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# Limit of sin(x)/x as x approaches 0

In this video, we prove that the limit of sin(θ)/θ as θ approaches 0 is equal to 1. We use a geometric construction involving a unit circle, triangles, and trigonometric functions. By comparing the areas of these triangles and applying the squeeze theorem, we demonstrate that the limit is indeed 1. This proof helps clarify a fundamental concept in calculus.

## Want to join the conversation?

• how did one come up with such a complex proof ?
i mean how did one guess that we must first take a part of circle cut it into triangles and make an inequality
also the whole proof was making no sense until we reached the last step ......

sorry if my question is stupid or if i am missing some basic skills to come up with a proof like this!
(171 votes)
• whoever did this really clever theorem didn't made it by accident or just because he wanted to know whats the limit of sinx/x without any previous knowledge. It is enough to see the graph of the function to see that sinx/x could be 1. NOW, that's the first step. then, the mathematician must figure a formal and irrefutable theorem for the limit to be commonly accepted. It must be in fact hard to find some relationship like this but these people are just committed with their work but they are normal just like you.
(92 votes)
• It seems clear that the area of the wedge is greater than the red triangle, but how do we know they can be equal? Same with the blue triangle being greater than or equal to the wedge.
(51 votes)
• As θ approaches 0, all the three areas approach each other, try to think about it visually. Maybe that is the reason we put an equality sign there.
(35 votes)
• why does Sal only consider the first and fourth quadrant? he doesn't do an restrict the values of theta.
(36 votes)
• Sal was trying to prove that the limit of sin x/x as x approaches zero. To prove this, we'd need to consider values of x approaching 0 from both the positive and the negative side. So, for the sake of simplicity, he cares about the values of x approaching 0 in the interval (-pi/2, pi/2), which approach 0 from both the negative (-pi/2, 0) and the positive (0, pi/2) side. The interval (-pi/2, pi/2) spans only the first and fourth quadrants. This is why Sal discards the rest.
(81 votes)
• To Sal Khan,

I believe that more important than learning a concept itself is learning how it came to be, learning how it was discovered, learning how to THINK like that. You can just teach the proof just because you have learnt the proof. But please, if possible, teach how to think like a mathematician. Teach how we can, by ourselves, prove something like this.
I mean, that proof made no sense in the beginning. Half the time, I didn't have a single clue to what he was doing or more importantly, 'why' he was doing it.

Nobody's a genius mathematician by birth (well except for some maybe). You can't just think of proofs like that out of thin air! Please tell me, I want to know, of some way to actually think of doing bizarre things like that to prove what we want. I think that remembering the proof after someone else has taught me is kinda useless.

A reply is requested!
(35 votes)
• Math major here.

Your objection is well-placed, and you're correct that proof-writing is a skill that is mostly improved by practice. Before you continue, there are two unfortunate truths to keep in mind:
1. The United States math curriculum places almost zero emphasis on proof writing or proof comprehension (and the things that pass for proofs in geometry are a joke). Sal's primary target audience is U.S. grade-schoolers and high-schoolers, and so the videos on Khan Academy cater to these admittedly bad standards. Before Common Core became widespread, it was common to see KA videos presenting the exact same topics in the exact same order as your own teacher.

2. The culture in the mathematics community dictates that once you've written a proof, you 'polish' it to make it as short and concise as possible. The onus is very much put on the reader of the proof to slog through it word-by-word, and not on the writer to be clear.

All that said, the only thing that Sal really pulled out of a hat was the idea to compare the different areas in the figure. To draw the figure in the first place was fairly natural, because how else can we interpret the sine function if not by the unit circle? And once he had made the decision to compare areas, the proof was fairly straightforward algebra.

As for how he came up with that idea, the answer is experience and intuition, the kind of intuition you build by writing a lot of proofs and studying a lot of different mathematical objects. If you really want to learn this type of thinking, a standard place to start is the book 'How to Solve It' by George Pólya.
(70 votes)
• Why were the inequalities switched?
(24 votes)
• Because he took the reciprocal of each function/number. The same way the reciprocal of three is 1/3, the reciprocals of each function (other than one) were less than the original value. Thus, the inequalities needed to be switched. Think of how 1 is less than three, which is less than four, but if you take the reciprocals you switch the order of inequalities. (1 is greater than 1/3, which is greater than 1/4). Hope this helps!
(65 votes)
• Why were the absolute value signs necessary for inclusion in the beginning and middle parts of the process but discarded at the end? At he mentions that they are discarded because we're only concerned with the 1st and 4th quadrants but why is that? Please explain.
(11 votes)
• Sal uses absolute values because we're dealing with areas and areas must be positive. In the instance that θ is negative, we'd have negative area since sine and tangent of θ would be negative. So, he took the absolute values of sine and tangent so that the areas would be positive regardless of θ.

