Converging sequence and its limit Show that the sequence $x_n$ given by
$$x_0=0, x_1=1 \ \ \text{ and} \ \  x_{n+1}=\sqrt{\frac{1}{4}x^{2}_{n}+\frac{3}{4}x^{2}_{n-1}}$$ for $n\in\mathbb N$
converges and find its limit.
My progress so far:
This is an intertwining sequence, and I am trying to show $I_{n+1}\subseteq I_n$ where interval $I_{n+1}=[x_{n}, x_{n-1}]$ by mathematical induction, but I failed to prove that case $n+1$ is true given that case $n$ is true. And I have no idea with the steps thereafter.
 A: Clearly
$x_{n+1}^2-x_n^2=-\frac{3}{4}(x_n^2-x_{n-1}^2)$ and hence $x_{n+1}^2-x_n^2=(-\frac{3}{4})^n$. Thus
$$ x_{n+1}^2=\sum_{k=0}^n(x_{k+1}^2-x_k^2)=\sum_{k=0}^n(-\frac{3}{4})^n\to \frac{1}{1-(-\frac34)}=\frac{4}{7} $$
and hence
$$ \lim_{n\to\infty}x_n=\frac{2}{\sqrt 7}. $$
A: The square root is making this difficult.  Let's square both sides of your recursive expression:
$$x_{n+1}^2 = \frac{1}{4}x_n^2 + \frac{3}{4}x_{n-1}^2$$
let's consider the sequence $(y_n) = (x_n^2)$.  Since each term in $(x_n)$ is nonnegative (by virtue of the square root), we have that $(x_n) = (\sqrt{y_n})$, so if $(y_n)$ converges to $y$, then $(x_n)$ will converge to $\sqrt{y}$.  
We now have the linear recurrence relation
$$y_{n+1} = \frac{1}{4}y_n + \frac{3}{4}y_{n-1}$$
which we can rewrite in matrix form as
$$\left[\begin{array}{c} y_{n+1} \\ y_n\end{array}\right] = \left[\begin{array}{cc} \frac{1}{4} & \frac{3}{4} \\ 1 & 0\end{array}\right] \left[\begin{array}{c} y_{n} \\ y_{n-1}\end{array}\right]$$
so
$$\left[\begin{array}{c} y_{n+1} \\ y_n\end{array}\right] = \left[\begin{array}{cc} \frac{1}{4} & \frac{3}{4} \\ 1 & 0\end{array}\right]^n \left[\begin{array}{c} y_1 \\ y_0\end{array}\right]$$
If find that the eigenvectors of the matrix are $\left[\begin{array}{cc} 1 & 1\end{array}\right]^T$ and $\left[\begin{array}{cc} 1 & -\frac{4}{3}\end{array}\right]^T$, with eigenvalues of $1$ and $-\frac{3}{4}$, respectively.  Thus, if we could write
$$\left[\begin{array}{c} y_1 \\ y_0\end{array}\right] = a \left[\begin{array}{c} 1 \\ 1\end{array}\right] + b \left[\begin{array}{c} 1 \\ -\frac{4}{3}\end{array}\right]$$
Then we would have
$$\left[\begin{array}{c} y_{n+1} \\ y_n\end{array}\right] = \left[\begin{array}{cc} \frac{1}{4} & \frac{3}{4}\\ 1 & 0\end{array}\right]^n \left[\begin{array}{c} y_1 \\ y_0\end{array}\right] = a(1)^n \left[\begin{array}{c} 1 \\ 1\end{array}\right] + b\left(-\frac{3}{4}\right)^n \left[\begin{array}{c} 1 \\ -\frac{4}{3}\end{array}\right]$$
as $n\to\infty$, the second term vanishes, so we would have $(y_n) \to a$, and thus $(x_n) \to \sqrt{a}$.  I'll leave it to you to fill in the details.
