TITLE: If $x_n \geq 0$ for all n $\in N$ and $lim((-1)^nx_n)$ exists. Show that $x_n$ converges.
QUESTION [8 upvotes]: If $x_n \geq 0$ for all n $\in N$ and $lim((-1)^nx_n)$ exists. Show that $x_n$ converges.
Let $lim((-1)^nx_n)=l$
therefore,
for $\epsilon>0$ $\exists k\in N $  such that 
$|(-1)^nx_n - l|<\epsilon/2$            $\forall n\geq k$
$=>|x_n + l| < \epsilon/2 $            $\forall n\geq k$ & n is odd
$-\epsilon/2 - l<x_n<\epsilon/2-l$       $\forall n\geq k$ & n is odd --> eq 1
also,
$|x_n - l| < \epsilon/2 $            $\forall n\geq k$ & n is even
$=>-\epsilon/2 +l<x_n<\epsilon/2 + l$ $\forall n\geq k$ & n is even --eq 2
from 1 and 2,
$-\epsilon  < x_n < \epsilon$   $\forall n\geq k$
$=> |x_n - 0| < \epsilon $  $\forall n\geq k$
Hence $lim(x_n) = 0 $
Is this argument correct?

REPLY [4 votes]: I think this approach might be better (As also pointed in the comments by @ ΘΣΦGenSan above):
Use triangle inequality at the very early step.
$||x_n|-|l|| \le |(-1)^n x_n-l| \lt \frac {\epsilon}2 \; \forall \; n \ge k$.
But since $x_n \ge 0 \; \forall \; n \in \Bbb N$ is given to us in hypothesis.
$\therefore |x_n-|l|| \lt \frac {\epsilon}2 \lt \epsilon \; \forall \; n \ge k$.
From here, we can conclude that $x_n$ converges.

EDIT: $x_n$ indeed converges to $0$ but not by adding ($1$) and ($2$). My answer was only meant to show that $x_n$ is convergent.
To show that it converges to $0$ specifically, first note that $l$ is a real number.
Assume $l \gt 0$. Then by your ($1$) and using $x_n \ge 0 \; \forall \; n \in \Bbb N$, we get $0 \le x_n \lt \frac {\epsilon}2-l \; \forall \; n \ge k$. Hence for very small $\epsilon$, we get absurdity.
Assume $l \lt 0$. Then your ($2$) and the property $x_n \ge 0 \; \forall \; n \in \Bbb N$ yields $0 \le x_n \lt \frac {\epsilon}2 +l \; \forall \; n \ge k$. Again for very small $\epsilon$ values this produces absurdity.
Therefore by Trichotomy property of real numbers, $l=0$.
(You can check that for $l=0$, $0 \le x_n \lt \frac {\epsilon}2 \; \forall \; n \ge k$).