If $x_n \geq 0$ for all n $\in N$ and $\lim((-1)^nx_n)$ exists. Show that $x_n$ converges. 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
$| (-1)^nx_n - l|<\epsilon/2$ $\forall n\geq k$
$\implies |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 $\hspace{5mm} (1) $
also,
$|x_n - l| < \epsilon/2 $ $\forall n\geq k$ & n is even
$\implies-\epsilon/2 +l<x_n<\epsilon/2 + l$ $\forall n\geq k$ & n is even $\hspace{5mm} (2)$
from (1) and (2),
$-\epsilon  < x_n < \epsilon$ $\forall n\geq k$
$\implies |x_n - 0| < \epsilon $ $\forall n\geq k$
Hence $\lim(x_n) = 0 $
Is this argument correct?
 A: 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 $l=0$, $0 \le x_n \lt \frac {\epsilon}2 \; \forall \; n \ge k$).
A: Let $l=\lim_{n\rightarrow\infty}\left[(-1)^{n}x_{n}\right].$ We go
to prove that $l=0$. Prove by contradiction. Suppose the contrary
that $l\neq0$. Let $\varepsilon=\frac{|l|}{2}>0$, then there exists
$N$ such that $\left|(-1)^{n}x_{n}-l\right|<\varepsilon$ whenever $n\geq N$. Consider
two cases. Case 1: $l>0$. In this case, $(-1)^{n}x_{n}>l-\varepsilon=\frac{l}{2}$
whenever $n\geq N$. Choose an odd integer $n\geq N$, then $-x_{n}=(-1)^{n}x_{n}>\frac{l}{2}$,
which is a contradiction because it is given that $x_{n}\geq0$.
Case 2: $l<0$. In this case, $(-1)^{n}x_{n}<l+\varepsilon=\frac{l}{2}$
whenever $n\geq N$. Choose an even integer $n\geq N$, then $x_{n}=(-1)^{n}x_{n}<\frac{l}{2}<0$
which is a contradiction.

Now, it is clear that $\lim_{n\rightarrow\infty}x_{n}=0$ by observing
that $\lim_{n}x_{n}=\lim_{n\rightarrow\infty}(-1)^{n}\left[(-1)^{n}x_{n}\right]=0$.
Note that $(-1)^{n}$ is bounded while $(-1)^{n}x_{n}\rightarrow0$,
so the product converges to $0$.
A: I might be very late to this one, but consider this (Your method is correct, but I think I have a better one, and a very useful lemma):
Lemma: Every alternating sequence converges to 0, or else it diverges.
proof:
consider a sequence $(x_{n})$ which is alternating, but has a non zero limit $L$.
case 1: Limit $L>0$
$\forall \epsilon\in\mathbb{R^+}$ $\exists k\in\mathbb{N}$ such that $|x_{n}-L|<\epsilon$ $ \forall n\geq k$
$\implies L-\epsilon<x_{n}<L+\epsilon$ $\forall n\geq k$
choose $\epsilon=L $ (since L>0)
we get:
$0<x_{n}<2L$ $\forall n \geq k$
This contradicts the fact that $(x_{n})$ is an alternating sequence. Same thing can be done for the case of $L<0$ $\square$
now given an alternating sequence $(-1)^nx_{n}$, we are told it converges, and from the lemma we know it must converge to 0. hence,
$\forall \epsilon \in \mathbb{R^+}$ $\exists k \in \mathbb{N}$ such that $|(-1)^nx_{n}|<\epsilon$ $\forall n \geq k$ $\implies$ $|x_{n}|<\epsilon$ $\forall n \geq k$
We have shown now that the sequence $(x_{n})$ converges to 0
The lemma is useful in many places where an alternating sequence might come up. Instead of checking for an arbitrary L to be the limit, we simply have to check for 0 and it simplifies the problem by a lot.
