The Problem
Starting with the vertices $P_1(0,1), P_2(1,1), P_3(1,0), P_4(0,0)$ of a square, we construct further points as shown in the figure: $P_5$ is the midpoint of $P_1P_2$, $P_6$ is the midpoint of $P_2P_3$, $P_7$ is the midpoint of $P_3P_4$, and so on. The polygonal spiral path $P_1P_2P_3P_4P_5P_6P_7\cdots$ approaches a point $P$ inside the square. Find the coordinates of $P$.
[Calculus by James Stewart, $7^\text{th}$ edition, chapter $11$, problem plus, problem $18$]
What I Have Tried
First of all, let $P_n=(x_n,y_n)$. We will first consider only $x$-coordinates first because the $y$-coordinate of $P$ can be found similarly. I have proven using induction that $\tfrac12x_n+x_{n+1}+x_{n+2}+x_{n+3}=2$ for all $n\geq1$ as follows.
Base Case ($n=1$)
Since $x_1=0, x_2=1, x_3=1, x_4=0$, we have $\tfrac12x_1+x_2+x_3+x_4=\tfrac12\cdot0+1+1+0=2.$
Induction ($n>1$)
Assume that $\tfrac12x_n+x_{n+1}+x_{n+2}+x_{n+3}=2$ is true for $n=k$, then \begin{equation} x_{k+2}+x_{k+3}=2-\tfrac12x_k-x_{k+1}. \end{equation} Also observe that $x_{n+4}=\tfrac12(x_n+x_{n+1})$. Therefore \begin{equation} \begin{split} \tfrac12x_{k+1}+x_{k+2}+x_{k+3}+x_{k+4}&=\tfrac12x_{k+1}+(2-\tfrac12x_k-x_{k+1})+x_{k+4}\\\\ &=\tfrac12x_{k+1}+2-\tfrac12x_k-x_{k+1}+\tfrac12(x_k+x_{k+1})\\\\ &=2 \end{split} \end{equation} and so it is true for $n=k+1$.
So if $\lim_{n\to\infty}x_n$ exists, we can apply the limit as $n$ approaches $\infty$ to both sides of the equation to get (letting $\lim_{n\to\infty}x_n=L_x$)
$$\tfrac12L_x+3L_x=2$$
or
$$L_x=\tfrac47.$$
My Question
The problem arises: how do we prove that $\lim_{n\to\infty}x_n$ does exist? All I've been able to do is prove that $0\leq x_n\leq1$ for all $n\geq1$. I couldn't find anything on the web that could help.
Thanks in advance.
$$x_{n+4}=\tfrac12(x_n+x_{n+1})$$ $$u_n=x_{n+1}-x_n$$ $$u_{n+3}=\tfrac12(x_n+x_{n+1}-2x_{n+3})$$ $$=\tfrac12(-u_n-2u_{n+1}-2u_{n+2})$$ $$\vdots$$ $$u_{n+9}=\tfrac18(-u_n+2u_{n+2})$$ $$u_{n+10}=\tfrac18(-u_n-3u_{n+1}-2u_{n+2})$$ $$u_{n+11}=\tfrac18(u_n+u_{n+1}-u_{n+2})$$ $$|u_{n+9}|\leq\tfrac18(1+2)\max\{|u_n|,|u_{n+2}|\}$$ $$|u_{n+10}|\leq\tfrac18(1+3+2)\max\{|u_n|,|u_{n+1}|,|u_{n+2}|\}$$ $$|u_{n+11}|\leq\tfrac18(1+1+1)\max\{|u_n|,|u_{n+1}|,|u_{n+2}|\}$$ $$\max\{|u_{n+9}|,|u_{n+10}|,|u_{n+11}|\}\leq\tfrac68\max\{|u_n|,|u_{n+1}|,|u_{n+2}|\}$$ $$\max\{|u_{n+9}|,|u_{n+10}|,\cdots,|u_{n+17}|\}\leq\tfrac34\max\{|u_n|,|u_{n+1}|,\cdots,|u_{n+8}|\}$$ $$w_m=\max\{|u_{9m}|,|u_{9m+1}|,\cdots,|u_{9m+8}|\}\geq0$$ $$|u_n|\leq w_{\lfloor n/9\rfloor}$$ $$w_{m+1}\leq\tfrac34w_m$$ $$w_m\leq(\tfrac34)^mw_0$$ $$x_{n+k}-x_n=u_n+u_{n+1}+u_{n+2}+\cdots+u_{n+k-1}$$ $$|x_{n+k}-x_n|\leq|u_n|+|u_{n+1}|+\cdots+|u_{n+k-1}|$$ $$\leq w_{\lfloor n/9\rfloor}+w_{\lfloor(n+1)/9\rfloor}+\cdots+w_{\lfloor(n+k-1)/9\rfloor}$$ $$\leq\Big((\tfrac34)^{\lfloor n/9\rfloor}+(\tfrac34)^{\lfloor(n+1)/9\rfloor}+\cdots+(\tfrac34)^{\lfloor(n+k-1)/9\rfloor}\Big)w_0$$ $$\lfloor\tfrac n9\rfloor>\tfrac n9-1$$ $$|x_{n+k}-x_n|\leq\Big((\tfrac34)^{n/9-1}+(\tfrac34)^{(n+1)/9-1}+\cdots+(\tfrac34)^{(n+k-1)/9-1}\Big)w_0$$ $$=w_0\,(\tfrac34)^{n/9-1}\Big(1+(\tfrac34)^{1/9}+(\tfrac34)^{2/9}+\cdots+(\tfrac34)^{(k-1)/9}\Big)$$ $$=\tfrac43w_0\,\sqrt[9]{\tfrac34}^{\,n}\Big(1+\sqrt[9]{\tfrac34}+\sqrt[9]{\tfrac34}^{\,2}+\cdots+\sqrt[9]{\tfrac34}^{\,k-1}\Big)$$ $$=\tfrac43w_0\,\sqrt[9]{\tfrac34}^{\,n}\left(\frac{1-\sqrt[9]{\tfrac34}^{\,k}}{1-\sqrt[9]{\tfrac34}}\right)$$ $$\leq\tfrac43w_0\,\sqrt[9]{\tfrac34}^{\,n}\left(\frac{1}{1-\sqrt[9]{\tfrac34}}\right)$$ $$\to0,\quad n\to\infty$$