Vtyjkyuk

How to calculate the limit $limlimits_ntoinfty dfracx_ny_n$ where $0<x_0<y_0<dfracpi2$ and $x_n+1=sinx_n, y_n+1=siny_n$?

I have proved that the limit exists because $sinx$ is monotonically increasing over $(0,dfracpi2]$ and

$dfracsinx_nsiny_n>dfracx_ny_n Leftrightarrow dfracsinx_nx_n>dfracsiny_ny_n Leftrightarrow dfracsinxx$ is strictly monotonically decreasing over $(0,dfracpi2]$

Then the sequence $leftdfracx_ny_nright$ is monotonically increasing and bounded, thus converges.

However, I cannot find a recurrence relation to let $ntoinfty$ and then calculate the limit.

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):
$$lim fracx_ny_n = lim fracfrac1y_nfrac1x_n = lim fracfrac1y_n_1-frac1y_nfrac1x_n+1-frac1x_n = lim fracy_n-sin(y_n)x_n-sin(x_n) fracx_nx_n+1y_ny_n+1 =\ (lim fracx_ny_n)^2 limfracy_n^3/6+o(y_n^5)x_n^3/6+o(x_n^5) = (lim fracx_ny_n)^-1 $$
From this
$$lim fracx_ny_n = 1,lor limfracx_ny_n = -1 $$
but it obviously cannot be negative.

Hint Another way to answer the question would be to first show that $ undersetnrightarrow inftylim 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.

First, you can see easily that: $x_nto 0$ adn $y_nto 0$.

By Stolz theorem, we can get:$$x_nsimsqrtfrac3nsim y_n, textas ntoinfty.$$
So use this result, $$limlimits_ntoinfty dfracx_ny_n=1.$$
For details, it need to prove $limlimits_ntoinftynx^2_n=3.$
By Stolz theorem:$$lim_ntoinftynx^2_n=lim_ntoinftyfracnfrac1x^2_n=
lim_ntoinftyfrac1frac1x^2_n+1-frac1x^2_n=lim_ntoinftyfracx^2_nx^2_n+1x^2_n-x^2_n+1$$
$$=
lim_ntoinftyfracx^2_nsin^2x_nx^2_n-sin^2x_n=
lim_xto 0fracx^2sin^2xx^2-sin^2x=3.$$