Prove that if $\left\{ x_{n}\right\} $ converges then $\left\{ \left(x_{n}\right)^{2}\right\} $ converges. Good night. I have a problem with this problem. I tried the following:
Proof:
Let $\left\{ x_{n}\right\}   $ be a convergent sequence. By definition:
$\mid x_{n}-x\mid<\epsilon$
Then:
$\mid\mid x_{n}\mid-\mid x\mid\mid<\mid x_{n}-x\mid$
$(\mid\mid x_{n}\mid-\mid x\mid\mid)^{2}<(\mid x_{n}-x\mid)^{2}$
$\mid(\mid x_{n}\mid-\mid x\mid)^{2}\mid<(\mid x_{n}-x\mid)^{2}$
$\mid\mid x_{n}\mid^{2}-\mid\left(x_{n}\right)\left(x\right)\mid+\mid x\mid^{2}\mid<(\mid x_{n}-x\mid)^{2}$
I don't know how finish. Please help ):
 A: Suppose that $x_n\to x$, for some $x\in\mathbb{R}$. Also suppose you have $\epsilon>0$.
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 $x\le M$) and because $x_n\to x$  $\exists n_0\in\Bbb N:\forall n\geq n_0 \ \ |x_n-x|<\frac{\epsilon}{2M}.$
Then, for all $n\geq n_0$ we have that $|(x_n)^2-x^2|=|x_n-x||x_n+x|\le (|x_n|+|x|)|x_n-x|\le 2M|x_n-x|\le 2M\cfrac{\epsilon}{2M}=\epsilon$
and so $(x_n)^2\to x^2$.
A: Hint: if $a=\lim_{n\to \infty}a_n$ exists and $b=\lim_{n\to \infty}b_n$ exists, then $\lim_{n\to \infty}a_nb_n$ exists and equals $ab$.
A: We have $$\forall { \varepsilon  }_{ 1 }>0\quad ,\exists { n }_{ { \varepsilon  }_{ 1 } },n\ge { n }_{ { \varepsilon  }_{ 1 } }\quad \mid x_{ n }-x\mid <{ \varepsilon  }_{ 1 }$$ and we have to show $$\lim _{ n\rightarrow \infty  }{ { x }_{ n }^{ 2 } } =a $$ where $a$ is  equal to $x=\sqrt { a } ,$ $$\quad \mid { x }_{ n }^{ 2 }-a\mid =\left| \left( { x }_{ n }+\sqrt { a }  \right) \left( { x }_{ n }-\sqrt { a }  \right)  \right| =\left| { x }_{ n }+\sqrt { a }  \right| \left| { x }_{ n }-\sqrt { a }  \right| <{ \varepsilon  }_{ 1 }\left| { x }_{ n }+\sqrt { a }  \right| =\\ ={ \varepsilon  }_{ 1 }\left| { { x }_{ n }-\sqrt { a } +2\sqrt { a }  } \right| <{ \varepsilon  }_{ 1 }\left| { x }_{ n }-\sqrt { a }  \right| +2={ \varepsilon  }_{ 1 }^{ 2 }+2\left| { \varepsilon  }_{ 1 }\sqrt { a }  \right| $$
 and ${ \varepsilon  }_{ 2 }={ \varepsilon  }_{ 1 }^{ 2 }+2\left| { \varepsilon  }_{ 1 }\sqrt { a }  \right|  $ 
so we have $\forall { \varepsilon  }_{ 2 }>0\quad ,\exists { n }_{ { \varepsilon  }_{ 2 } },n\ge { n }_{ { \varepsilon  }_{ 2 } }$

$$|{ x }_{ n }^{ 2 }-a\mid <{ \varepsilon  }_{ 2 }$$

A: Suppose that the sequence $\{x_n\}$ converges to $a$. We claim that the squared sequence converges to $a^2$. Note that $$|x_n^2-a^2|=|x_n-a||x_n+a|.$$ We bound $|x_n+a|$ by requiring that $|x_n-a|<1$ so that $|x_n+a|=|x_n-a+2a|<1+2|a|$ . Given $\epsilon>0$ choose $N$ s.t. $n\geq N$ implies that $|x_n-a|<\min(1, \epsilon/(1+2|a|))$ and the result follows.
