How to calculate $\lim\frac{x_n}{y_n}$ where $0How to calculate the limit $\lim\limits_{n\to\infty} \dfrac{x_n}{y_n}$ where $0<x_0<y_0<\dfrac{\pi}{2}$ and $x_{n+1}=\sin{x_n}, y_{n+1}=\sin{y_n}$?

I have proved that the limit exists because $\sin{x}$ is monotonically increasing over $(0,\dfrac{\pi}{2}]$ and 
$\dfrac{\sin{x_n}}{\sin{y_n}}>\dfrac{x_n}{y_n} \Leftrightarrow \dfrac{\sin{x_n}}{x_n}>\dfrac{\sin{y_n}}{y_n} \Leftrightarrow \dfrac{\sin{x}}{x}$ is strictly monotonically decreasing over $(0,\dfrac{\pi}{2}]$
Then the sequence $\left\{\dfrac{x_n}{y_n}\right\}$ is monotonically increasing and bounded, thus converges. 
However, I cannot find a recurrence relation to let $n\to\infty$ and then calculate the limit. 
 A: Hint Another way to answer the question would be to first show that $ \underset{n\rightarrow \infty}{\lim} x_n = 0$. After that you could try to find an equivalent of $(x_n)$, for example by considering the sequence $z$ defined such as : 
$$ z_{n} = x_{n+1}^{-2} - x_n^{-2} $$
Doing the same for $y$ would lead you to the result.
A: Using Stolz theorem and Taylor expansion of sine, assuming that the limit is not $0$ (which it can't, because sequence $x_n/y_n$ is increasing):
$$\lim \frac{x_n}{y_n} = \lim \frac{\frac{1}{y_n}}{\frac{1}{x_n}} = \lim \frac{\frac{1}{y_{n_1}}-\frac{1}{y_n}}{\frac{1}{x_{n+1}}-\frac{1}{x_n}} = \lim \frac{y_n-\sin(y_n)}{x_n-\sin(x_n)} \frac{x_nx_{n+1}}{y_ny_{n+1}} =\\ (\lim \frac{x_n}{y_n})^2 \lim\frac{y_n^3/6+o(y_n^5)}{x_n^3/6+o(x_n^5)} = (\lim \frac{x_n}{y_n})^{-1} $$
From this
$$\lim \frac{x_n}{y_n} = 1\,\lor \lim\frac{x_n}{y_n} = -1 $$
but it obviously cannot be negative.
A: First, you can see easily that: $x_n\to 0$ adn $y_n\to 0$.
By Stolz theorem, we can get:$$x_n\sim\sqrt{\frac{3}{n}}\sim y_n, \ \text{as}\ n\to\infty.$$
So use this result, $$\lim\limits_{n\to\infty} \dfrac{x_n}{y_n}=1.$$
For details, it need to prove $\lim\limits_{n\to\infty}nx^2_n=3.$
By Stolz theorem:$$\lim_{n\to\infty}nx^2_n=\lim_{n\to\infty}\frac{n}{\frac{1}{x^2_n}}=
\lim_{n\to\infty}\frac{1}{\frac{1}{x^2_{n+1}}-\frac{1}{x^2_n}}=\lim_{n\to\infty}\frac{x^2_nx^2_{n+1}}{x^2_n-x^2_{n+1}}$$
$$=
\lim_{n\to\infty}\frac{x^2_n\sin^2x_n}{x^2_n-\sin^2x_n}=
\lim_{x\to 0}\frac{x^2\sin^2x}{x^2-\sin^2x}=3.$$
