Let \(\displaystyle\epsilon\) be any positive number

We need to show that \(\bigg|\frac{1}{n^2}-0\bigg|<\epsilon\)

Now take \(\displaystyle{N}\epsilon{N},\ {\frac{{{1}}}{{\epsilon}}}{<}{N}.\ \because{\frac{{{1}}}{{{N}}}}{<}\epsilon\)

\(\displaystyle\forall{n}{>}{N}{\left(\text{i.e.}{\frac{{{1}}}{{{n}}}}{<}{\frac{{{1}}}{{{N}}}}\right)}.\ {\frac{{{1}}}{{\epsilon}}}{<}{N}\)

and \(\bigg|\frac{1}{n^2}-0\bigg|=\bigg|\frac{1}{n^2}\bigg|=\frac{1}{n^2}<\frac{1}{N}<\epsilon; \ \bigg(\because\frac{1}{n^2} \ \text{is always positive}\bigg)\)

Hence \(\bigg|\frac{1}{n^2}-0\bigg|<\epsilon\)

\(\displaystyle\lim_{{{n}\to\infty}}{\frac{{{1}}}{{{n}^{{2}}}}}={0}\)