If a sequence has limit, have another two limit? Here's a nice question:
For a sequence $(a_n)_{n\geq0}$ we define the sequences $(x_n)_{n\geq0}$ and $(y_n)_{n\geq0}$
such that $x_n=\min(a_n,a_{n+1})$ and $y_n=\max(a_n,a_{n+1}),n\geq0$.
Prove, that if the sequence $(a_n)_{n\geq0}$ has limit, then the sequences $(x_n)_{n\geq0}$ and $(y_n)_{n\geq0}$ have limit. Is the reciprocal true?
 A: $$\min(a,b)=\frac{a+b-|a-b|}2$$
Therefore $\lim\limits_{n\to\infty}\min(a_n,b_n)=\min(\lim\limits_{n\to\infty}a_n,\lim\limits_{n\to\infty}b_n)$ if limits on the right hand side exist because the above operations preserve limits.
In your case $\lim\limits_{n\to\infty}\min(a_n,a_{n+1})=\min(A, A)=A$, where $A=\lim\limits_{n\to\infty}a_n=\lim\limits_{n\to\infty}a_{n+1}$.
The $\max$ solution is almost identical because $\displaystyle\max(a,b)=\frac{a+b+|a-b|}2$.
For the converse, just consider the sequence $a_n=(-1)^n$.
A: The reciprocal is not true $a_n=(-1)^n$ does not converge but here $x_n=1$, $y_n=-1$ so they do converge.
Also the first assumption is true since convergence implies cauchy convergence, leading $x_n$ and $y_n$ to converge, i.e as $n\rightarrow\infty$, $|a_n-a_{n+1}|\rightarrow0$
A: If $a_n \to l$, then for all $\epsilon > 0$ there exists $N > 0$ such that $\forall n \geq N, |a_n - l| < \epsilon$. Indeed you also have $|a_{n+1} - l| < \epsilon$, and so both $|x_n - l| < \epsilon$ and $|y_n - l| < \epsilon$. This proves that $x_n \to l$ and $y_n \to l$.
The reciprocal isn't true; for $a_n = (-1)^n$, $x_n = -1$ and $y_n = 1$ both converge but $a_n$ doesn't converge.
A: To add some insight with respect to the other (correct) answers. Generally speaking if $x_n$ is a sequence and $n_k$ is a non decreasing sequence of natural numbers such that $n_k\to \infty$, then $x_{n_k}$ has the same limit of $x_n$ (if $x_n$ has a limit). The converse is generally not true.
