Limit and Geometry 001

Let $\rm O_1$ be the center of the unit circle and let $\rm O_2$ be the center of the circle of radius $r(\theta)$ .

$\overline{\rm OO_1}=1$ , $\overline{\rm OO_2}=\sqrt{2}+r$ , $\overline{\rm O_1O_2}=1-r$

$\angle \rm O_1 O O_2=\angle \rm O_1 O A-\angle \rm P O A=\dfrac{\pi}{4}-\theta$

Apply cosine rule to triangle $\rm O O_1 O_2$ , and we get the following equation.

${\overline{\rm O_1O_2}}^2={\overline{\rm OO_1}}^2+{\overline{\rm OO_2}}^2-2{\overline{\rm OO_1}} \times {\overline{\rm OO_2}} \cos \left(\angle \rm O_1 O O_2\right)$

$\Rightarrow \left(1-r\right)^2=1+\left(\sqrt{2}+r\right)^2-2 \left(\sqrt{2}+r\right) \cos \left(\dfrac{\pi}{4}-\theta\right)$

$\Rightarrow r(\theta)=\dfrac{\sqrt{2} \cos\left(\dfrac{\pi}{4} - \theta \right)-1}{\sqrt{2}+1-\cos\left(\dfrac{\pi}{4} - \theta \right)}=\dfrac{\cos \theta + \sin \theta -1}{\sqrt{2}+1-\cos\left(\dfrac{\pi}{4} - \theta \right)}$

(Angle sum identity was used in the last equality.)

Now we can evaluate the limit.

$\lim \limits_{\theta \to 0^+} \dfrac{r(\theta)}{\theta} = \lim \limits_{\theta \to 0^+} \left(\dfrac{\sin \theta}{\theta}+\dfrac{\cos \theta -1}{\theta}\right) \times \lim \limits_{\theta \to 0^+} \dfrac{1}{\sqrt{2}+1-\cos \left(\dfrac{\pi}{4}-\theta\right)}$

$=\left(1+0\right) \times \dfrac{1}{\frac{\sqrt2}{2}+1}=2-\sqrt{2}$

Let $O$ be the origin $(0,0)$ of the $xy$ -plane. Then $OA$ and $OB$ are two sides of the square inscribed in the unit circle. Since $AB=2$ , $OA=OB=\sqrt 2=OP$ . The center of the unit circle is $Q\left(\frac 1{\sqrt 2}, \frac 1{\sqrt 2} \right)$ . Let the center of the circle tangent to the arc $AB$ and unit circle be $R$ . Then $OR=OP+PR = \sqrt 2 + r(\theta)$ and $R((\sqrt 2 + r)\cos \theta, (\sqrt 2+r)\sin \theta)$ . We note that $QS=1$ since it is a radius of the unit circle. Then $QR = QS-RS = 1 - r(\theta)$ . By Pythagorean theorem :

$\begin{aligned} \left((\sqrt 2+r)\cos \theta - \frac 1{\sqrt 2} \right)^2 + \left((\sqrt 2+r)\cos \theta - \frac 1{\sqrt 2} \right)^2 & = (1-r)^2 \\ (\sqrt 2+r)^2 - \sqrt 2 (\sqrt 2+r)(\sin \theta + \cos \theta) + 1 & = r^2 - 2r + 1 \\ r^2 + 2\sqrt 2 r + 2 - 2(\sqrt 2+r)\sin \left(\theta + \frac \pi 4 \right) & = r^2 - 2r \\ \left(\sqrt 2 + 1 - \sin \left(\theta + \frac \pi 4 \right)\right)r & = \sqrt 2 \sin \left(\theta + \frac \pi 4 \right) - 1 \\ \implies r(\theta) & = \frac {\sqrt 2 \sin \left(\theta + \frac \pi 4 \right)-1}{\sqrt 2 + 1 - \sin \left(\theta + \frac \pi 4 \right)} \end{aligned}$

Therefore

$\begin{aligned} \lim_{\theta \to 0^+} \frac {r(\theta)}\theta & = \lim_{\theta \to 0^+} \frac {\sqrt 2 \sin \left(\theta + \frac \pi 4 \right)-1}{\theta \left(\sqrt 2 + 1 - \sin \left(\theta + \frac \pi 4 \right)\right)} & \small \blue{\text{A 0/0 case, L'Hôpital's rule applies.}} \\ & = \lim_{\theta \to 0^+} \frac {\sqrt 2 \cos \left(\theta + \frac \pi 4 \right)}{\sqrt 2 + 1 - \sin \left(\theta + \frac \pi 4 \right)-\theta \cos \left(\theta + \frac \pi 4 \right)} & \small \blue{\text{Differentiate up and down w.r.t. }\theta.} \\ & = \frac 1{\sqrt 2+1 - \frac 1{\sqrt 2}} = \frac {\sqrt 2}{2 + \sqrt 2 - 1} = \frac {\sqrt 2}{\sqrt 2 +1} \\ & = \frac {\sqrt 2(\sqrt 2-1)}{(\sqrt2+1)(\sqrt2-1)} = \boxed{2-\sqrt 2} \end{aligned}$

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Reference: L'Hôpital's rule