If $a_{n+2} = \frac{1}{3}\left(a_{n+1}+\frac{1}{a_{n}}\right)$, then $\lim_{n\to\infty}a_{n} = ?$ 
If $\displaystyle a_{n+2} = \frac{1}{3}\left(a_{n+1}+\frac{1}{a_{n}}\right),a_{n}>0$, then $\lim_{n\rightarrow \infty}a_{n} = ?$

My try:
It seems that when $n \rightarrow \infty$ then we can write $a_{n}=a_{n+1}=a_{n+2} = l$ (finite number), but I did not understand how I can prove that the sequence is strictly increasing. Help required. Thanks.
 A: Replace $a_n$ $(n\geq0)$ by the new variables $x_n$ and $y_n$ $(n\geq1)$, defined by
$$x_n:=\sqrt{2}a_n,\qquad y_n:={1\over\sqrt{2}a_{n-1}}={1\over x_{n-1}}\ .$$Then
$$x_{n+1}=\sqrt{2}a_{n+1}={1\over3}\left(\sqrt{2}a_n+{2\over\sqrt{2}a_{n-1}}\right)={1\over3}x_n+{2\over3}y_n\ .$$
In this way we obtain the first order recursion
$$\left.\eqalign{x_{n+1}&={1\over3}x_n+{2\over3}y_n\cr y_{n+1}&={1\over x_n}\cr}\right\}\qquad(n\geq1)\ .$$
The point $p:=(1,1)$ is obviously a fixed point of the implied transformation
$$f(x,y):=\left({1\over3}x+{2\over3}y, \ {1\over x}\right)\ .$$
In order to check whether $p$ is attracting we have to look at
$$df(p)=\left[\matrix{{1\over3}&{2\over3}\cr -1&0\cr}\right]\ .$$
The singular values of ths matrix are  $1.08554$ and $0.61413$, which is not sufficient. But if we now look at
$$\bigl(df(p)\bigr)^2=\left[\matrix{-{5\over9}&{2\over9}\cr -{1\over3}&-{2\over3}\cr}\right]$$
then we get the singular values $0.74985$ and $0.59271$. This allows to conclude that $p$ is attracting after all. It follows that there is some neighborhood $U$ of $p$ such that all initial points $(x_0,y_0)\in U$ are attracted to $p$ under iteration of $f$.
Returning to the $a_n$ in the orginal problem this means that there is a neighborhood $U'$ of the point $\bigl({1\over\sqrt{2}},{1\over\sqrt{2}}\bigr)$ such that for all initial values $(a_0,a_1)\in U$ one has $\lim_{n\to\infty} a_n={1\over\sqrt{2}}$.
The $f$ defined in $(1)$ is easily seen to be injective on the open first quadrant. The following figure shows in blue the ellipse $\epsilon$ with center $(1,1)$ and semi-axes $0.2$, $0.28$, and in olive the image of $\epsilon$ under $f^{\circ2}$. From this picture it becomes plausible that the iterated images of the region bounded by $\epsilon$ converge to the point $(1,1)$.

