# Sequences with two other sequences

Let $$(z_n) \subset \mathbb{R}$$ be a sequence defined, for each $$n \in \mathbb{N}$$, by:

$$z_n=min\{x_n,y_n\}$$

With $$(x_n)$$ and $$(y_n)$$ being two convergent sequences in $$\mathbb{R}$$ with $$\lim{x_n}=a$$ and $$\lim{y_n}=b$$.

Show that $$(z_n)$$ is also convergent and calculate the limit.

First I've separated in two cases: $$a=b$$ and $$a \not= b$$. For the first one, we have:

• $$\forall \varepsilon > 0, \exists n_0 \in \mathbb{N} ; n > n_0 \Rightarrow |x_n - a|<\varepsilon$$
• $$\forall \varepsilon > 0, \exists n_1 \in \mathbb{N} ; n > n_1 \Rightarrow |y_n - a|<\varepsilon$$

So, if $$n_\delta = max\{n_0,n_1\}$$, we have: $$n > n_\delta \Rightarrow |z_n-a|<\varepsilon$$ By the defnition of $$z_n$$ sequence. (Right?)

But what about the second case? I can't come up with anything. Leads?

Hint: If $$a \neq b$$, without loss of generality, assume $$a < b$$. So, as usual, there are $$n_1, n_2 \in \Bbb N$$ such that $$\Bbb N \ni n > n_1$$ implies $$|x_n - a| < \frac{b - a}{2} \implies x_n - a < \frac{b - a}{2}$$ and $$\Bbb N \ni n > n_2$$ implies $$|y_n - b| < \frac{b - a}{2} \implies -\frac{b - a}{2} < y_n - b$$. So what can you say about $$x_n$$ and $$y_n$$ for all $$\Bbb N\ \ni n > \max\{n_1, n_2\}$$, after some rearrangement?
Hint. Taking $$\lim{x_n}=a$$ and $$\lim{y_n}=b$$ for second case let assume, that $$a. This means that $$\exists N \in \mathbb{N}$$ such that for $$n>N$$ $$z_n=\min\{x_n,y_n\} = x_n$$.