more important than anything else. Do you really mean that? or is that a misguided attempt at click-bait? I guess it doesn't really matter because either way it costs you credibility.
What exactly do you think sin(x)/x means? We know that sin(x) is a ratio, but the denominator is an angle. So it would make no sense to divide by degrees because sin(x) is defined in terms of radians. Yes, I know that degrees can be converted to radians, BUT once you find the ratio, that is, sin(x), you can only divide it by *radians* , not by degrees! For otherwise, you would have units/deg which is nonsense. I think you agree with me so far. At least I hope you do! Newton's greatest accomplishment was to find a series for sin(x). He did this *entirely* without the use of any calculus. The series is given by: sin (x) = x - x^3/3! + x^5/5! - ... So, sin(x)/x = [ x - x^3/3! + x^5/5! - ...]/x = 1 - x^2/3! + x^4/5! - ... = f(x) 1=f(0). As you can see, there is nothing about taking limits anywhere there. Therefore, your claim that sin(x)/x requires limit is obviously false, yes? This incorrect thinking of mainstream math academics stems from their inability and failure to understand the concept of number. The Ancient Greeks rejected the concept of zero magnitude. For one thing, most of the propositions in Book V which establish the arithmetic operations of subtraction, addition, division and multiplication would be untrue if "zero magnitudes" were included. Take proposition 12 of Book V: 9/12 = (9-3)/(12-4) = 6/8 = (6-3)/(8-4) = 3/4 =? (3-3)/(4-4) = 0/0 ? See, that would be nonsense. Therefore, no zero magnitude is allowed. The expression sin(x)/x is not in irreducible form which is given by f(x) = 1 - x^2/3! + x^4/5! - ... This same errant thinking is evident in expressions such as (x^2 - 1)/(x-1) whose irreducible form is x+1. It's preposterous to even assume that one can contemplate 0 as part of actual numbers, because 0 is not a true number. While 0 is extremely useful, students need to be informed of these facts which are usually waived over by the circular nonsense of limit theory. dy/dx is a symbolic fraction and it is entirely correct to write dy/dx = dy/dt * dt/dx. In the first place, we could not write dy/dx unless both dy and dx refer to numbers and 0 is merely a place-holder, not a number. So, it's very foolish to even talk about sin(x)/x if x is 0, unless sin(x)/x is in irreducible form as explained above. Otherwise, all you are doing is punching a hole in the function f(x) = 1 - x^2/3! + x^4/5! - ... which is clearly well defined when x=0, that is, you're removing x=0 from the domain which is very silly indeed.
I was quite relieved when you pointed out that proving l'hôpital's rule requires proving this limit from definitions to avoid circular argument. Also quite impressed by the correct use of the phrase "begging the question" Have a like
@@fix5072 okay, but find me a proof that d/dx sin(x)=cos(x) that doesn't use L'Hôpital's rule. If we use a result that is a consequence of L'Hôpital's rule to prove L'Hôpital's rule, that's a tautology and is therefore not a proof. It is an example of a formal logical fallacy called "petitio principii" which directly translates from Latin as "begging of the principle", gradually bastardized in English to "begging the question"
@@fix5072 the problem is not in fhe proof of l'Hospital rule. The problem is how do you know what is the derivative of sin(x) equal to which you need to know to apply l'Hospital's rule.
@@fix5072 ofcourse without l'Hospital - as it is just in the video. The problem is how you prove that without that limit - and using/calculating the limit without the knowledge of sin'(x).
Amazing eloquence and clarity. I was hitting my head on a brick wall trying to explain this and prove this limit to my daughter when I searched on youtube and landed on your playlist. My prayer was answered. Thanks a lot Dr. Trefor! May God bless you for spreading your knowledge!
My favorite way: Inscribe a regular n-gon in the unit circle. The length of one side is 2sin(π/n), so the perimeter is 2n*sin(π/n). As n -> infinity, this should approach the circumference, 2π. Thus sin(π/n)/(π/n) -> 1. Letting x = π/n, we conclude sin(x)/x -> 1 as x -> 0.
Nice wrong proof haha. Showing that there is a particular sequence such that the limit works doesn't imply that it will work for every sequence. Take the sequence (-1^n), of course I could go to infinity by "letting x = 2k" (k natural) and them of course it would be "proved" that lim (-1^n) = 1 right?
The best thing I like about your videos are that they are short n having graphs n pictures that fix the concepts into long term memory.Your passion is infectious.
I remember using this trick a dozen times in my electrical engineering classes. In complicated electrical systems it pays a lot of dividends to ignore terms right out, so pretty often we'd have an equation for an AC system in the form of: lim x>0 (______ / x ) * (sin(θ) / x) or lim x>0 ________ + cos(θ) where the sin or cos term describes the alternating current/voltage. Since we needed to do transient math (e.g. what happens in the first half cycle when you turn on this motor circuit at 60 Hz?) the sinusoidal nature of the source really didn't matter. We could just cross those trig terms out and just look at the transient math.
Great video! I often find myself looking for a correct and accessible proof of a theorem to show someone, but so many proofs on TH-cam are full of handwaving and mistakes. Your videos are always spot on.
Since we devide by sin(x) you have to consider not only one, but two cases: 1) x > 0 then sin(x)> 0; 2) x < 0 then sin(x) < 0. The sign of sin(x) changes the two inequalities, but in the end the proof is the same. Also we have to mention why from the left and right limit having the same value (in this equal to 1) we would have that the limit also has that value. Your videos are very inspiring! I don't want to judge you, but help to make them even better! :)
I suppose you can hack it by making a geometric argument that sin(x) < 0 if x is small and x < 0, i.e. the signs of x and sin(x) are the same (for small x).
Wanted to bring up, in case nobody else has, that when I looked into this limit through a video on blackpenredpen's channel, he goes over how this limit is actually used in the development of the definition of the derivative of sin(x), so we actually can't use L'Hopital's rule to work out that it's 1. I suppose that since it still gives us the correct answer to use that logic, it's still technically viable to use it as a calc student, since every calc student knows d/dx (sin(x) = cos(x), and it's just simpler to use L'Hopital's rule than to work out this entire proof. Still, very excellent explanation here, glad to find yet another great math channel to watch content from! I've tutored math for most of my life, and watching content like this and seeing the many ways to explain these kinds of concepts helps me become better at tutoring. Keep up that enthusiasm, you're helping inspire people like me!
I feel that this proof is more elegant since it only relies on a basic property of length and some elementary geometry, whereas the standard proof uses area, algebra, formulas related to the measurement of circles.
Such a great video! Just learned about paraxial approximations for optical systems and small angle approximations for calculating vertical distances of bands in single-slit diffraction in my first year under-grad physics courses! The rigor in proof of the small-angle approx you showed here makes me feel more confident that these approximations weren't ad hoc (even though I knew they had to come from somewhere!)
