Prove that $a_n \to 1$ implies $a_n^2 \to 1$ using the definition of convergence Suppose that $(a_{n})$$_{n \in \mathbb{N}}$ is convergent, with limit 1. Show, directly from the definition, that $(a^2_{n})$$_{n \in \mathbb{N}}$ is convergent, with limit 1.
My Attempt:
Let $\epsilon > 0$ be given. We want to find $N \in \mathbb{N}$ such that $\forall n>N, |a^2_{n} - 1| < \epsilon$. 
$|a^2_{n} - 1| = |a_{n} - 1||a_{n} + 1|$.
As $(a_{n})$$_{n \in \mathbb{N}}$ is convergent with limit 1, $\exists M \in \mathbb{N}$ such that $\forall m>M, |a_{n} - 1| < \frac{\epsilon}{2}$.
Then for $N>M, |a_{n} - 1||a_{n} + 1| < \frac{\epsilon}{2} (\frac{\epsilon}{2} + 2) = \epsilon + \frac{\epsilon^2}{4} $
A bit stuck from here, would appreciate some help.
 A: Well since $a_n$ is convergent then there exists $N_1$ such that for all $n>N_1$ you have $|a_n - 1| < 1$ so $$|a_n + 1| = |a_n + 1 - 1 + 1|\leq |a_n - 1| + 2 < 1+2 = 3.$$
But also there is an $N_0$ such that for all $n > N_0$ you have $|a_n - 1| < \frac{\epsilon}{3}$.
So if you pick $N = \max(N_0, N_1)$ then for all $n > N$ we have $$|a^2_n - 1| = |a_n-1||a_n + 1| < \frac{\epsilon}{3} \cdot 3 = \epsilon$$

Handwaving Meta: the thing with your approach is that you start by assuming $|a_n - 1| < \epsilon$ and then find an upper bound for $|a_n + 1|$ in terms of $\epsilon$. But the optimal approach is bounding $|a_n + 1|$ and then finding an appropriate $\epsilon'$ satisfying $|a_n - 1| < \epsilon'$ and then $|a_n + 1||a_n - 1| < |a_n + 1|\epsilon' < \epsilon$. 
This allows you to use all the freedom you have, since you can make $|a_n - 1|$ as small as you like it's best to work out how big everything else is and then decide how small $|a_n - 1|$ must be to counteract the size of everything else (in this case $|a_n +1|$). Whilst what you do is pick a size $|a_n - 1| < \epsilon$ and then find out how big everything else is, this throws away your freedom to making $|a_n - 1|$ as small as required. 
A: We know $a_n$ converges to $1$. We know that for each $\delta>0$, there's an $N$ such that for all $n>N$, we must have 
$$|a_n-1|<\delta$$
Let $\epsilon>0$. Take $\delta=-1+\sqrt{1+\epsilon}$ (and see $\delta>0$). 
Now note that 
$$|a_n+1|\leq |a_n|+|1|=|a_n|-|1|+2\leq |a_n-1|+2<2+\delta$$
so that
$$|a_n^2-1|=|a_n-1||a_n+1|< \delta(2+\delta)=\epsilon$$
which proves the required statement.
