Limit and Geometry 002

Set up coordinates with $A$ at the origin and $B$ on the $x-$ axis. Then the coordinates of the points are $A(0,0)$ , $B(\sin\theta,0)$ , $C(\sin\theta,1)$ , $D(0,1)$ and $E(\cos \theta \cdot \sin \theta,\sin^2 \theta)$ .

The area of $ABCE$ is the sum of the areas of $ABE$ and $BCE$ ; that is $S(\theta)=\frac12 \sin^3 \theta+\frac12(\sin \theta-\cos \theta \cdot \sin \theta)$

Using the small angle approximations $\sin \theta \approx \theta$ and $\cos \theta \approx 1-\frac12 \theta^2$ , this becomes $S(\theta) \approx \frac12 \theta^3+\frac12 \left(\theta-\theta+\frac12 \theta^3 \right)=\frac34 \theta^3$

Hence the required limit is $\boxed{\frac34}$ .

Note that $S(\theta) = [ABCE] = [ABCD]-[AECD]$ . To find $[AECD]$ , draw $EF || AD$ . Then:

$\begin{aligned} S(\theta) & = [ABCD] - [AECD] \\ & = [ABCD] - [AEFD] - [CEF] \\ & = \overline{AD} \times \overline{AB} - \frac {(\overline{AD}+\overline{EF})\times \overline{DF}}2 - \frac {\overline{CF}\times\overline{EF}}2 \\ & = \overline{AD} \times \overline{AB} - \frac {\overline{AD}\times \overline{DF}}2 - \frac {\overline{CD}\times\overline{EF}}2 \\ & = \sin \theta - \frac {\sin \theta \cos \theta}2 - \frac {\sin \theta(1-\sin^2 \theta)}2 \\ & = \frac {\sin \theta(1 - \cos \theta + \sin^2 \theta)}2 \\ & = \frac {2\sin \theta - \sin 2\theta}4 + \frac {\sin^3 \theta}2 \end{aligned}$

Therefore,

$\begin{aligned} \lim_{\theta \to 0^+} \frac {S(\theta)}{\theta^3} & = \blue{\lim_{\theta \to 0^+} \frac {2\sin \theta - \sin 2\theta}{4 \theta^3}} + \lim_{\theta \to 0^+} \frac 12 \left(\frac {\sin\theta} \theta \right)^3 & \small \blue{\text{A 0/0 case, L'Hôpital's rule applies.}} \\ & = \blue{\lim_{\theta \to 0^+} \frac {2\cos \theta - 2 \cos 2\theta}{12 \theta^2}} + \frac 12 \cdot 1^3 & \small \blue{\text{After differentiation, again 0/0}} \\ & = \blue{\lim_{\theta \to 0^+} \frac {-2\sin \theta + 4 \sin 2\theta}{24 \theta}} + \frac 12 & \small \blue{\text{Again a 0/0 case}} \\ & = \blue{\lim_{\theta \to 0^+} \frac {-2\cos \theta + 8 \cos 2\theta}{24}} + \frac 12 & \small \blue{\text{Differentiate up and down w.r.t. }\theta} \\ & = \blue{\frac 6{24}} + \frac 12 = \boxed{\frac 34} \end{aligned}$

Draw a line segment from E till it intersects AB at F, which is perpendicular to AB. Draw a line segment from E till it intersects BC at G, which is perpendicular to BC. Then, the shaded area is the following:

$S = \mathrm{Area} (\triangle AEF) + \mathrm{Area} ( \triangle EGC) +\mathrm{Area} ( EFBG)$

$EFBG$ is a rectangle. Let $\sin{\theta}=r$ . Then the expression comes out to be:

$S = \frac{r^2}{2}\sin{\theta}\cos{\theta} + \frac{1}{2}(r - r\cos{\theta})(1-r\sin{\theta}) + (r\sin{\theta})((r - r\cos{\theta})$

This expression can be simplified to obtain the following:

$S = \frac{\sin^3{\theta}\cos{\theta} + \sin{\theta}(1 - \cos{\theta})(1-\sin^2{\theta}) + 2\sin^2{\theta}(1 - \cos{\theta})}{2}$

$S = \frac{\sin^3{\theta}\cos{\theta} + \sin{\theta}(1 - \cos{\theta}) \cos^2{\theta} + 2\sin^2{\theta}(1 - \cos{\theta})}{2}$

Therefore:

$\lim_{\theta \to 0^{+}} \frac{S}{\theta^3} =\lim_{\theta \to 0^{+}} \frac{\sin^3{\theta}\cos{\theta} + \sin{\theta}(1 - \cos{\theta}) \cos^2{\theta} + 2\sin^2{\theta}(1 - \cos{\theta})}{2\theta^3}$

$\lim_{\theta \to 0^{+}} \frac{S}{\theta^3} = \left(\lim_{\theta \to 0^{+}}\frac{\sin^3{\theta}\cos{\theta}}{2\theta^3} \right)+\left(\lim_{\theta \to 0^{+}}\frac{\sin{\theta}(1-\cos{\theta})\cos^2{\theta}}{2\theta^3}\right) +\left( \lim_{\theta \to 0^{+}}\frac{2\sin^2{\theta}(1 - \cos{\theta})}{2\theta^3}\right)$

$\lim_{\theta \to 0^{+}} \frac{S}{\theta^3} = \frac{1}{2} + \frac{1}{4} + 0$

$\boxed{\lim_{\theta \to 0^{+}} \frac{S}{\theta^3} = \frac{3}{4}}$

Each limit can be solved in more than one way. I have left out these evaluations.