At first I thought that you could omit the geometry by applying Euler's formula for the complex exponential and the Taylor-Maclaurin series for that function plus a few symmetry properties for sine and cosine but then you will sadly end up with a divergent Laurent-series for the limit to zero. Anyway, a very nice piece of geometric proof: elegant, straighforward and nowhere near complicated! Very suitable for a first year's calculus course in college, I reckon. 👍
I'm glad you mentioned the limit definition of cosine after mentioning L'hopital's Rule. Too often, students rely on L'hopital as proof, though they're relying on circular logic to find a solution. Just stating "we know the limit as x->0 of sin(x)/x =1" is sufficient for most problems. Thanks for the enjoyable video!
I used Maple waaaay back in my undergrad years but I ended up using Matlab and Mathematica much more. That said ... this calculator app is awesome!! Thanks for sharing! And, as always, great math lesson!
I am from India and I did my graduation from IIT and when I was solving calculus I used to think about this proof and I tried a lot geometrically and ended up with nothing but I came to know the proof now and sadly I did the same back then many times but didn't come with the final answer thanks a lot Sir for remembering my JEE days
You actually can't use lhopital's rule in this case, because the derivative of sin(x) at x=0 is lim x->0 (sin(x)-sin(0))/(x-0) = lim x->0 sin(x)/x, making the argument circular.
You can actually use lhopitals rule. It's like how you can have a recursive function as long as you have a base case. If at some point in your circular argument you use the other proof for this limit, you can substitute it back in and resolve this circular "recursive" proof.
@@gamingbutnotreally6077 so your argument is that we can use l'hopital's rule to prove that lim x->0 sin(x)/x = lim x->0 sin(x)/x, then use some other proof to prove that lim x->0 sin(x)/x=1? I mean, technically you're correct, but step 1 still does nothing
I thought that in lhopital you have the limit of a fraction (where both numerator and denominator converge to 0) and apply the derivative to both numerator and denominator, giving you here lim x->0 cos(x)/1, which is simply 1.
Dr TB this was a fantastic video. I love the explanation as it clear and to the point without being too verbose. Love it and thank you for sharing your knowledge and empowering others to learn.
As an alternative, you could use the Taylor series for sin(x) and divide by x. In the limit as x -> 0, the lead term (x/x = 1) is the only term to survive and the limit is 1.
True, but the standard way to derive the taylor series for sin(x) involves differentiating, which involves knowing this limit to begin with so we fall into the same circular trap. There are lots of ways to build up a calculus framework, and you can solve a lot of problems by permuting the order of how you build things up, but ultimately geometric arguments need to be input somewhere.
@@DrTrefor we in germany, in real analysis (We dont really have calculus classes), usually define e^x, sinx, cosx by its taylorseries so it can be valid ^^
@@philippg6023 Regardless of where you start your definition of the trig functions, at some point you need to make a connection between any power series definition of sin(x), and the usual geometric one (whether the y-coordinate on the unit circle of an angle CCW from +x, or whether it is the opposite over hypotenuse). So some proof equivalent to the proof of this limit will be needed, somewhere in the process to prove this connection.
You can write sin(x) - as exponential function then take the derivative of exponential. Since sin(x) = (e^ix - e^-ix)/(2i) it will solve the issue. Then you can go back and find the limit either by definition or L'Hôpitals.
Are there ways of proving Euler's formula, that don't already depend on this limit? Because that could ultimately be seen as circular reasoning. The most common proof of Euler's formula uses the Taylor series of e^x, and evaluates it at x=i*theta, and compares it to the Taylor series of sin(theta) and cos(theta). Since this requires using the derivatives of trigonometry that both depend on this limit to be established, it would be circular reasoning to use Euler's formula unless there were an independent proof of it.
The geometric proof was SO helpful. I'm more of a visual learner, and my teacher's explanation made no sense. Thanks for helping me not fail honors calc👍
my favorite way of proving this limit is to, instead of using the OCD triangle, draw the arc of radius cos x and length x*cos x from A to the hypotenuse of the right triangle. From there, you can make the (kinda sus) claim that x*cos x ≤ sin x ≤ x ⇒ cos x ≤ sin x/x ≤ 1, as x →0, 1 ≤ sin x/x ≤ 1. More rigorously, you could use the areas of the sectors and the right triangle, since it's evidently true from the geometry that the area of the triangle is in-between the areas of the internal and external sectors. Therefore x/2* cos^2 x ≤ 1/2*cos x*sin x ≤ x/2 ⇒ cos x ≤ sin x/x ≤ 1/cos x, as x → 0, 1 ≤ sin x/x ≤ 1
it actually depends how you define sin(x). it's common in analysis to define sin(x) as it's Maclaurin expansion and then use the simplification of dividing all the terms by x just leaving 1 + O(x) which goes to 1 as x goes to 0. it then becomes an exercise to prove that the Maclaurin series actually has the geometric properties of sin.
Sure. At some point you have to make a geometric argument to connect the analytic and geometric sides of sin together, but there are multiple starting spots to do that.
It's actually really funny that in Baby Rudin, pi is defined as 2 times the smallest value for which cos is 0. Crazy backwards from the historical order of discovery.
You can use L'Hospital on sin(x)/x, just have to use a different approach for the proof of L'Hospital. Saying you cannot use L'Hospital on sin(x)/x may lead to a false understanding on the rule itself
The equations sin^2t + cos^2t = 1 and (dsint)^2 + (dcost)^2 = (dt)^2 give us (sin'(0))^2 = 1. We can conclude the correct value is the positive one because the derivative is clearly positive at zero.
I like to simply use the definition of a radian, which is an arc length of one radius along the perimeter of the circle. Since sin(x) is the height of a point along the circle as a function of the angle, and at very small angles, we can see that the edge of the circle is almost completely vertical, we can intuitively infer that for very small angles, the height of any point along the circle is approximately equal to the distance you have moved along the circle from an angle of 0, since for very small angles, you will be moving close to vertically. The distance you have travelled along the circle then also happens to be equivalent to the angle, as long as you use radians as the angle and measure height by number of radii(which for small numbers should be a small portion of the radius).
6:27 it's this part here that's the reason the limit only holds for sin measured in radians. And consequently, why d sin(x)/dx = cos(x) only holds in radians.
Fun fact: the formula at 2:20 already uses a small angle approximation. Which one? Well, the supernova remnant is approximately a sphere. Those lines are tangent to the sphere, and it's a well known fact from geometry that: tangent lines are perpendicular to radii. A little parallel line reasoning reveals that the base of that triangle in the diagram cannot be the diameter of the sphere. What's the correct formula? I can't draw a diagram in a TH-cam comment, but it's: tan(delta/2) = (d/2) / sqrt(D^2 - (d/2)^2), or sin(delta/2) = (d/2) / D.
