Limit of sequence (by definiton)

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

$$\lfloor \frac{2}{\sqrt{\epsilon}} \rfloor +1$$

We have that

$$\frac{-2}{n^2}\le\frac{(-1)^n-3}{n^2}\le \frac{-4}{n^2} $$

and then

$$\frac{2}{n^2}\le\left|\frac{(-1)^n-3}{n^2}\right|\le \frac{4}{n^2} $$

therefore

$$\left|\frac{(-1)^n-3}{n^2}\right|\le \frac{4}{n^2} <\epsilon \implies n^2 > \frac{4}{\epsilon} \implies n>\frac{2}{\sqrt\epsilon}$$

therefore it suffices to take

$$n_0=\overbrace{\color{blue}{\left[\frac{2}{\sqrt\epsilon}\right]}}^{\color{red}{\text{integer part}}}+1$$