Suppose that $(x_n)$ and $(y_n)$ are convergent sequences and let un=min{xn,yn}. Prove that (un) is a convergent sequence Suppose that $(x_n)$ and $(y_n)$ are convergent sequences and let $u_n=min${$x_n,y_n$}. Prove that $(u_n)$ is a convergent sequence
I feel like this should be handled by cases but it seems to be the more roundabout way of solving.  You could take cases such that $(x_n) < (y_n), (x_n) = (y_n)$ or $(x_n) > (y_n)$. However, I feel like I'm missing a case (such as when $(x_n)$ crosses $(y_n) $ or one is between the other)?
 A: Hint: $\min\{a,b\}=\frac{a+b}{2}-\frac{|a-b|}{2}$
A: We can do it from basics. Let the sequence $(x_n)$ converge to $x$, and the sequence $(y_n)$ converge to $y$. There are three possibilities: (i) $x= y$; (ii) $x\lt y$; (iii) $y\lt x$. By symmetry we need only consider cases (i) and (ii).
Case (i): Let $u_n=\min(x_n,y_n)$. We show that the sequence $(u_n)$ has limit $x$.
We need to show that for any $\epsilon\gt 0$, there is an $N$ such that if $n\gt N$ then $|u_n-x|\lt \epsilon$.
There is an $N_x$ such that if $n\gt N_x$, then $|x_n-x|\lt \epsilon$. There is an $N_y$ such that if $n\gt N_y$ then $|y_n-x|\lt \epsilon$. Let $N=\max(N_x,N_y)$. If $n\gt N$, then we have $|x_n-x|\lt \epsilon$ and $|y_n-x|\lt \epsilon$. Since $u_n=x_n$ or $u_n=y_n$, we have $|u_n-x|\lt \epsilon$.  
Case (ii): Let $d=y-x$. There is an $N_x$ such that if $n\gt N_x$, then $|x_n-x|\lt \min(d/2,\epsilon)$. Similarly, there is a an $N_y$ such that if $n\gt N_y$ then $|y_n-y|\lt \min(d/2,\epsilon)$. Let $N=\max(N_x,N_y)$. If $n\gt N$, then both inequalities hold. It follows that if $n\gt N$, we have $u_n=x_n$ and therefore $|u_n-x|\lt \epsilon$. 
A: HINT:
The function $(x,y) \mapsto \min(x,y)$ is continuous.
A: Sketch: $$|x_n-X| < \epsilon, \forall n \geq N_1$$ $$|y_n-Y|<\epsilon, \forall n \geq N_2$$
Taking $N=max\{N_1,N_2\}$ we have convergence for both when $n \geq N$.
In the end the only thing that will matter is the $max\{X,Y\}$. WLOG suppose $X>Y$, then there exists $m$ such that: $x_m>y_n$ $\forall m >n \Rightarrow (u_n) \to Y.$ 
A: Call $x$ and $y$ the limits of $x_n$ and $y_n$ respectively. You need to check what happen only is two cases:


*

*$x < y$ (the case $y < x$ can be of course treated in the same way)

*$x = y$


You don't care if the sequences cross or whatever, what matters is what happen definitely.
The first case is easy: if $x < y$ then there exists a positive (possibly small) number $\eta$ such that $|x - y| \ge \eta$. Now pick $N_1$ and $N_2$ such that 


*

*$n \ge N_1$ we have $|x - x_n| < \eta/2$

*$n \ge N_2$ we have $|y - y_n| < \eta/2$
(this can be done since we know that the sequences are converging to $x$ and $y$)


Then for $n \ge \max\{N_1,N_2\}$ we have $\min\{x_n,y_n\} = x_n \to x$.
For the second case you can prove directly that the $\epsilon$-$\delta$ definition of the limit holds.
I hope this helps :)
