Prove that the limit of $(-1)^n x_n$ is $0$ I'm given a sequence $x_n$ which is $\geq 0$ for all $n\in\mathbb N$ and that the limit $$\lim_{n\to\infty}(-1)^n x_n$$ exists.
I have to prove that the limit $$\lim_{n\to\infty}(-1)^n x_n$$ is $0$. 
I think I can use the properties of subsequences here?
 A: Yes actually you almost gave the answer with your sentence: "I think I can use the properties of subsequences here?".
Let us wrtie $u_n = (-1)^n x_n$ for $n$ a positive integer. We denote $l \in \mathbb R$ the limit of the sequence $(u_n)_n$. Since $(u_{2n})_n$ and $(u_{2n+1})_n$ are subsequence of $(u_n)_n$, they also converge to $l$. Now remark that for every $n$, $u_{2n}=x_{2n} \geq 0$. Thus $l = \lim\limits_{n \to \infty} u_{2n} \geq 0$. In the same way we deduce that $l = \lim\limits_{n \to \infty} u_{2n+1} \leq 0$. Therefore $l = 0$.
Another way to prove this is to assume $l \neq 0$ and get a contradiction from the definition of $l$ is the limit of $(u_n)_n$.
I hope I have made myself clear enough.
A: Let $L=\lim_{n\to \infty} (-1)^n x_n$
Let $L\neq 0$.
CASE I:Let $L>0$ .Then we can choose $\epsilon=\frac{L}{2}$ such that forall $n\ge m$ $\frac{L}{2}<(-1)^n x_n<\frac{3L}{2}$ which is false as $x_n\ge 0$
CASE  II: Let  $L<0$ .Then..
A: Hint:


*

*Half of the elements of the sequence $(-1)^nx_n$ are smaller than or equal to $0$, and hald are greater than or equal to $0$.

*You can use the fact above to construct a proof by contradiction. Assume that the limit is $>0$ and construct a contradiction, then assume that the limit is $<0$ and do the same.

