Compute the area of regions enclosed by a circular sector and tangent lines

In the figure below let $S(X)$ be the area of the shaded region and $T(x)$ the area of the triangle $ABC$ . Compute

$T(x)$ ;
$S(x)$ ;
$\displaystyle{ \lim_{x \to 0^+} \frac{T(x)}{S(x)}}$ .

We have the following diagram, with some additional markings from the diagram in Apostol,

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First, to calculate the area of the triangle $ABC$ we calculate the area of the polygon $OACB$ and subtract the area of the triangle $OAB$ . Since the polygon $OACB$ is two congruent triangles (since the line from $O$ to $C$ bisects the polygon into two equal sized triangles) we have
$\begin{align*} \operatorname{Area}(OABC) &= 2 \cdot \operatorname{Area}(OAC) \\ &= 2 \cdot \left( \frac{1}{2} \right) (1) \left( \tan \frac{x}{2} \right) \\ &= \tan \frac{x}{2}. \end{align*}$

To calculate the area of the triangle $OAB$ we use the auxiliary point $D$ . The line from $B$ to $D$ is the height of the triangle $OAB$ and it has length $\sin x$ . Therefore,

$\operatorname{Area}(OAB) = \left( \frac{1}{2} \right) (1) (\sin x) = \frac{1}{2} \sin x.$

Hence,

$\operatorname{Area}(ABC) = \tan \frac{x}{2} - \frac{1}{2} \sin x.$
The area of the shaded region $S(x)$ is the area of circular sector minus the area of the triangle $OAB$ (which we calculated in part (a) as $\frac{1}{2} \sin x$ ). The area of the sector is given by
$\pi r^2 \cdot \frac{x}{2 \pi} = \frac{1}{2} x.$

Hence,

$S(x) = \frac{1}{2}x - \frac{1}{2} \sin x.$
For this we take the limit using our expressions for $T(x)$ and $S(x)$ obtained in parts (a) and (b) and use L’Hopital’s,
$\begin{align*} \lim_{x \to 0^+} \frac{T(x)}{S(x)} &= \lim_{x \to 0^+} \frac{ \tan \frac{x}{2} - \frac{1}{2} \sin x}{\frac{1}{2} x - \frac{1}{2} \sin x} \\[9pt] &= \lim_{x \to 0^+} \frac{ \frac{1}{2} \sec^2 \frac{x}{2} - \frac{1}{2} \cos x}{\frac{1}{2} - \frac{1}{2} \cos x} \\[9pt] &= \lim_{x \to 0^+} \frac{ \frac{1}{\cos^2 \frac{x}{2}} - \cos x}{1 - \cos x}. \end{align*}$

Then, we note that $\cos^2 \frac{x}{2} = \frac{1}{2} (\cos x + 1)$ (from the trig identity $\cos (2x) = \cos^2 x - 1$ ). Therefore,

$\begin{align*} \lim_{x \to 0^+} \frac{T(x)}{S(x)} &= \lim_{x \to 0^+} \frac{ \frac{1}{\cos^2 \frac{x}{2}} - \cos x}{1 - \cos x} \\[9pt] &= \lim_{x \to 0^+} \frac{ \frac{2}{\cos x + 1} - \cos x}{1 - \cos x} \\[9pt] &= \lim_{x \to 0^+} \frac{2 - \cos^2 x - \cos x}{(1+\cos x)(1- \cos x)} \\[9pt] &= \lim_{x \to 0^+} \frac{1 - \cos^2 x + 1 - \cos x}{(1+\cos x)(1- \cos x)} \\[9pt] &= \lim_{x \to 0^+} \frac{(1- \cos x)(1+\cos x) + (1 - \cos x)}{(1+\cos x)(1 - \cos x)} \\[9pt] &= \lim_{x \to 0^+} \frac{1 + \cos x + 1}{1 + \cos x} \\[9pt] &= \lim_{x \to 0^+} \frac{2+ \cos x}{1+ \cos x} \\[9pt] &= \frac{3}{2}. \end{align*}$