Show that $\lim (A_n\cdot x_n )=0$. Let  $(a_n)_n $ a sequence and $A_n=\max \{a_1,\dots,a_n\}$ for any $n\in \mathbb {N} $.
If $(x_n)_n$ is decreasing, $\lim x_n=0$ and $\lim a_n\cdot x_n=0$ show that $\lim A_n\cdot x_n =0$.
I tried to apply the definition of limit with $\varepsilon $ but I don't know why it's important the monotony and the convergence of $(x_n)_n. $
 A: Let $\varepsilon > 0$.
Since $\lim_n a_nx_n = 0$, there exists $n_0 \in \mathbb{N}$ such that $n \ge n_0 \implies |a_nx_n| < \frac{\varepsilon}2$.
Since $\lim_n x_n = 0$, there exists $n_1 \in \mathbb{N}$ such that $$n \ge n_1 \implies x_n \le \frac{\varepsilon}{2 \max\{|a_1|, \ldots, |a_{n_0}|\}}$$
For $n \ge \max\{n_0, n_1\}$ we have:
$$|A_nx_n| \le x_n \max\{|a_1|, \ldots, |a_{n}|\} \le x_n \max\{|a_1|, \ldots, |a_{n_0}|\} + x_n \max\{|a_{n_0+1}|, \ldots, |a_{n}|\}$$
Because $n \ge n_1$, we have that $x_n \max\{|a_1|, \ldots, |a_{n_0}|\} < \frac\varepsilon2$.  
Also, since $n \ge n_0$ and $(x_n)_n$ is decreasing, we have:
$$x_n |a_{n_0 + 1}| \le x_{n_0 + 1} |a_{n_0 + 1}| < \frac\varepsilon2$$
$$x_n |a_{n_0 + 2}| \le x_{n_0 + 2} |a_{n_0 + 2}| < \frac\varepsilon2$$
$$\vdots$$
$$x_{n}|a_{n-1}|\le x_{n-1}|a_{n-1}|<\frac\varepsilon2$$
$$x_n |a_{n}| < \frac\varepsilon2$$
Thus, $x_n \max\{|a_{n_0+1}|, \ldots, |a_{n}|\} < \frac{\varepsilon}2$ so we obtain:
$$|A_nx_n| \le \frac{\varepsilon}2 + \frac{\varepsilon}2 = \varepsilon$$
Therefore, $\lim_n A_nx_n  = 0$.