The easiest way to see this is to recognize that the tangent line of a continuous function at a given point is a fantastic “local” estimate of that function. Thus, sine of x at zero has a slope of one and the line y=x is tangent to sine of x at that point. We might as well say that the limit as x goes to a of any differentiable function of one variable (sans some possible pathologies) has the property that f(a)/(f’(a)(x-a)+f(a))=1
Draw a unit circle and a random angle θ Let A be the point on the unit circle where the OA has a angle θ from the x axis. Let B be the shadow of line A, and C be the point (1,0). We have OAB(triangle)
@@Nino-eo8ey the shadow of a point on the xy plane is the point S which has the same x coordinate but 0 as the y coordinate. So, the shadow of A(3,2) is S(3,0). Basically the point on the x-axis directly underneath or above our chosen point.
Do you perhaps study in Russia or a slavic country? I'm studying in Russia and we too call it the First Wonderous Limit, or in russian Первый замечательный предел
@@shernader i studied in Mongolia. I’m Mongolian. But it’s amazing to know that you call this the same there too. Probably we translated the name from Russian textbooks.
When I read the title, I considered a unit circle at the origin. The angle x would be the length of an arc starting at (1,0), and sin(x) would be the length of a vertical line from the x-axis to end of that arc. As the angle gets smaller, the difference between the two lengths gets smaller. At the limit, the arc doesn’t diverge from the vertical line. I was surprised that the video didn’t mention this (especially since it said “geometric proof”).
Well his 'pizza slice' construction is a sector of the unit circle, and he basically spends most of his time spelling out why the ratio of the arc to the vertical line is 1. Noting that they both get closer doesn't tell you anything about the limit of their ratio because they're both approaching 0, so the ratio 'at the limit' is 0/0. That's why he goes to the extra effort.
hi, I am currently in my first yr of A-level, and we(me and my physics class) were looking at young's double slit experiment last week Friday and we came across the d2/D for calculations and were told tan x ≈ sin x for very small angles; but, we weren't told why. I thought it was because, when x --> 0, cos x =1 , sin x = 0 tan x = sin x / cos x = 0/1 = 0 ∴ sin x = tan x , when x is a very small angle this is what I was thinking in my head in that lesson but graphically based on the CAST diagram from trig in my maths lesson i think this is what you were explaining in a more cool and fun way, I think (I didn't fully understand so i'm gonna rewatch this to understand and visualise this concept in my head.) also i think your videos are amazing and they don't make me feel like I need to wait until uni and 'out of my calibre' gutted feeling. (I wish i found out about this during gcse) Thank you! 😊😊
Fun Fact: there is another way to approach the derivative of sin(x) without taking a limit, and by that I mean take the half derivative of sin(x) twice which leads to the first derivative of sin(x) in this case we find sin(x+1/2Pi) which is the same as cos(x), this approach uses the gamma function which uses the integrals instead of limits.
Using trig identities and the knowledge of the limit of sin x/x as x tends to zero, the limit (cos h - 1)/h as h tends to 0 can be found without a geometric argument
This reminds me of my university interview! One of the things they asked me was if I could draw a graph of y=(sin x)/x (Now I see why it's so important!) and I had to work it out a step at a time using the graphs of y=sin x and y=x, so I worked it out geometrically. I don't remember if I used calculus to work out the value of y at x=0 or the small angle approximation. By the way it would have been good if you mentioned angle conversion factors for the astonomical example because if the small angle is in minutes or seconds of arc it needs to be converted to radians. Wow, I didn't know you could get calculators or calculator apps that can do calculus! I have a programmable calculator so it can be used for integration using a numerical approximation, which gets tedious with too many iterations for smaller 'dx' intervals Have you done a video on integrating the length of a curve (of any given equation)? I was curious to see how to work out the length of the line drawing out a sine wave, but it gets really complicated and I could only do it with a numerical approximation on Excel.
This such a fantastic argument/proof. I feel like the way it was explained here was tailored for me 😂 I tried to do this myself (not realising the geometric/equality trick described here)...i was unsuccessful 😅
most importemt limit is the seiries for e to the x as it basicly proves everything. the whole sin thing while cute is mostly unesscery once you intruduce complex numbers andthe diffrentiol equtions in phisics that use sine could just be solved with complex exponentiols
personally the easiest way to prove this formula is just to open excel, put x as 0.1...push x closer and closer to 0, and look at the trend of sinx/x. it will definitely be 1. Sometimes, i have no idea why they teach this in school when we can easily just use excel
I was doing a proof of this in Calc I when a hand went up at the back. The question was: How do you know that the area of an enclosed region is less than or equal to the enclosing area? I thought for a moment and could not come up with the explanation at that time, so I told them that I wanted to say it was obvious, but that you can't use that in a mathematical proof. I told them to just assume that it was true so we could finish the proof, and I would supply the answer the next day. The answer is that this is inherent in the definition of area. Though I did not say it to them, it is the sub additivity of a measure, and area is a measure . Technically, one has to assume that both regions have a measure, but that is beyond Calc I concepts. I also said to the person that he might think of math as a major, since he was thinking like a mathematician. This incident was also prominent in my evaluations that semester, in that I did not just dismiss the question, but treated it as I would from a colleague.
It's a really nice proof, thank you for making this video and introducing us to maple app. Just there is something that is not quite right in the physic example. We are not able to determinate d unless we have found D. And to find D (the distance earth-supernova) we use a method called parallax, that consist of taking the triangle vertex from 2 different places on earth (and for some cases 2 mesures from the same place on earth but the second 1 is taken after 6 months, so we can maximize the distance between these 2 points to 2x sun-earth distance). So now we know the distance between our 2 points and the angle delta, now we can make the calculations as you told us to.
For the derivative of sin(x), you don't need to use sinx/x to evaluate it. You can use the definition of sine, move the angle for a small amount and observe the change. th-cam.com/video/S0_qX4VJhMQ/w-d-xo.html Go to 14:51
Fun fact on that proof: it's a bit misleading to call (let t = theta here) dt to be around the circumference, since it should be a change in where the angle t is, however as the change in angle by dt in the correct location results in a change of dt times the radius of the circle in the circumference and said radius is 1 it works. If the same proof was done using a circle of radius 2 it'd be a triangle of hypotenuse 2dt and the other side 2d(sin(t)).
If I become a math professor or tutor, I will prove the derivative first, then show the limit and the angle-sum identities. The proof (via the function definition of arc length) is as follows: Imagine a helix around the x-axis, turning forward as x increases. Dilate it so the radius is 1, and then move it to the right so that it touches (0,1,0) and put a tangent line there. Now let f(x)=[y,z]. As you move to the right, the helix rotates, and so its position and derivative rotate at the same rate without changing in magnitude. Therefore, f’(x) is just f(x) rotated and scaled. If you let x=0, y is at its peak (think of the unit cylinder: y^2+z^2=0): y=1, y’=0, z=0, z’=some constant k, inversely proportional to the value of one rotation (because if we stretch the graph on the x-axis by s, the derivative divides by s). Also if you move until z is at its peak, z=1, z’=0, y=0, y’=-k. Therefore the angle between f(x) and f’(x) is a right angle. This means that the derivative of sin x is k*cos x and the derivative of cos x is -k*sin x, but how do you prove that k=1 when x is in radians? sin^2 x+cos^2 x=1. If k=1, then d/dx cos x=cos(x+1τ/4)=-sin x=-sqrt(1-cos^2 x) for sin x>=0, which holds for 0 (which is arccos 1)
Sir, some teachers argue that we should not use L' Hospital's Rule here because this limit is used to find derivative of cosx. This becomes a circular arguement then.
