Limit of sequence (by definiton) I have this assignment:
$$\lim_{n\to \infty }\frac{(-1)^n-3}{n^2}=0$$
I have to prove that limit by definition, but I am stuck with 
$$|a_n - 0|< \epsilon$$
Dont know how to deal with that absolute value.
The solution is
$$n_0=\left[\frac{2}{\sqrt\epsilon}\right]+1$$

Given $\epsilon>0 $, we should find $N$ such that if $n\ge N$ then $$|\frac{(-1)^n-3}{n^2}|<\epsilon$$
but
$$|\frac{(-1)^n-3}{n^2}|\le |\frac{(-1)^n}{n^2}|+|\frac{3}{n^2}|\le \frac{4}{n^2}$$
thus it is sufficient to find one $N$ such that,
$$n\ge N \;\; \implies \;\; \frac{4}{n^2}<\epsilon$$
or
$$n\ge N \;\;  \implies \;\; n^2 > \frac{4}{\epsilon}$$
or
$$n\ge N \;\; \implies \;\; n > \frac{2}{\sqrt{\epsilon}}$$
So, each  $N$  satisfying $N>\frac{2}{\sqrt{\epsilon}}$ will work.
the smallest of these $N$ is
$$\lfloor \frac{2}{\sqrt{\epsilon}} \rfloor   +1$$
