Single Variable Calculus, Chapter 3, 3.4, Section 3.4, Problem 50

Find $\displaystyle \lim\limits_{\theta \to 0^+} \frac{A(\theta)}{B(\theta)}$ for a semicircle with diameter $PQ$ that sits on a isosceles triangle $PQR$ to form a region shaped like a two dimensional ice cream cone, as shown in the figure. Where $A(\theta)$ is there area of the semicircle and $B(\theta)$ is the area of the triangle.




For the area of semicircle $A(\theta)$. We let $r$ be the radius forming this triangle





$\begin{equation} \begin{aligned} \sin \frac{\theta}{2} &= \frac{r}{10}\\ \\ r &= 10 \sin \frac{\theta}{2} \end{aligned} \end{equation}$

Solving for $A(\theta)$

$\begin{equation} \begin{aligned} A(\theta) &= \frac{\pi r^2}{2} &&; \text{where } r = 10 \sin \frac{\theta}{2}\\ \\ A(\theta) &= \frac{\pi(10 \sin \frac{\theta}{2})^2}{2}\\ \\ A(\theta) &= 50 \pi \left( \sin^2 \left( \frac{\theta}{2}\right)\right) \end{aligned} \end{equation}$


For the area of triangle $B(\theta)$ given two sides and an included angle we have,

$\begin{equation} \begin{aligned} B(\theta) &= \frac{1}{2} ab \sin \theta\\ \\ B(\theta) &= \frac{1}{2} (10)(10) \sin \theta\\ \\ B(\theta) &= 50 \sin \theta \end{aligned} \end{equation}$


Thus,

$\begin{equation} \begin{aligned} \lim\limits_{\theta \to 0^+} \frac{A(\theta)}{B(\theta)} &= \lim\limits_{\theta \to 0^+} \frac{\cancel{50} \left( \sin \frac{\theta}{2}\right)^2}{\cancel{50}\sin \theta} && \text{We can introduce a factor } \frac{\theta}{\theta} \text{ and } \frac{\frac{\theta}{2}}{\frac{\theta}{2}} \text{ to use the property of limit.}\\ \\ \lim\limits_{\theta \to 0^+} \frac{A(\theta)}{B(\theta)} &= \lim\limits_{\theta \to 0^+} \frac{\pi \left( \sin \frac{\pi}{2}\right)^2}{\sin \theta} \left( \frac{\theta}{\cancel{\theta}} \right) \left( \frac{\frac{\cancel{\theta}}{2}}{\frac{\theta}{2}}\right)\\ \\ \lim\limits_{\theta \to 0^+} \frac{A(\theta)}{B(\theta)} &= \lim\limits_{\theta \to 0^+} \frac{\pi}{2} \left[ \left( \frac{\theta}{\sin \theta} \right) \left( \frac{\sin \frac{\theta}{2}}{\frac{\theta}{2}}\right) \left( \sin \frac{\theta}{2}\right)\right] && \text{recall that } \lim\limits_{\theta \to 0 } \frac{\sin \theta}{\theta} =1\\ \\ \lim\limits_{\theta \to 0^+} \frac{A(\theta)}{B(\theta)} &= \lim\limits_{\theta \to 0^+} \frac{\pi}{2} (1) (1) \left( \sin \frac{\theta}{2}\right)\\ \\ \lim\limits_{\theta \to 0^+} \frac{A(\theta)}{B(\theta)} &= \frac{\pi}{2} \left( \sin \frac{\theta}{2}\right)\\ \\ \lim\limits_{\theta \to 0^+} \frac{A(\theta)}{B(\theta)} &= \frac{\pi}{2} \sin \frac{0}{2}\\ \\ \lim\limits_{\theta \to 0^+} \frac{A(\theta)}{B(\theta)} &= 0 \end{aligned} \end{equation}$