I really like the way you plotted graph. Can you please make a tutorial on how to install this software for plotting multiple graphs together. It's a request 🙏🙏
Ah, but what is the definition of the derivative of sinx at 0? Why, that's (sin(x)-sin(0))/(x-0) as x approaches 0. sin(0) is 0, so we're left with the derivative being sinx/x as x approaches 0. Do you see why that might not be a very good argument here?
Piggybacking on Athhenri's comment, consider 1. f (x) = [(cos x - 1) / x] , and , when x = 0, the numerator is 0, the denominator is 0, and the ratio, by your argument, is 1. But lim x->0 f(x) = 0. So no, one can't simply plug in the numbers. 2. g (x) = ln (3*x + 1)/ x. When x = 0, 3*x + 1 = 1, so ln (3*x+1) = 0. so, by your argument, 0/0 = 1. But, lim x-> 0 of g(x) = 1/(3*0+1)*3 (by L'hopital) = 3. Not 1. Need you more examples? The power, and beauty of calculus.
The million dollar question: Can you use L'Hopital's Rule for this limit? For instance, Michael Penn regularly uses it, but blackpenredpen was very emphatic NOT to use L'Hopital's Rule because according to him, it is circular reasoning since to determine the derivative of sin(x), it uses the lim of sin(x)/x. What are your thoughts on this issue?
I must take issue with your statement(at 8:25) that "you could just prove that lim[x->0] (sin x/x) = 1 using l'Hosptal's rule." We would have to know that the derivative of sin x is cos x. But how did we learn that? Did we not need this limit to carry out the proof? We can prove that the derivative is sin x in either of two "styles". (It's not two ways, just two styles): If y = sin x, then (view it in Courier or other fixed-size font) sin(x + h) - sin x dy/dx = lim[h->0] ------------------- h sin x cos h + cos x sin h - sin x dy/dx = lim[h->0] --------------------------------- h sin x cos h - sin x cos x sin h dy/dx = lim[h->0] -------------------- + ------------- h h (cos h - 1) cos x sin h dy/dx = lim[h->0] sin x ------------ + lim ------------- h h Now, since lim[h->0] sin h/h = 1, supposedly already proved, and, since lim[h->0] (cos h - 1)/h = 0, already proved, using that same limit, dy/dx simplifies to cos x. OR do it like this: [I remember my Calc I teacher using this style] sin r - sin x dy/dx = lim[r->x] -------------- r - x Using a "sums-to-products" identity" 2 cos((r+x)/2) sin((r-x)/2) dy/dx = lim[r->x] ---------------------------- r - x sin((r-x)/2) dy/dx = lim[r->x] cos((r+x)/2) lim[r->x] ------------ (r-x)/2 When r->x, the first factor is just cos x. the second is the same limit (equals 1)
Approximations trigger my OCD =P What really blows my mind is how calculator works. At its base it's 1s and 0s. So it operates basically with +1 and -1. And you can calculate multiplication, roots, powers, sin\cos, integrals, EVERYTHING translating it into +1s and -1s. All the math that we know and that we can calculate via computer goes down to simple clicks of relays. On and off. Absolutely insane. I have no idea how it works and I'm afraid I'd lose my mind if I ever figure it out. So I prefer to believe it's magic. Just kidding. No, really, it's amazing to me.
Hope you all enjoyed! Don't forget to check out the sponsor of the video:
►Get my favorite, free calculator app for your phone or tablet: MAPLE CALCULATOR: www.maplesoft.com/products/maplecalculator/download.aspx?p=TC-9857
►Check out MAPLE LEARN for your browser to make beautiful graphs and much more: www.maplesoft.com/products/learn/?p=TC-9857
yes, very good explanation. I will have justified the Tangent part a little bit more, but the video was already long. You got a new subscriber.
#AD
I am really curious about what you have to say regarding this limit, but can't bring myself to watch the video, because I disagree per se with the premise that anything in math is more important than anything else. Do you really mean that? or is that a misguided attempt at click-bait? I guess it doesn't really matter because either way it costs you credibility.
Maple calculator fails to graph sin(x)/x when x is close to zero. Sponsor fail 😵💫
What exactly do you think sin(x)/x means? We know that sin(x) is a ratio, but the denominator is an angle. So it would make no sense to divide by degrees because sin(x) is defined in terms of radians. Yes, I know that degrees can be converted to radians, BUT once you find the ratio, that is, sin(x), you can only divide it by *radians* , not by degrees! For otherwise, you would have units/deg which is nonsense.
I think you agree with me so far. At least I hope you do!
Newton's greatest accomplishment was to find a series for sin(x). He did this *entirely* without the use of any calculus. The series is given by:
sin (x) = x - x^3/3! + x^5/5! - ...
So, sin(x)/x = [ x - x^3/3! + x^5/5! - ...]/x = 1 - x^2/3! + x^4/5! - ... = f(x)
1=f(0).
As you can see, there is nothing about taking limits anywhere there. Therefore, your claim that sin(x)/x requires limit is obviously false, yes?
This incorrect thinking of mainstream math academics stems from their inability and failure to understand the concept of number. The Ancient Greeks rejected the concept of zero magnitude. For one thing, most of the propositions in Book V which establish the arithmetic operations of subtraction, addition, division and multiplication would be untrue if "zero magnitudes" were included. Take proposition 12 of Book V:
9/12 = (9-3)/(12-4) = 6/8 = (6-3)/(8-4) = 3/4 =? (3-3)/(4-4) = 0/0 ?
See, that would be nonsense. Therefore, no zero magnitude is allowed.
The expression sin(x)/x is not in irreducible form which is given by
f(x) = 1 - x^2/3! + x^4/5! - ...
This same errant thinking is evident in expressions such as (x^2 - 1)/(x-1) whose irreducible form is x+1. It's preposterous to even assume that one can contemplate 0 as part of actual numbers, because 0 is not a true number. While 0 is extremely useful, students need to be informed of these facts which are usually waived over by the circular nonsense of limit theory.
dy/dx is a symbolic fraction and it is entirely correct to write dy/dx = dy/dt * dt/dx.
In the first place, we could not write dy/dx unless both dy and dx refer to numbers and 0 is merely a place-holder, not a number.
So, it's very foolish to even talk about sin(x)/x if x is 0, unless sin(x)/x is in irreducible form as explained above. Otherwise, all you are doing is punching a hole in the function f(x) = 1 - x^2/3! + x^4/5! - ... which is clearly well defined when x=0, that is, you're removing x=0 from the domain which is very silly indeed.
