If $\lim(x_n) \rightarrow x$ then show that $\lim(x_n)^2 \rightarrow x^2$ If  $\lim(x_n) \rightarrow x$ then for all $\epsilon > 0$ I know there is an $ \geq N \text{ s.t. } |(x_n) - x|< \epsilon$
So I'm starting with $|(x_n)^2-x^2|=|(x_n)-x||(x_n)+x|<\epsilon $
But I'm stuck here, can anyone help, please?
Thanks!

$$
|x_n^2-x^2| = |x_n-x||x_n+x|
$$
The first factor on the right can be made small by making $n$ big enough.  The problem is what to do with the second term.
If $|x_n-x|<\varepsilon \le 1$ then $x-1<x_n<x+1$, so
$$|x_n+x|\le |x_n|+|x| \le (|x|+1)+|x|=2|x|+1.$$
Now given any $\varepsilon>0$, there exists $N$ so large that if $n\ge N$ then
\begin{align}
|x_n-x| & < \frac\varepsilon{2|x|+1} \\[10pt]
\text{and } |x_n-x| & < 1.
\end{align}
Therefore
$$
|x_n^2-x^2| =|x_n-x||x_n+x| \le |x_n-x| \cdot (2|x|+1) < \frac\varepsilon{2|x|+1} \cdot (2|x|+1) = \varepsilon.
$$
Thus
$$
|x_n^2-x^2| < \varepsilon.
$$
