TwistedSquares 2

First, we need to find how the sides shrink based on the angle. $\tan{\theta}=\frac{1-x}{x}$ so $x=\frac{1}{\tan{\theta}+1}$ . Also, $\cos{\theta}=\frac{x}{y}$ .

Put these together and you get $y=\frac{1}{cos{\theta}(\tan{\theta}+1)}=\frac{1}{sin{\theta}+\cos{\theta}}$

So this is the ratio from one square to the next smaller. Working in radians $\theta=\frac{\pi}{2n}$ and the ratio is applied $n$ times so we seek

$\lim_{n\to\infty}(\sin{\frac{\pi}{2n}}+\cos{\frac{\pi}{2n})^{-n}}$

Which WolframAlpha tells me is $e^{-\pi/2} \approx \boxed{0.2078795764}$

Relevant wiki: L'Hopital's Rule - Basic

Let the side lengths of the largest square to the smallest square be $a_0$ , $a_1$ , $a_2$ ... $a_n$ respectively. We note that

$\begin{aligned} a_0 & = a_1\sin \frac \pi{2n} + a_1\cos \frac \pi{2n} \\ \implies a_1 & = \frac {a_0}{\sin \frac \pi{2n} + \cos \frac \pi{2n}} & \small \color{#3D99F6} \text{Similarly,} \\ a_2 & = \frac {a_1}{\sin \frac \pi{2n} + \cos \frac \pi{2n}} = \frac {a_0}{\left(\sin \frac \pi{2n} + \cos \frac \pi{2n}\right)^2} \\ a_3 & = \frac {a_0}{\left(\sin \frac \pi{2n} + \cos \frac \pi{2n}\right)^3} \\ \cdots & = \cdots \\ \implies a_n & = \frac {a_0}{\left(\sin \frac \pi{2n} + \cos \frac \pi{2n}\right)^n} \\ a & = \frac {a_n}{a_0} = \frac 1{\left(\sin \frac \pi{2n} + \cos \frac \pi{2n}\right)^n} \end{aligned}$

$\begin{aligned} \lim_{n \to \infty} a & = \lim_{n \to \infty} \frac 1{\left(\sin \frac \pi{2n} + \cos \frac \pi{2n}\right)^n} & \small \color{#3D99F6} \text{A }1^\infty \text{ case, the following applies} \\ & = \exp \left( \lim_{n \to \infty} -n \left(\sin \frac \pi{2n} + \cos \frac \pi{2n} - 1 \right) \right) & \small \color{#3D99F6} \implies \lim_{x \to \infty} (f(x))^{g(x)} = e^{\lim_{x\to \infty} g(x)(f(x)-1)} \\ & = \exp \left( \lim_{x \to 0} - \frac \pi 2 \left(\frac {\sin x + \cos x - 1}x \right) \right) & \small \color{#3D99F6} \text{Let }x = \frac \pi{2n} \\ & = \exp \left(-\frac \pi 2 \left(\lim_{x \to 0} \frac {\sin x}x +\color{#3D99F6} \lim_{x \to 0} \frac {\cos x - 1}x \right) \right) & \small \color{#3D99F6} \text{A 0/0 case, L'Hôpital's rule applies} \\ & = \exp \left(-\frac \pi 2 \left(1 +\color{#3D99F6} \lim_{x \to 0} \frac {-\sin x}1 \right) \right) & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. }x \\ & = \exp \left(-\frac \pi 2 \left(1 +\color{#3D99F6} 0 \right) \right) \\ & = e^{-\frac \pi 2} \approx \boxed{0.208} \end{aligned}$

---

Reference: Limit for $1^\infty$ case, see method 2 .

As the other solutions show, we have $a = \left ( \sin \dfrac{\pi}{2n} + \cos \dfrac{\pi}{2n} \right )^{-n},$ and we want to find $L = \displaystyle \lim_{n \rightarrow \infty} a.$ In this present form, we cannot use L'Hopital's rule to find the limit, but the $f(n)^n$ form of $a$ suggests that taking the natural logarithm of the limit might yield some results. So we proceed:

$\begin{aligned} L &= \lim_{n \rightarrow \infty} \left ( \sin \dfrac{\pi}{2n} + \cos \dfrac{\pi}{2n} \right )^{-n} \\ \ln L &= \lim_{n \rightarrow \infty} \ln \left ( \sin \dfrac{\pi}{2n} + \cos \dfrac{\pi}{2n} \right )^{-n} \\ \ln L &= \lim_{n \rightarrow \infty} -n \ln \left ( \sin \dfrac{\pi}{2n} + \cos \dfrac{\pi}{2n} \right ). \\ \end{aligned}$