I was quite relieved when you pointed out that proving l'hôpital's rule requires proving this limit from definitions to avoid circular argument. Also quite impressed by the correct use of the phrase "begging the question"
Have a like
There are multiple proofs of L'Hospitals rule, also ones not using that limit, so it's completely fine to use L'Hospital on sin(x)/x
@@fix5072 okay, but find me a proof that d/dx sin(x)=cos(x) that doesn't use L'Hôpital's rule. If we use a result that is a consequence of L'Hôpital's rule to prove L'Hôpital's rule, that's a tautology and is therefore not a proof. It is an example of a formal logical fallacy called "petitio principii" which directly translates from Latin as "begging of the principle", gradually bastardized in English to "begging the question"
@@fix5072 the problem is not in fhe proof of l'Hospital rule. The problem is how do you know what is the derivative of sin(x) equal to which you need to know to apply l'Hospital's rule.
@@nasserallammah2035 you can prove sin(x)'=cos(x) without L'Hospital. I can give you the proof of you want
@@fix5072 ofcourse without l'Hospital - as it is just in the video. The problem is how you prove that without that limit - and using/calculating the limit without the knowledge of sin'(x).
Mathematics is beautiful, but not everyone is able to reveal that beauty. Thanks 👍 for this beautiful brain teaser
Absolutely right❤️I love maths very much
Amazing eloquence and clarity. I was hitting my head on a brick wall trying to explain this and prove this limit to my daughter when I searched on youtube and landed on your playlist. My prayer was answered. Thanks a lot Dr. Trefor! May God bless you for spreading your knowledge!
My favorite way: Inscribe a regular n-gon in the unit circle. The length of one side is 2sin(π/n), so the perimeter is 2n*sin(π/n). As n -> infinity, this should approach the circumference, 2π. Thus sin(π/n)/(π/n) -> 1. Letting x = π/n, we conclude sin(x)/x -> 1 as x -> 0.
ooh wow!
oooh I like that!!
My teacher told me in his class, hehe
Great😀😀
Nice wrong proof haha. Showing that there is a particular sequence such that the limit works doesn't imply that it will work for every sequence.
Take the sequence (-1^n), of course I could go to infinity by "letting x = 2k" (k natural) and them of course it would be "proved" that lim (-1^n) = 1 right?
The best thing I like about your videos are that they are short n having graphs n pictures that fix the concepts into long term memory.Your passion is infectious.
I remember using this trick a dozen times in my electrical engineering classes. In complicated electrical systems it pays a lot of dividends to ignore terms right out, so pretty often we'd have an equation for an AC system in the form of:
lim x>0 (______ / x ) * (sin(θ) / x) or lim x>0 ________ + cos(θ)
where the sin or cos term describes the alternating current/voltage. Since we needed to do transient math (e.g. what happens in the first half cycle when you turn on this motor circuit at 60 Hz?) the sinusoidal nature of the source really didn't matter. We could just cross those trig terms out and just look at the transient math.
These kind of videos are so interesting; the tricks, intuitions, and algebraic manipulations are so beneficial!
Great video! I often find myself looking for a correct and accessible proof of a theorem to show someone, but so many proofs on TH-cam are full of handwaving and mistakes. Your videos are always spot on.
Since we devide by sin(x) you have to consider not only one, but two cases: 1) x > 0 then sin(x)> 0; 2) x < 0 then sin(x) < 0. The sign of sin(x) changes the two inequalities, but in the end the proof is the same. Also we have to mention why from the left and right limit having the same value (in this equal to 1) we would have that the limit also has that value.
Your videos are very inspiring! I don't want to judge you, but help to make them even better! :)
I suppose you can hack it by making a geometric argument that sin(x) < 0 if x is small and x < 0, i.e. the signs of x and sin(x) are the same (for small x).
you are the most passionate math educator in youtube.
Wanted to bring up, in case nobody else has, that when I looked into this limit through a video on blackpenredpen's channel, he goes over how this limit is actually used in the development of the definition of the derivative of sin(x), so we actually can't use L'Hopital's rule to work out that it's 1. I suppose that since it still gives us the correct answer to use that logic, it's still technically viable to use it as a calc student, since every calc student knows d/dx (sin(x) = cos(x), and it's just simpler to use L'Hopital's rule than to work out this entire proof.
Still, very excellent explanation here, glad to find yet another great math channel to watch content from! I've tutored math for most of my life, and watching content like this and seeing the many ways to explain these kinds of concepts helps me become better at tutoring. Keep up that enthusiasm, you're helping inspire people like me!
Yup! If developing calculus formally, this limit needs to be proven before derivatives and before l’hospitals rule
He said that later.
If you were to join the straight line BC, a triangle would be formed in which the angle BAC is right. Therefore AB
I feel that this proof is more elegant since it only relies on a basic property of length and some elementary geometry, whereas the standard proof uses area, algebra, formulas related to the measurement of circles.
Such a great video! Just learned about paraxial approximations for optical systems and small angle approximations for calculating vertical distances of bands in single-slit diffraction in my first year under-grad physics courses! The rigor in proof of the small-angle approx you showed here makes me feel more confident that these approximations weren't ad hoc (even though I knew they had to come from somewhere!)
At first I thought that you could omit the geometry by applying Euler's formula for the complex exponential and the Taylor-Maclaurin series for that function plus a few symmetry properties for sine and cosine but then you will sadly end up with a divergent Laurent-series for the limit to zero.
Anyway, a very nice piece of geometric proof: elegant, straighforward and nowhere near complicated! Very suitable for a first year's calculus course in college, I reckon.
👍
I'm glad you mentioned the limit definition of cosine after mentioning L'hopital's Rule. Too often, students rely on L'hopital as proof, though they're relying on circular logic to find a solution. Just stating "we know the limit as x->0 of sin(x)/x =1" is sufficient for most problems. Thanks for the enjoyable video!
Man, your channel is refreshing, Im going over many things that I went over in my undergrad thinking about how much details were just glossed over
I used Maple waaaay back in my undergrad years but I ended up using Matlab and Mathematica much more. That said ... this calculator app is awesome!! Thanks for sharing! And, as always, great math lesson!
I am from India and I did my graduation from IIT and when I was solving calculus I used to think about this proof and I tried a lot geometrically and ended up with nothing but I came to know the proof now and sadly I did the same back then many times but didn't come with the final answer thanks a lot Sir for remembering my JEE days
Haha that was me in my first year, where did you study?
You actually can't use lhopital's rule in this case, because the derivative of sin(x) at x=0 is lim x->0 (sin(x)-sin(0))/(x-0) = lim x->0 sin(x)/x, making the argument circular.
I think he talked about this in the last segment of the video
@@SzanyiAtti I made it most of the way through, saw him use lhopital's rule, and got triggered. He definitely does point this out. Thank you
You can actually use lhopitals rule. It's like how you can have a recursive function as long as you have a base case. If at some point in your circular argument you use the other proof for this limit, you can substitute it back in and resolve this circular "recursive" proof.
