Calculate the limit $\frac{(2n-1)x_1+(2n-3)x_2+\dots +3x_{n-1}+x_n}{n^2}$ when $x_n\to x$. 
Let $\{x_n\}$ be a sequence in $\Bbb R$ and $x_n\to x$ as $n\to \infty$.
Then $$\frac{(2n-1)x_1+(2n-3)x_2+\dots +3x_{n-1}+x_n}{n^2}\to x$$.

Do anyone know how to solve this kind of problem efficiently? I think I need to estimate $$\left|\frac{(2n-1)x_1+(2n-3)x_2+\dots +3x_{n-1}+x_n}{n^2}-x\right|$$ Is there some other ways to calculate the limit directly?
Thanks for any comments.

This limit problem looks very much like intentionally designed for the use of the Stolz-Cesàro theorem:
So, let
$$a_n =\sum_{k=1}^n(2n-(2k-1))x_k = \sum_{k=1}^n(2(n-k)+1)x_k$$
and $$b_n = n^2$$
So, $$a_{n+1}-a_n = \sum_{k=1}^{n+1}(2(n+1-k)+1)x_k - \sum_{k=1}^n(2(n-k)+1)x_k$$$$ = x_{n+1}+2\sum_{k=1}^n x_k$$
Hence,
$$\frac{a_{n+1}-a_n}{b_{n+1}-b_n}= \frac{x_{n+1}+2\sum_{k=1}^n x_k}{2n}=\underbrace{\frac{x_{n+1}}{2n}}_{\stackrel{n\to\infty}{\longrightarrow}0} + \underbrace{\frac{\sum_{k=1}^n x_k}{n}}_{\stackrel{n\to\infty}{\longrightarrow}x}$$
where the last limit is just another application of Stolz-Cesàro.
