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I came across the following problem that says:

Let the sequence $\{x_n\} $where $n \in N$ of real numbers converge to a non-zero real number $a$ and let $y_n=a-x_n$.Then max{$x_n,y_n$} converge to which of the following?
$1.a$ always
$2.0$ always
$3.$max{a,0}
$4.$min{a,0}

If I take $x_n=a+1/n$ then $x_n \rightarrow a$ where as $y_n \rightarrow 0$ as $n \rightarrow \infty.$ But from hereon,I can not progress.Can someone point me in the right direction? Thanks in advance for your time.

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1 Answer 1

up vote 1 down vote accepted

There are three cases to examine: $a$ positive, $a$ negative, and $a=0$.

Note that the sequence $(y_n)$ always converges to $0$. Let $z_n=\max(x_n,y_n)$.

Let $a$ be positive. Then if $n$ is large enough, $x_n\gt a/2$, and $y_n\lt a/2$, so $\max(x_n,y_n)=x_n$. So the sequence $(z_n)$ converges to $a$.

Let $a$ be negative. By a similar argument, if $n$ is large enough, $x_n\lt a/2$, while $y_n\gt a/2$, so $\max(x_n,y_n)=y_n$. So the sequence $(z_n)$ converges to $0$.

Finally, if $a=0$, then the sequence $(z_n)$ converges to $0$, by the Comparison Test, since $|z_n|\le |x_n|+|y_n|$.

Thus in all cases, the sequence $(z_n)$ converges to $\max(a,0)$.

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