From here, we can turn the limit into an $\dfrac{0}{0}$ indeterminate form by writing $-n$ as $\dfrac{1}{-\frac{1}{n}},$ which we can apply L'Hopital's on:

$\begin{aligned} \ln L &= \lim_{n \rightarrow \infty} \dfrac{\ln \left ( \sin \frac{\pi}{2n} + \cos \frac{\pi}{2n} \right )}{-\frac{1}{n}} \\ \ln L &= \lim_{n \rightarrow \infty} \dfrac{\frac{1}{\sin \frac{\pi}{2n} + \cos \frac{\pi}{2n}} \cdot \left ( \cos \frac{\pi}{2n} - \sin \frac{\pi}{2n} \right ) \cdot \left ( -\frac{\pi}{2n^2} \right ) }{\frac{1}{n^2}}. \end{aligned}$

(See the Addendum at the bottom for the calculation of the derivative of $\ln \left ( \sin \dfrac{\pi}{2n} + \cos \dfrac{\pi}{2n} \right ).$ )

Cleaning up the expression, we finally have

$\begin{aligned} \ln L &= \lim_{n \rightarrow \infty} \dfrac{\pi}{2} \dfrac{\sin \frac{\pi}{2n} - \cos \frac{\pi}{2n}}{\sin \frac{\pi}{2n} + \cos \frac{\pi}{2n}} \\ \ln L &= \dfrac{\pi}{2} \cdot \dfrac{0 - 1}{0 + 1} \\ \ln L &= -\dfrac{\pi}{2} \\ L &= \boxed{e^{-\pi / 2}}, \end{aligned}$

which is approximately $\boxed{0.208}$ to three decimal places. $\blacksquare$

---

Addendum : To find the derivative of $\ln \left ( \sin \dfrac{\pi}{2n} + \cos \dfrac{\pi}{2n} \right ),$ we have to apply the chain rule twice: once on the expression in the natural log, and once on the sine and cosine expressions.

$\begin{aligned} \dfrac{d}{dn} \left [ \ln \left ( \sin \dfrac{\pi}{2n} + \cos \dfrac{\pi}{2n} \right ) \right ] &= \dfrac{1}{\sin \frac{\pi}{2n} + \cos \frac{\pi}{2n}} \cdot \dfrac{d}{dn} \left [ \sin \dfrac{\pi}{2n} + \cos \dfrac{\pi}{2n} \right ] \\ &= \dfrac{1}{\sin \frac{\pi}{2n} + \cos \frac{\pi}{2n}} \cdot \left ( \cos \dfrac{\pi}{2n} - \sin \dfrac{\pi}{2n} \right ) \cdot \dfrac{d}{dn} \left [ \dfrac{\pi}{2n} \right ] \\ &= \dfrac{1}{\sin \frac{\pi}{2n} + \cos \frac{\pi}{2n}} \cdot \left ( \cos \dfrac{\pi}{2n} - \sin \dfrac{\pi}{2n} \right ) \cdot \left ( -\dfrac{\pi}{2n^2} \right ). \end{aligned}$

If $x=\dfrac{\pi}{2n}$ for large $n$ , then the series expansion of $Sin(x)+Cos(x)$ is

$Sin(x)+Cos(x)=1+x-\dfrac{1}{2}x^2-...$ , so that

$\displaystyle \lim_{x\to 0} \left( Sin(x) + Cos(x) \right)^{-\frac{1}{x}}= \displaystyle \lim_{x\to 0} \left (1+x\right)^{-\frac{1}{x}}=e^{-1}$

Letting $x=\dfrac{\pi}{2n}$ , we have

$\displaystyle \lim_{n\to \infty} \left( Sin \left( \dfrac{\pi}{2n} \right) + Cos \left(\dfrac{\pi}{2n} \right ) \right)^{-\dfrac{2n}{\pi}} =e^{-1}$

or

$\displaystyle \lim_{n\to \infty} \left( Sin \left( \dfrac{\pi}{2n} \right) + Cos \left(\dfrac{\pi}{2n} \right ) \right)^{-n} =e^{-\frac{\pi}{2}}$