Now, why did he discard the absolute values?

1. In the case of |sin(θ)|/|θ|
Well,
sin(θ)/θ = sin(-θ)/-θ = -sin(θ)/-θ

So, |sin(θ)|/|θ| = sin(θ)/θ

He can therefore discard the absolute values.

2. In the case of cos(θ)

Since -π/2 < θ < π/2,

cos(θ) is always positive and so for this constraint, |cos(θ)| = cos(θ).

So the absolute values can be discarded
(6 votes)
• How about lim of sin(x)/x as x approaches infinity?
(5 votes)
• The numerator is bounded between 1 and -1, while the denominator goes to infinity, so the limit is 0.
(13 votes)
• At why is the limit of 1 is 1 when theta approaches zero?
(5 votes)
• The lim(1) when Θ→0 means: on the graph y=1, what does the y-coordinate approach when the x-coordinate (or in this case Θ) approach 0. But on the graph y=1, the y-coordinate is always 1 no matter what the x-coordinate is. In other words, lim(k) as Θ→n = k, where k,n are any real numbers.
(11 votes)
• At why does Sal multiply theta/2 pi by pi
(6 votes)
• He multiplies theta/2pi by pi because he was multiplying by the area of the entire circle to find the area of the wedge, the area of the entire circle is pi*r^2 and since the radius is one and one squared is one, the area of the entire circle is pi.
(3 votes)
• So 0 over 0 is one?
(1 vote)
• The limit of 𝑥 ∕ 𝑥 as 𝑥 → 0 is equal to 1.

But the value of 𝑥 ∕ 𝑥 for 𝑥 = 0 is undefined.
(12 votes)

