How to find the limit of this matrix function 
Let $A$ be $n\times n$ real symmetric matrix that is positive definite. Let $x\in\mathbb{R^n}, \space x\ne 0$. Prove that the following limit
  $$
\lim_{m\to\infty}\dfrac{x^TA^{m+1}x}{x^TA^{m}x}
$$
  exists and find that limit.

My thought: since $A$ is symmetric, its eigenvalues are real. The solution should be related to eigenvalues of $A$. But I don't know how to do it and need help.
 A: As a real symmetric positive definite matrix, it is orthogonally diagonalizable with real positive eigenvalues. Changing basis to the diagonal form, write $A = PD P^{\top}$ where $P^{\top} = P^{-1}$ is orthogonal. It follows that 
$$
\lim_{m \to \infty} \frac{x^{\top} A^{m+1} x}{x^{\top} A^m x} = \lim_{m \to \infty} \frac{(Px)^{\top} D^{m+1} (Px)}{(Px)^{\top} D^m (Px)}. 
$$
Write $x = \sum_{i=1}^n \alpha_i p_i$ where $Pp_i = e_i$ (here $\{e_1,\cdots,e_n\}$ is the standard basis of $\mathbb R^n$) and $De_i = \lambda_i e_i$. We re-write : 
$$
\lim_{m \to \infty} \frac{(Px)^{\top} D^{m+1} (Px)}{(Px)^{\top} D^m (Px)} = \lim_{m \to \infty} \frac{\sum_{i=1}^n \alpha_i^2 \lambda_i^{m+1}}{\sum_{i=1}^n \alpha_i^2 \lambda_i^m}.
$$
Let $I = \{ i \in \{1,\cdots,n\} \, | \, \alpha_i \neq 0 \}$ and $\lambda = \max\{\lambda_i \, | \, i \in I \}$. We have
$$
\frac{\sum_{i=1}^n \alpha_i^2 \lambda_i^{m+1}}{\sum_{i=1}^n \alpha_i^2 \lambda_i^m} = \frac{\sum_{i \in I} \alpha_i^2 \lambda_i (\lambda_i/\lambda)^m}{\sum_{i \in I} \alpha_i^2 (\lambda_i/\lambda)^m}. 
$$
Let $J \subseteq I$ be the set of indices for which $\lambda_i = \lambda$. We conclude
$$
\frac{\sum_{i \in I} \alpha_i^2 \lambda_i (\lambda_i/\lambda)^m}{\sum_{i \in I} \alpha_i^2 (\lambda_i/\lambda)^m} \underset{m \to \infty}{\longrightarrow} \frac{\sum_{j \in J} \alpha_j^2 \lambda_j}{\sum_{j \in J} \alpha_j^2} =  \frac{\sum_{j \in J} \alpha_i^2 \lambda}{\sum_{j \in J} \alpha_j^2} 
= \lambda.
$$
In other words, the final answer is 
$$
\max \{ \lambda_i \, | \, i \in \{1,\cdots,n\}, \langle x, P^{\top}e_i \rangle = \alpha_i \neq 0 \}
$$
where $\alpha_i$ is the coefficient above. Which in some sense is obvious when considering a diagonal matrix $A$ with positive elements on the diagonal ; the answer should depend on $x$. 
EDIT : ronno raised a good point. I decided to edit the above instead of re-writing the same arguments. 
Hope that helps,
A: Hint: $A$ has an orthonormal basis of eigenvectors.
A: By Spectral Theorem, $A$ is orthogonally similar to a diagonal matrix, i.e
$$
P^{T}AP=\pmatrix{\lambda_1 \\ & \ddots \\ && \lambda_n}
$$
where $\lambda_i>0$ is eigenvalue of $A$, and $\space P^{T}P=P^{-1}P=I$.
So we have
$$
P^{T}A^{m}P=\pmatrix{\lambda_1^m \\ & \ddots \\ && \lambda_n^m}
$$
Let $x=Py$. Then
\begin{align}
\dfrac{x^TA^{m+1}x}{x^TA^{m}x}&=\dfrac{y^TP^TA^{m+1}Py}{y^TP^TA^{m}Py}
\\
&=\dfrac{\sum\limits_{k=1}^n\lambda_k^{m+1}y_k^2}{\sum\limits_{k=1}^n\lambda_k^{m}y_k^2}
\\
&=\dfrac{\sum\limits_{k=1}^n\lambda_k(\lambda_k/\lambda_{max})^{m}y_k^2}{\sum\limits_{k=1}^n(\lambda_k/\lambda_{max})^my_k^2}
\end{align}
where $\lambda_{max}=\max\{\lambda_1,\cdots,\lambda_n\}$.
Suppose $\lambda_{k_i}=\lambda_{max}$. Then
$$
\lim_{m\to\infty}\dfrac{x^TA^{m+1}x}{x^TA^{m}x}=\dfrac{\sum\limits_{i=1}^l\lambda_{k_i}y_{k_i}^2}{\sum\limits_{i=1}^ly_{k_i}^2}=\lambda_{max}
$$
A: Consider the eigenvalues $\lambda_1 > \lambda_2 > \cdots > \lambda_n$ of $A$. Expand $x = \sum_{i} x_i e_i$ where $e_i$ are the orthonormal eigenvectors basis of $A$ (pointed out by Robert Israel), you will find your limit is
$$
\lim_{m \to \infty} \frac{\sum_{i} x_i^2 \lambda_i^{m+1}}{\sum_{i} x_i^2 \lambda_i^{m}} = \lambda_1
$$
try to see why.
