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The figure shows a circular arc of length $ s $ and $ a $ chord of length $ d, $ both subtended by a central angle $ \theta $. Find

$ \displaystyle \lim_{\theta \to 0+} \frac {s}{d} $

$\lim _{\theta \rightarrow 0^{+}} \frac{s}{d}=1$

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Missouri State University

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Idaho State University

Yeah. The figure shows a circular arc of like us and a quarter lengths. You want to find the limit as data approaches zero from the right of us divided by D. So this question is challenging the understanding of limits in particular system. An understanding of how to construct functions using things in the circles and then how to have a limit from there. So what this means is first we need to buy DNS has function the data if you want to be able to value the women above. So we can think of D. Using the data including so each of the two black legs in the figure. Either the or here are the radius of the circle are thus for data over to where this angle is bisected by the lines in particular, candy, we have that over to our side data over two. That's the equal to our idea that we know that this line bisects the angle and E. Over G. Giving us the original on either side simply because these two sides are equivalent length. Therefore these angles are equivalent. Therefore, we know the line connecting these 5 60 simply because of geometry principles with here. So now that we understand why how we constructed G equal to our side data over to we notice I think that some of the circumference of the circle is two pi R. As is simply the fractional, you know, over two pi times the circumference. Thus, C gives us equal our data. So from here we can put us over D. As data divided by two side data over to where the Rh equation cancel out one right? That's over D. That we have a clear given the black on the right. We see that getting towards the right hand limit data approaching zero for the right but the function approaches value want thus, from the graph, which is the limit, as data approaches zero for the right after everybody is equal to what we want. So