## Video transcript

- [Instructor] What we're going to do in this video is prove that the limit as theta approaches zero of sine of theta over theta is equal to one. So let's start with a little bit of a geometric or trigonometric construction that I have here. So this white circle, this is a unit circle, that we'll label it as such. So it has radius one, unit circle. So what does the length of this salmon-colored line represent? Well, the height of this line would be the y-coordinate of where this radius intersects the unit circle. And so by definition, by the unit circle definition of trig functions, the length of this line is going to be sine of theta. If we wanted to make sure that also worked for thetas that end up in the fourth quadrant, which will be useful, we can just insure that it's the absolute value of the sine of theta. Now what about this blue line over here? Can I express that in terms of a trigonometric function? Well, let's think about it. What would tangent of theta be? Let me write it over here. Tangent of theta is equal to opposite over adjacent. So if we look at this broader triangle right over here, this is our angle theta in radians. This is the opposite side. The adjacent side down here, this just has length one. Remember, this is a unit circle. So this just has length one, so the tangent of theta is the opposite side. The opposite side is equal to the tangent of theta. And just like before, this is going to be a positive value for sitting here in the first quadrant but I want things to work in both the first and the fourth quadrant for the sake of our proof, so I'm just gonna put an absolute value here. So now that we've done, I'm gonna think about some triangles and their respective areas. So first, I'm gonna draw a triangle that sits in this wedge, in this pie piece, this pie slice within the circle, so I can construct this triangle. And so let's think about the area of what I am shading in right over here. How can I express that area? Well, it's a triangle. We know that the area of a triangle is 1/2 base times height. We know the height is the absolute value of the sine of theta and we know that the base is equal to one, so the area here is going to be equal to 1/2 times our base, which is one, times our height, which is the absolute value of the sine of theta. I'll rewrite it over here. I can just rewrite that as the absolute value of the sine of theta over two. Now let's think about the area of this wedge that I am highlighting in this yellow color. So what fraction of the entire circle is this going to be? If I were to go all the way around the circle, it would be two pi radians, so this is theta over to two pis of the entire circle and we know the area of the circle. This is a unit circle, it has a radius one, so it'd be times the area of the circle, which would be pi times the radius square, the radius is one, so it's just gonna be times pi. And so the area of this wedge right over here, theta over two. And if we wanted to make this work for thetas in the fourth quadrant, we could just write an absolute value sign right over there 'cause we're talking about positive area. And now let's think about this larger triangle in this blue color, and this is pretty straightforward. The area here is gonna be 1/2 times base times height. So the area, and once again, this is this entire are, that's going to be 1/2 times our base, which is one, times our height, which is the absolute value of tangent of theta. And so I can just write that down as the absolute value of the tangent of theta over two. Now, how would you compare the areas of this pink or this salmon-colored triangle which sits inside of this wedge and how do you compare that area of the wedge to the bigger triangle? Well, it's clear that the area of the salmon triangle is less than or equal to the area of the wedge and the area of the wedge is less than or equal to the area of the big, blue triangle. The wedge includes the salmon triangle plus this area right over here, and then the blue triangle includes the wedge plus it has this area right over here. So I think we can feel good visually that this statement right over here is true and I'm just gonna do a little bit of algebraic manipulation. Let me multiply everything by two so I can rewrite that the absolute value of sine of theta is less than or equal to the absolute value of theta which is less than or equal to the absolute value of tangent of theta, and let's see. Actually, instead of writing the absolute value of tangent of theta, I'm gonna rewrite that as the absolute value of sine of theta over the absolute value of cosine of theta. That's gonna be the same thing as the absolute value of tangent of theta. And the reason why I did that is we can now divide everything by the absolute value of sine of theta. Since we're dividing by a positive quantity, it's not going to change the direction of the inequalities. So let's do that I'm gonna divide this by an absolute value of sine of theta. I'm gonna divide this by an absolute value of the sine of theta and then I'm gonna divide this by an absolute value of the sine of theta. And what do I get? Well, over here, I get a one and on the right-hand side, I get a one over the absolute value of cosine theta. These two cancel out. So the next step I'm gonna do is take the reciprocal of everything. And so when I take the reciprocal of everything, that actually will switch the inequalities. The reciprocal of one is still going to be one but now, since I'm taking the reciprocal of this here, it's gonna be greater than or equal to the absolute value of the sine of theta over the absolute value of theta, and that's going to be greater than or equal to the reciprocal of one over the absolute value of cosine of theta is the absolute value of cosine of theta. We really just care about the first and fourth quadrants. You can think about this theta approaching zero from that direction or from that direction there, so that would be the first and fourth quadrants. So if we're in the first quadrant and theta is positive, sine of theta is gonna be positive as well. And if we're in the fourth quadrant and theta's negative, well, sine of theta is gonna have the same sign. It's going to be negative as well. And so these absolute value signs aren't necessary. In the first quadrant, sine of theta and theta are both positive. In the fourth quadrant, they're both negative, but when you divide them, you're going to get a positive value, so I can erase those. If we're in the first or fourth quadrant, our X value is not negative, and so cosine of theta, which is the x-coordinate on our unit circle, is not going to be negative, and so we don't need the absolute value signs over there. Now, we should pause a second because we're actually almost done. We have just set up three functions. You could think of this as f of x is equal to, you could view this as f of theta is equal to one, g of theta is equal to this, and h of theta is equal to that. And over the interval that we care about, we could say for negative pi over two is less than theta is less than pi over two, but over this interval, this is true for any theta over which these functions are defined. Sine of theta over theta is defined over this interval, except where theta is equal to zero. But since we're defined everywhere else, we can now find the limit. So what we can say is, well, by the squeeze theorem or by the sandwich theorem, if this is true over the interval, then we also know that the following is true. And this, we deserve a little bit of a drum roll. The limit as theta approaches zero of this is going to be greater than or equal to the limit as theta approaches zero of this, which is the one that we care about, sine of theta over theta, which is going to be greater than or equal to the limit as theta approaches zero of this. Now this is clearly going to be just equal to one. This is what we care about. And this, what's the limit as theta approaches zero of cosine of theta? Well, cosine of zero is just one and it's a continuous function, so this is just gonna be one. So let's see. This limit is going to be less than or equal to one and it's gonna be greater than or equal to one, so this must be equal to one and we are done.