Prove that if $\left\{ x_{n}\right\} $ converges then $\left\{ \left(x_{n}\right)^{2}\right\} $ converges One person explain me this prove, but i dont understand this:
Why $2M\mid x_{n}-x\mid\leq2M\frac{\epsilon}{2M}=\epsilon$? 
The proof of the exercise:
Suppose that $x_n\to x$, for some $x\in\mathbb{R}$. Also suppose you have $\epsilon>0$.
Then $\exists n_0\in\Bbb N:\forall n\geq n_0 \ \ |x_n-x|<\epsilon.$
Now, because $\{x_n\}$ is a convergent sequence we have that ${x_n}$ is bounded and so $\exists M\in\Bbb {R^+}:\forall n\in\Bbb N \ \ |x_n|\le M$ (and so $x\le M$).
Then, $|(x_n)^2-x^2|=|x_n-x||x_n+x|\le (|x_n|+|x|)|x_n-x|\le 2M|x_n-x|\le 2M\cfrac{\epsilon}{2M}=\epsilon$
and so $(x_n)^2\to x^2$.
Now, I have a question. Why $2M\mid x_{n}-x\mid\leq2M\frac{\epsilon}{2M}=\epsilon$ ??? Thanks!
 A: You should fix a bound $M$ from the start. Then we can say that $\lvert x_n - x \rvert < \frac{\epsilon}{2M}$ for $n$ large enough.
A: First notice that $\epsilon/2M>0$, where $M\ge |x_n|$ for all $n\in\Bbb N$, i.e. the sequence $\{x_n\}$ is bounded due that is convergent. 
Then cause $\{x_n\}$ converges exists some $N$ such that
$$|x_n-x|<\epsilon/2M$$
when $n\ge N$. You can call $\epsilon/2M=\gamma$ if you want... the point is that exists some $N$ for this undefined but positive quantity.
Then from here you shows that $|x_n^2-x|<\epsilon$ in a chain of inequalities.

An "alternative" proof: exists some $N$ such that $|x_n-x|<\epsilon$ when $n\ge N$, cause $\{x_n\}$ converges.
Then we have that
$$|x_n^2-x^2|=|x_n+x||x_n-x|$$
Because $\{x_n\}$ is bounded, by example by $M$, then $|x_n+x|\le 2M$, then
$$|x_n^2-x^2|=|x_n+x||x_n-x|\le 2M|x_n-x|<2M\epsilon$$
then
$$|x_n^2-x|<2M\epsilon,\quad\forall n\ge N$$
and notice that $2M\epsilon$ is an undefined and arbitrarily small quantity (because $\epsilon$ is arbitrarily small), so the proof is done. What I want to show here is that you dont need to end the proof showing explicitly that $|x_n^2-x|<\epsilon$ because if $2M$ is a positive constant then $2M\epsilon$ is completely undefined too.
A: Here is a one-line proof using a trick.
$\lim_{n\to\infty} {x_n}^2 = \lim_{n\to\infty} (x_n-L) \times \lim_{n\to\infty} (x_n+L) + L^2 = 0 \times ( \lim_{n\to\infty} x_n + L ) + L^2 = L^2$.