@@gamingbutnotreally6077 so your argument is that we can use l'hopital's rule to prove that lim x->0 sin(x)/x = lim x->0 sin(x)/x, then use some other proof to prove that lim x->0 sin(x)/x=1? I mean, technically you're correct, but step 1 still does nothing
I thought that in lhopital you have the limit of a fraction (where both numerator and denominator converge to 0) and apply the derivative to both numerator and denominator, giving you here lim x->0 cos(x)/1, which is simply 1.
Dr TB this was a fantastic video. I love the explanation as it clear and to the point without being too verbose. Love it and thank you for sharing your knowledge and empowering others to learn.
Glad you enjoyed it!
gotta love that proof, so beautiful! haha i've shown this proof to several classmates but they weren't quite so thrilled.
lol people don't appreciate a nice geometric proof well enough:D
Best teacher ever.
Thank you for a clear and understandable treatment.
Outstanding! Former high school math teacher.
This is the best video about Lim x->0 (sin x)/x. Well explained sir.
As an alternative, you could use the Taylor series for sin(x) and divide by x. In the limit as x -> 0, the lead term (x/x = 1) is the only term to survive and the limit is 1.
That assumes the value of the limit being 1 in the first place.
True, but the standard way to derive the taylor series for sin(x) involves differentiating, which involves knowing this limit to begin with so we fall into the same circular trap. There are lots of ways to build up a calculus framework, and you can solve a lot of problems by permuting the order of how you build things up, but ultimately geometric arguments need to be input somewhere.
@@DrTrefor we in germany, in real analysis (We dont really have calculus classes), usually define e^x, sinx, cosx by its taylorseries so it can be valid ^^
@@philippg6023
Regardless of where you start your definition of the trig functions, at some point you need to make a connection between any power series definition of sin(x), and the usual geometric one (whether the y-coordinate on the unit circle of an angle CCW from +x, or whether it is the opposite over hypotenuse). So some proof equivalent to the proof of this limit will be needed, somewhere in the process to prove this connection.
You can write sin(x) - as exponential function then take the derivative of exponential. Since sin(x) = (e^ix - e^-ix)/(2i) it will solve the issue. Then you can go back and find the limit either by definition or L'Hôpitals.
Or u can just see that sin(x)/x= [sin(x)-0]/x-0 which converges towards sin’(0)=cos(0)=1
Are there ways of proving Euler's formula, that don't already depend on this limit? Because that could ultimately be seen as circular reasoning.
The most common proof of Euler's formula uses the Taylor series of e^x, and evaluates it at x=i*theta, and compares it to the Taylor series of sin(theta) and cos(theta). Since this requires using the derivatives of trigonometry that both depend on this limit to be established, it would be circular reasoning to use Euler's formula unless there were an independent proof of it.
Thanks for showing Maple Calculator. It rocks.
Dr Trefor Bazett, thank you so much. You are a gift to mankind.
The geometric proof was SO helpful. I'm more of a visual learner, and my teacher's explanation made no sense. Thanks for helping me not fail honors calc👍
If you construct the first triangle OBC. The area in 1/2*sinx. It makes things slightly simpler.
my favorite way of proving this limit is to, instead of using the OCD triangle, draw the arc of radius cos x and length x*cos x from A to the hypotenuse of the right triangle. From there, you can make the (kinda sus) claim that x*cos x ≤ sin x ≤ x ⇒ cos x ≤ sin x/x ≤ 1, as x →0, 1 ≤ sin x/x ≤ 1. More rigorously, you could use the areas of the sectors and the right triangle, since it's evidently true from the geometry that the area of the triangle is in-between the areas of the internal and external sectors. Therefore x/2* cos^2 x ≤ 1/2*cos x*sin x ≤ x/2 ⇒ cos x ≤ sin x/x ≤ 1/cos x, as x → 0, 1 ≤ sin x/x ≤ 1
I don't like using the area of sector formula, you can show that using some simple geometry too
it actually depends how you define sin(x). it's common in analysis to define sin(x) as it's Maclaurin expansion and then use the simplification of dividing all the terms by x just leaving 1 + O(x) which goes to 1 as x goes to 0. it then becomes an exercise to prove that the Maclaurin series actually has the geometric properties of sin.
Sure. At some point you have to make a geometric argument to connect the analytic and geometric sides of sin together, but there are multiple starting spots to do that.
It's actually really funny that in Baby Rudin, pi is defined as 2 times the smallest value for which cos is 0. Crazy backwards from the historical order of discovery.
The graph portion was so amazingly intuitive I can't believe I never encountered that in all these years of learning calculus.
I find it crazy how many teachers skip that...so easy and explains so much!
@@DrTrefor Indeed. A picture really is worth a thousand words 😃
I really freaked out when you told me to use L'Hospital's rule for this limit. Thanks for revealing the issue and not misleading students XD
You can use L'Hospital on sin(x)/x, just have to use a different approach for the proof of L'Hospital. Saying you cannot use L'Hospital on sin(x)/x may lead to a false understanding on the rule itself
Beautiful explanation. Sir, my mathematics teacher can't explain topology in an understandable way so please explain me like this awsome video.
Your videos are awesome. Thank you for the enthusiasm!
Very thanks for Maple calculator. You are really genius.
Very nicely done! Many thanks for the explanation!
It was actually good for explaining for school students because they don't know about liner approximation
this is the best sinx/x explanation I've ever seen
You need grouping symbols: sin(x)/x at least.
this is the explanation i needed at school 7 years ago
0_o
haha "just in time" :D
@@DrTrefor anyway sometimes (actually every time(night)) i'm so glad to watch random math video at 3 am. (i'm sure, i'm not alone at this. LOL)
Oh Colleage you are a great math teacher👌👏
The equations sin^2t + cos^2t = 1 and (dsint)^2 + (dcost)^2 = (dt)^2 give us (sin'(0))^2 = 1. We can conclude the correct value is the positive one because the derivative is clearly positive at zero.
I like to simply use the definition of a radian, which is an arc length of one radius along the perimeter of the circle.
Since sin(x) is the height of a point along the circle as a function of the angle, and at very small angles, we can see that the edge of the circle is almost completely vertical, we can intuitively infer that for very small angles, the height of any point along the circle is approximately equal to the distance you have moved along the circle from an angle of 0, since for very small angles, you will be moving close to vertically. The distance you have travelled along the circle then also happens to be equivalent to the angle, as long as you use radians as the angle and measure height by number of radii(which for small numbers should be a small portion of the radius).
8:02 this is one of the most perfect things I ever saw
6:27 it's this part here that's the reason the limit only holds for sin measured in radians. And consequently, why d sin(x)/dx = cos(x) only holds in radians.
