This has been answered before on this site but I don't really understand those answers enough so I am asking again.
Prove that if $\{a_n\}$ converges to $l\in\mathbb{R}$, then $\{a_n^2\}$ converges to $l^2$.
Pf.
Suppose $\{a_n\}$ converges to $l\in\mathbb{R}$.
This means $\forall\epsilon>0,\exists N_1>0, s.t,\forall n\in\mathbb{N},\text{ if } n>N_1,\text { then } |a_n-l|<\epsilon$
I want to show that $\forall\epsilon>0,\exists N_2>0, s.t,\forall n\in\mathbb{N},\text{ if } n>N_2,\text { then } |a_n^2-l^2|<\epsilon$
Note: $a_n^2$ is the entire sequence, squared, evidently the same as $(a_n)^2$.
Consider $|a_n^2-l^2| = |a_n-l||a_n+l|$.
If $n>N_1$, then $|a_n-l||a_n+l|<\epsilon|a_n+l|$.
How do I continue from here?
Best Answer
Same approach but slightly shorter.
Suppose $\lim_{n\to\infty}a_n=l$. Let $B=\sup\{\vert a_n+l\vert\}$ for all $n$.
Let $\epsilon>0$ and let $N>0$ such that if $n>N$ then
$$ \vert a_n-l\vert<\frac{\epsilon}{B}$$
Then
$$ \vert a_n-l\vert\cdot\vert a_n+l \vert<\frac{\epsilon}{B}\cdot B$$
$$ \vert a_n^2-l^2\vert<\epsilon$$