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Let $P(n)$ be the perimeter of an $n$ -gon inscribed in a unit circle (Figure 8).(a) Explain, intuitively, why $P(n)$ approaches 2$\pi$ as $n \rightarrow \infty$(b) Show that $P(n)=2 n \sin \left(\frac{\pi}{n}\right)$(c) Combine ( a) and (b) to conclude that $$\lim _{n \rightarrow \infty} \frac{n}{\pi} \sin \left(\frac{\pi}{n}\right)=1$$(d)Use this to give another argument that$$\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1$$

a) $2 \sin \left(\frac{\pi}{n}\right)$b) $2 \pi$c) 1

Calculus 1 / AB

Chapter 2

LIMITS

Section 7

Limits at Infinity

Limits

Derivatives

Continuous Functions

Harvey Mudd College

Baylor University

University of Michigan - Ann Arbor

Idaho State University

Lectures

03:09

In mathematics, precalculu…

31:55

In mathematics, a function…

10:39

Graphical Reasoning Consid…

02:24

(a) Let $ A_n $ be the are…

09:50

(a) Let $A_{n}$ be the are…

06:04

A geometric limit Let $f(\…

08:13

A regular polygon with $n$…

0:00

The figure shows a sector …

04:39

The perimeter of a regular…

03:08

Show that $\sec \left(\fra…

02:51

Show that $\csc \left(\fra…

00:59

Limits of Sequences If the…

Okay, so this problem is a bit of a loved one. So we're going to go through it part by far a through D and solve each part into the Julian with each part building off. So this question size let Pete and be the perimeter oven n gon inscribed in the US circle and the figure is talking about is right here. So if you can look at it, we have hex a gone and Nana gone, and it goes up. So you have all of these shapes inscribed inside us circle. So basically, all saying is that the radius is one to make it so a exploit intuitively wide PM and approaches to pie as end of purchasing. So let's think about that for a second as an approaches infinity. That just means that there's an infinite number of sides to this. Holly are. So when we have the shape and the sides are becoming an infinite amount, it starts to resemble just a circle because, as you can see as the sun size growth from 6 to 9 to 12 it starts to resemble more of a circle. So intuitively, when the sides approach infinity, the protocol approach to pie because it's approaching the perimeter of the unis circle, and that is just too Hi are. And because our is just one here, our perimeter approaches 25 and that's always asking you to do so we go to be show y. P even equals to end, sign high over. And now this is a more tricky problem to solve, and I think the best way to solve it if we just draw out a polygon to use as an example, and I'm just going to draw out, draw out. And often this is a very poorly drawn octagon, but it's really just to get the idea. So we have a center point right here and we have the radius. You draw to radio I just to show, but it's one and we're gonna manage. This is inscribed in a circle, and really, that's all we got. Way really need is the information that is described. A circle is that the radius is or one here, and they reached a corner points, of course, as he should. So we need to show the perimeter equals two inside high over, and I think the easiest way to do This is to use some trig that we could recall from pre calculus. So if we haven't angle right here, call it speed up. No data. We have to know its data Peoples to pie over end Now, Why is that? Because the angles inside of a circle. Of course, if you cut off the circle, always angle. They're going back to two pie. So we know that if we have these radius the radio and if we cut up this shape into all the angles, all the angles are going out of the two pi. And of course, in this situation, because we do have eight sides that there that would be too kind Ray or Hyo four. But we don't need that information because, really, all we're using this after going as just a shape to fewer problems. So we know that they are depending on how many sides are to this polygon is to pile around Now we can use a trick to figure out what this silent right here is the easiest way I think to do this this this wouldn't have perpendicular to do it by sector, meaning that we're going to split this angle in half and it's gonna come straight down and we're gonna call both sides right here. X now, using our trade identities. We know that sign well, from Kolkata, as we memorize a while ago, sine is the opposite over hype Linus and hear the opposite of the angle data over to now because we're just using half this angle. His part is X, and we know that our high partners right here is just one. So we can say that sign of data over to which is just high over end equals X, which is our opposite, or our partners, which is just one. Now we know that X ignoring the one equals sign of pie over and and we know that two X is our silence to our polygon. So two exes, just two sign high over. And now when we have one side length and we have insides to find the perimeter, all we had to do is multiply our side length with the amount of size that we have. And if we multiply this five end, we get it to end sign hi over it. And that's just be no. Our next part, see, is asking for us to combine what we've learned from a B to conclude thus and approaches infinity. The limit becomes one. So how do we do this? It's crazy. We basically just have to look at the fact that when we concluded that p of n equals two and signed pipe end, we were able to prove that this is correct, that this is this is the perimeter. Now if we divide both sides of this by a certain something, we get what is stated right here and over Pi to get this over pie right here. All we had to do is divine both sides off. This of this equals to pie as and riches infinity as we as we did before because we use their intuition and a and then we really calculated out. And b we combine that together because we know as an approaches infinity, the perimeter becomes two eyes we can set that equals to buy and to get to this and over pie on the left side, Right here. All we have to do is divide both sides by two pipe and if we divide both sides by two pi, we get the born on this side and we get and over pi sign of piranhas because want but works out perfectly because as we divided well size by two pi, we get exactly what we need to prove right here. Now the last part of this question says to use this to give another argument that as data approaches, zero signed data over theta equals one. This will require some thinking as well. This's pretty complex problem. What we can think about is this right here. You could rewrite it just like that. How? By saying that Seita here, peoples Hi over end. Now, how is it that we're allowed to say that? Because as n approaches infinity, Saito will approach zero. Think about it as we drew this picture right here, even though we said data equals to power and there was just an arbitrary value we could think about. It just is this angle as an approaches infinity. This angle is going to grow smaller, smaller and smaller, and it's going eventually. It's going to start approaches. Zero, as in gets bigger and bigger so we can say as an approaches infinity. They'd approaches zero if they don't people's high over end Now all we have to do is take what we have right here and replace everything with the end of the pies with fatal. So if you know, data equals power and we have the limit data for a cheesy room and replace everything and over pie. If we have St Michael's pyre end, it's just one over data and we have signs of pie over and and that's just sign Ophelia in that is exactly. But this identity says right here, and that's really all this problem is asking you to solve. It's just to show how, as a limited approaches infinity. We could think about it in our heads, especially in this terms of polygon. As it's it's like the circle is becoming a circle because as the size of Russian unity, it starts to look like this when it's really all that I was asking for you to look at. Yeah, that's it

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