Prove that if $\{a_n\}$ converges to $a$, that $\{(a_n)^2\}$ converges to $a^2$ by definition

Question

Prove that if $\{a_n\}$ converges to $a$, that $\{(a_n)^2\}$ converges to $a^2$ by definition.

Pf.

Suppose $\{a_n\}$ converges to $a$.

This means: $$\forall \epsilon >0, \exists N>0, \text{s.t, if} \text{ }n>N, |a_n-a|< \epsilon$$

I want to show that: $$\forall \epsilon >0, \exists N>0, \text{s.t, if} \text{ }n>N, |a_n^2-a^2|< \epsilon$$

How do I use the given info? I don't know how I can manipulate it to get $a_n^2$

Answer 1

Write $a_n^2-a^2 =(a_n-a)(a_n+a) $ and assume that $a_n$ is not too far from $a$.

My answer here shows that this works for $a_n^m$ for any $m$:

Show that $\lim_{x \to c} x^{3}=c^{3}$