Limit of quotient sequence Let $\{x_n\}_{n\in\mathbb{N}}$ be a sequence with $\lim\limits_{n\to\infty}x_n = a$, $\{t_n\}_{n\in\mathbb{N}}$ be a sequence with $\lim\limits_{n\to\infty}t_1+t_2+\ldots+t_n = +\infty$.
Prove that
$$\lim_{n\to\infty}\frac{t_1x_1+t_2x_2+\ldots+t_nx_n}{t_1+t_2+\ldots+t_n} = a$$
My idea is use that 
$$(t_1+\ldots+t_n)\min\{x_n: n\in\mathbb{N}\}\leq t_1x_1+\ldots+t_nx_n \leq (t_1+\ldots+t_n)\max\{x_n: n\in\mathbb{N}\}$$
But I'm not sure if $\lim\limits_{n\to\infty}\min\{x_n: n\in\mathbb{N}\} = \lim\limits_{n\to\infty}\max\{x_n: n\in\mathbb{N}\} = \lim\limits_{n\to\infty}x_n$.
 A: From your idea, I think you assume that $t_n>0$ when $n$ is large enough. 
Then, we can use Stolz Theorem:
$$\lim_{n\to\infty}\frac{t_1x_1+t_2x_2+\ldots+t_nx_n}{t_1+t_2+\ldots+t_n} = \lim_{n\to\infty}\frac{t_nx_n}{t_n}. $$
A: For the argument below, we need to assume that $t_n>0$ eventually. 
Fix $\varepsilon>0$. Then there exists $n_0$ such that $|x_n-a|<\varepsilon$ for all $n>n_0$. Then
$$\left|\frac{t_1x_1+t_2x_2+\ldots+t_nx_n}{t_1+t_2+\ldots+t_n} -a\right|=\left|\frac{t_1(x_1-a)+\cdots +t_n(x_n-a)}{t_1+\cdots+t_n}\right|\leq\left|\frac{t_1(x_1-a)+\cdots +t_{n_0}(x_{n_0}-a)}{t_1+\cdots+t_n}\right|+\left|\frac{t_{n_0+1}(x_{n_0+1}-a)+\cdots +t_n(x_n-a)}{t_1+\cdots+t_n}\right| \\
\leq\left|\frac{t_1(x_1-a)+\cdots +t_{n_0}(x_{n_0}-a)}{t_1+\cdots+t_n}\right|+
\frac{t_{n_0+1}|x_{n_0+1}-a|+\cdots +t_n|x_n-a|}{t_1+\cdots+t_n}\\
\leq\left|\frac{t_1(x_1-a)+\cdots +t_{n_0}(x_{n_0}-a)}{t_1+\cdots+t_n}\right|+
\varepsilon \,\frac{t_{n_0+1}+\cdots +t_n}{t_1+\cdots+t_n}\\
\leq \frac{(|a|+\max\{|x_n|\})(|t_1|+\cdots+|t_{n_0}|)}{t_1+\cdots+t_n}+\varepsilon.
$$
Taking $\limsup$ we get
$$
\limsup_n\left|\frac{t_1x_1+t_2x_2+\ldots+t_nx_n}{t_1+t_2+\ldots+t_n} -a\right|
\leq\varepsilon.
$$
As $\varepsilon$ was arbitrary, we conclude that 
$$
\lim_n\left|\frac{t_1x_1+t_2x_2+\ldots+t_nx_n}{t_1+t_2+\ldots+t_n} -a\right|=0.
$$