Fun fact: the formula at 2:20 already uses a small angle approximation. Which one? Well, the supernova remnant is approximately a sphere. Those lines are tangent to the sphere, and it's a well known fact from geometry that: tangent lines are perpendicular to radii. A little parallel line reasoning reveals that the base of that triangle in the diagram cannot be the diameter of the sphere. What's the correct formula? I can't draw a diagram in a TH-cam comment, but it's:
tan(delta/2) = (d/2) / sqrt(D^2 - (d/2)^2), or
sin(delta/2) = (d/2) / D.
The easiest way to see this is to recognize that the tangent line of a continuous function at a given point is a fantastic “local” estimate of that function. Thus, sine of x at zero has a slope of one and the line y=x is tangent to sine of x at that point. We might as well say that the limit as x goes to a of any differentiable function of one variable (sans some possible pathologies) has the property that f(a)/(f’(a)(x-a)+f(a))=1
Just Awesome ... Thank you sir .
The calculator app is awesome, and the video is interesting too. Thank you, sir.
Love you sir😋
Gee, I was actually able to follow your discussion,
Draw a unit circle and a random angle θ
Let A be the point on the unit circle where the OA has a angle θ from the x axis. Let B be the shadow of line A, and C be the point (1,0).
We have OAB(triangle)
Thanks, but this answer is a little bit confusing, I'm not entirely sure what the shadow of line A is supposed to mean.
@@Nino-eo8ey the shadow of a point on the xy plane is the point S which has the same x coordinate but 0 as the y coordinate. So, the shadow of A(3,2) is S(3,0). Basically the point on the x-axis directly underneath or above our chosen point.
Amazing !!! Good content, very clear and well done!
In my country, we call this limit “first wondrous limit”. It’s beautiful. Made me study maths in uni.
Do you perhaps study in Russia or a slavic country? I'm studying in Russia and we too call it the First Wonderous Limit, or in russian Первый замечательный предел
@@shernader i studied in Mongolia. I’m Mongolian. But it’s amazing to know that you call this the same there too. Probably we translated the name from Russian textbooks.
When I read the title, I considered a unit circle at the origin. The angle x would be the length of an arc starting at (1,0), and sin(x) would be the length of a vertical line from the x-axis to end of that arc. As the angle gets smaller, the difference between the two lengths gets smaller. At the limit, the arc doesn’t diverge from the vertical line. I was surprised that the video didn’t mention this (especially since it said “geometric proof”).
Well his 'pizza slice' construction is a sector of the unit circle, and he basically spends most of his time spelling out why the ratio of the arc to the vertical line is 1.
Noting that they both get closer doesn't tell you anything about the limit of their ratio because they're both approaching 0, so the ratio 'at the limit' is 0/0. That's why he goes to the extra effort.
hi, I am currently in my first yr of A-level, and we(me and my physics class) were looking at young's double slit experiment last week Friday and we came across the d2/D for calculations and were told tan x ≈ sin x for very small angles; but, we weren't told why. I thought it was because,
when x --> 0,
cos x =1 , sin x = 0
tan x = sin x / cos x = 0/1 = 0
∴ sin x = tan x , when x is a very small angle
this is what I was thinking in my head in that lesson but graphically based on the CAST diagram from trig in my maths lesson
i think this is what you were explaining in a more cool and fun way, I think (I didn't fully understand so i'm gonna rewatch this to understand and visualise this concept in my head.)
also i think your videos are amazing and they don't make me feel like I need to wait until uni and 'out of my calibre' gutted feeling. (I wish i found out about this during gcse)
Thank you!
😊😊
Wow this was really well explained! Thanks so much!
Glad you enjoyed it!
Fun Fact: there is another way to approach the derivative of sin(x) without taking a limit, and by that I mean take the half derivative of sin(x) twice which leads to the first derivative of sin(x) in this case we find sin(x+1/2Pi) which is the same as cos(x), this approach uses the gamma function which uses the integrals instead of limits.
thanks a lot for the helpful proof.
this is a rly cool proof
Awesome video! Thank you!
I'm a math teacher and I didn't know about this astronomy application. Very interesting ! I will use it to explain it now :D !!
Using trig identities and the knowledge of the limit of sin x/x as x tends to zero, the limit (cos h - 1)/h as h tends to 0 can be found without a geometric argument
Beautiful!
This reminds me of my university interview! One of the things they asked me was if I could draw a graph of y=(sin x)/x (Now I see why it's so important!) and I had to work it out a step at a time using the graphs of y=sin x and y=x, so I worked it out geometrically. I don't remember if I used calculus to work out the value of y at x=0 or the small angle approximation.
By the way it would have been good if you mentioned angle conversion factors for the astonomical example because if the small angle is in minutes or seconds of arc it needs to be converted to radians.
Wow, I didn't know you could get calculators or calculator apps that can do calculus!
I have a programmable calculator so it can be used for integration using a numerical approximation, which gets tedious with too many iterations for smaller 'dx' intervals
Have you done a video on integrating the length of a curve (of any given equation)?
I was curious to see how to work out the length of the line drawing out a sine wave, but it gets really complicated and I could only do it with a numerical approximation on Excel.
This such a fantastic argument/proof.
I feel like the way it was explained here was tailored for me 😂
I tried to do this myself (not realising the geometric/equality trick described here)...i was unsuccessful 😅
Very cool geometric proof. When I saw the thumbnail the sandwich theorem just popped in my mind.
Well done.. nice proof
I loved it… Thank you
I'm glad you liked it!
most importemt limit is the seiries for e to the x as it basicly proves everything.
the whole sin thing while cute is mostly unesscery once you intruduce complex numbers andthe diffrentiol equtions in phisics that use sine could just be solved with complex exponentiols
personally the easiest way to prove this formula is just to open excel, put x as 0.1...push x closer and closer to 0, and look at the trend of sinx/x. it will definitely be 1. Sometimes, i have no idea why they teach this in school when we can easily just use excel
I was doing a proof of this in Calc I when a hand went up at the back. The question was: How do you know that the area of an enclosed region is less than or equal to the enclosing area? I thought for a moment and could not come up with the explanation at that time, so I told them that I wanted to say it was obvious, but that you can't use that in a mathematical proof. I told them to just assume that it was true so we could finish the proof, and I would supply the answer the next day. The answer is that this is inherent in the definition of area. Though I did not say it to them, it is the sub additivity of a measure, and area is a measure . Technically, one has to assume that both regions have a measure, but that is beyond Calc I concepts. I also said to the person that he might think of math as a major, since he was thinking like a mathematician. This incident was also prominent in my evaluations that semester, in that I did not just dismiss the question, but treated it as I would from a colleague.
It's a really nice proof, thank you for making this video and introducing us to maple app.
Just there is something that is not quite right in the physic example.
We are not able to determinate d unless we have found D. And to find D (the distance earth-supernova) we use a method called parallax, that consist of taking the triangle vertex from 2 different places on earth (and for some cases 2 mesures from the same place on earth but the second 1 is taken after 6 months, so we can maximize the distance between these 2 points to 2x sun-earth distance). So now we know the distance between our 2 points and the angle delta, now we can make the calculations as you told us to.
