Sequences with two other sequences

Question

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?

Answer 1

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?

Answer 2

Hint. Taking $\lim{x_n}=a $ and $\lim{y_n}=b$ for second case let assume, that $a<b$. This means that $\exists N \in \mathbb{N}$ such that for $n>N$ $z_n=\min\{x_n,y_n\} = x_n$.