$x_n \in \mathbb R, \quad x_n \to A \implies \max \{x_n,A-x_n\} \to ?$ I am stuck on the following problem that is as follows:  

Let $\{x_n\}_{n \in \mathbb N}$ be a sequence of real numbers converging to $A \neq 0.$ Let $y_n=\max \{x_n,A-x_n\}$.Then the limit of the sequence  $\{y_n\}_{n \in \mathbb N}$ is always   



*

*$0$  

*$ A  $

*$\max \{A,0\}$  

*$-\min \{A,0\}$   


I do not know how to progress with the problem.Can someone point me in the right direction?Thanks in advance for your time.
 A: Hint: if $f$ is a continuous function and $x_n\to A$ then $f(x_n)\to f(A)$.
A: Try examples and think of the general case:
1) A monotone descending sequence
$$\frac{n+1}{n}\xrightarrow[n\to\infty]{}1$$
2) A monotone increasing sequence
$$\frac{n}{n+1}\xrightarrow[n\to\infty]{} 1$$
3) Non monotone sequence
$$\frac{(-1)^n}{n}\xrightarrow[n\to\infty]{}0$$
4) Non-monotone sequence converging to $\,A\neq 0\,$:
$$1+\frac{(-1)^n}{n}\xrightarrow[n\to\infty]{}1\ldots$$
A: Since $x_{n}\to A$, then $A-x_{n}\to 0$. Since $A\neq 0$, then you find $\varepsilon>0$ so that $B(A,\varepsilon)\cap B(0,\varepsilon)=\emptyset$. By definition of convergence we find $n_{\varepsilon}\in\mathbb{N}$ so that $x_{n}\in B(A,\varepsilon)$ and $A-x_{n}\in B(0,\varepsilon)$ for all $n\geq n_{\varepsilon}$. 


*

*If $A<-\varepsilon$, then $y_{n}=\max\{x_{n},A-x_{n}\}=A-x_{n}$ for all $n\geq n_{\varepsilon}$, whence $y_{n}\to 0=\max\{A,0\}$. 

*And if $A>\varepsilon$, then $y_{n}=\max\{x_{n},A-x_{n}\}=x_{n}$ for all $n\geq n_{\varepsilon}$, whence $y_{n}\to A=\max\{A,0\}$.
Since $A\notin B(0,\varepsilon)$ then 1. and 2. cover all the possible cases. Hence $y_{n}\to \max\{A,0\}$.