There's no way around it, math is so cool!
You can also use the squeeze theorem!
For the derivative of sin(x), you don't need to use sinx/x to evaluate it. You can use the definition of sine, move the angle for a small amount and observe the change. th-cam.com/video/S0_qX4VJhMQ/w-d-xo.html Go to 14:51
Fun fact on that proof: it's a bit misleading to call (let t = theta here) dt to be around the circumference, since it should be a change in where the angle t is, however as the change in angle by dt in the correct location results in a change of dt times the radius of the circle in the circumference and said radius is 1 it works. If the same proof was done using a circle of radius 2 it'd be a triangle of hypotenuse 2dt and the other side 2d(sin(t)).
You wrote it wrong. sin(x)/x
If I become a math professor or tutor, I will prove the derivative first, then show the limit and the angle-sum identities. The proof (via the function definition of arc length) is as follows:
Imagine a helix around the x-axis, turning forward as x increases. Dilate it so the radius is 1, and then move it to the right so that it touches (0,1,0) and put a tangent line there. Now let f(x)=[y,z]. As you move to the right, the helix rotates, and so its position and derivative rotate at the same rate without changing in magnitude. Therefore, f’(x) is just f(x) rotated and scaled. If you let x=0, y is at its peak (think of the unit cylinder: y^2+z^2=0): y=1, y’=0, z=0, z’=some constant k, inversely proportional to the value of one rotation (because if we stretch the graph on the x-axis by s, the derivative divides by s). Also if you move until z is at its peak, z=1, z’=0, y=0, y’=-k.
Therefore the angle between f(x) and f’(x) is a right angle.
This means that the derivative of sin x is k*cos x and the derivative of cos x is -k*sin x, but how do you prove that k=1 when x is in radians?
sin^2 x+cos^2 x=1. If k=1, then d/dx cos x=cos(x+1τ/4)=-sin x=-sqrt(1-cos^2 x) for sin x>=0, which holds for 0 (which is arccos 1)
2:59: Maple? Such an eco-friendly name!
Thanks for these vidéo.... this limit is OK if X in Radian ! ! Don’t forget it ! Soon....Fred
Used begging the question correctly!
This is the only one i remember from that class
Sir, some teachers argue that we should not use L' Hospital's Rule here because this limit is used to find derivative of cosx. This becomes a circular arguement then.
This is mesmerising though I didn't understand a thing. 🤔
I do not know whose proof this geometric proof is, but I indeed loved it!
I actually don't know who the original person was either, but it is a great one!
@@DrTrefor Thanks for the video!
I really like the way you plotted graph. Can you please make a tutorial on how to install this software for plotting multiple graphs together. It's a request 🙏🙏
The identity function is the first term in the Taylor expansion for sin(x)
Both function at numerator sin x and numerator x have value of zero at zero, then the limit is the ratio of derivatives. sin x --> cos x, and x --> 1
Ah, but what is the definition of the derivative of sinx at 0?
Why, that's (sin(x)-sin(0))/(x-0) as x approaches 0.
sin(0) is 0, so we're left with the derivative being sinx/x as x approaches 0. Do you see why that might not be a very good argument here?
Piggybacking on Athhenri's comment, consider
1. f (x) = [(cos x - 1) / x] , and , when x = 0, the numerator is 0, the denominator is 0, and the ratio, by your argument, is 1. But lim x->0 f(x) = 0. So no, one can't simply plug in the numbers.
2. g (x) = ln (3*x + 1)/ x. When x = 0, 3*x + 1 = 1, so ln (3*x+1) = 0. so, by your argument, 0/0 = 1. But, lim x-> 0 of g(x) = 1/(3*0+1)*3 (by L'hopital) = 3. Not 1.
Need you more examples?
The power, and beauty of calculus.
Zero over zero is indeterminate, it can result in any number
Very nice. I was going to point out how you can't use L'Hopital's rule but looks like you went on to do it in the video. 🙂
I'd argue that the limit producing e is the most important limit in calculus
The million dollar question: Can you use L'Hopital's Rule for this limit? For instance, Michael Penn regularly uses it, but blackpenredpen was very emphatic NOT to use L'Hopital's Rule because according to him, it is circular reasoning since to determine the derivative of sin(x), it uses the lim of sin(x)/x. What are your thoughts on this issue?
I think you have to prove it the first time geometrically, but after you know you have proven derivative of sin(x) then feel free after that
Very interesting
I know the geometric proof but have you ever thought about using the Maclaurin Series for sin(x) ? Using the power series is a nice method
Ah true, but for that you need to know derivatives and for the derivative you need this limit!
@@DrTrefor That is true Dr, thanks for the reply
thank you
TH-cam algorithm is crazy I was thinking about proof of sin x = x(approx) and this video showed up in my recommendation (I didn’t even search for it)
I swear it can mind read sometimes
@@DrTrefor sure it does thanks for the awesome explanation
I must take issue with your statement(at 8:25) that "you could just prove that lim[x->0] (sin x/x) = 1 using l'Hosptal's rule."
We would have to know that the derivative of sin x is cos x. But how did we learn that? Did we not need this limit to carry out the proof?
We can prove that the derivative is sin x in either of two "styles". (It's not two ways, just two styles):
If y = sin x, then (view it in Courier or other fixed-size font)
sin(x + h) - sin x
dy/dx = lim[h->0] -------------------
h
sin x cos h + cos x sin h - sin x
dy/dx = lim[h->0] ---------------------------------
h
sin x cos h - sin x cos x sin h
dy/dx = lim[h->0] -------------------- + -------------
h h
(cos h - 1) cos x sin h
dy/dx = lim[h->0] sin x ------------ + lim -------------
h h
Now, since lim[h->0] sin h/h = 1, supposedly already proved,
and, since lim[h->0] (cos h - 1)/h = 0, already proved, using that same limit, dy/dx simplifies to cos x.
OR do it like this: [I remember my Calc I teacher using this style]
sin r - sin x
dy/dx = lim[r->x] --------------
r - x
Using a "sums-to-products" identity"
2 cos((r+x)/2) sin((r-x)/2)
dy/dx = lim[r->x] ----------------------------
r - x
sin((r-x)/2)
dy/dx = lim[r->x] cos((r+x)/2) lim[r->x] ------------
(r-x)/2
When r->x,
the first factor is just cos x.
the second is the same limit (equals 1)
Cos(0)/1 = 1, very neat
Approximations trigger my OCD =P
What really blows my mind is how calculator works. At its base it's 1s and 0s. So it operates basically with +1 and -1. And you can calculate multiplication, roots, powers, sin\cos, integrals, EVERYTHING translating it into +1s and -1s. All the math that we know and that we can calculate via computer goes down to simple clicks of relays. On and off. Absolutely insane. I have no idea how it works and I'm afraid I'd lose my mind if I ever figure it out. So I prefer to believe it's magic. Just kidding. No, really, it's amazing to me.