Prove that $\lim_{n\to\infty}a_n^2=a^2$ when ${a_n}$ is a sequence in $\mathbb{R}$ with $\lim_{n\to\infty}a_n=a$ Question:
Let ${a_n}$ be a sequence in $\mathbb{R}$ with $\lim_{n\to\infty}a_n=a$
Prove that
$\lim_{n\to\infty}a_n^2=a^2$
Attempt:
Without using any properties of limits (this is a question in a section of the book that came before any list of any limit theorems)
By definition:
$|a_n^2-a^2|<\epsilon$ 
$|a_n^2-a^2|$ $\leq$ $|(a_n+a)(a_n-a)|$ $\leq$ $|(a_n+a)| |(a_n-a)|$ $\leq$ 
$(|a_n | + |a|)(|a_n -a|)$ $<$ $\epsilon$
And since I know that $\lim_{n\to\infty}a_n=a$, I'll subsitute that in for $|a_n|^2$ as $a^2$ 
$\therefore$ $a^2-|a|^2 < \epsilon$ which is true since $a^2-|a|^2$ in my case.
I wonder if this is a possible solution?
 A: You have written down the intuition towards a formal proof.
You are given an $\epsilon \gt 0$, and want to show there is an $n$ such that if $n\gt N$ then $|a_n^2-a^2|\lt \epsilon$.
By your calculation, 
$$|a_n^2-a^2|\le (|a_n|+|a|)(|a_n-a|).$$
We want to make the left side "small." That will be the case if $|a_n-a|$ is small, unless $|a_n|+|a|$ is big. So we first want to show that after a while, $|a_n|+|a|$ cannot be big. 
There are two cases, $a\ne 0$ and $a=0$.
If $a\ne 0$, there is an $N_1$ such that if $n\gt N_1$ than $|a_n-a|\lt \frac{|a|}{2}$. Thus if $n\gt N_1$, then $|a_n|\lt \frac{3|a|}{2}$, and therefore 
$$|a_n|+|a|\lt \frac{5|a|}{2}.$$
Since the sequence $(a_n)$ has limit $a$, there is an $N_2$ such that if $n\gt N_2$, then 
$$|a_n-a|\lt \frac{2\epsilon}{5|a|}.$$
Let $N=\max(N_1,N_2)$. If $n\gt N$ then $|a_n^2-a^2|\lt \epsilon$.
The argument for the case $a=0$ goes along similar lines, but is technically simpler. We leave it to you. 
Remark: We gave the full "$\epsilon$-$N$" argument. More informally, the sequence $(|a_n|+|a|)$ is bounded, and $|a_n-a|\to 0$, so $|a_n^2-a^2|\to 0$. 
