Answer
By limit definition: $\mathop {\lim }\limits_{n \to \infty } {n^{ - 2}} = 0$
Work Step by Step
The limit definition requires us to find, for every $\varepsilon > 0$, a number $M$ such that
(1) ${\ \ \ \ \ }$ $\left| {{n^{ - 2}} - 0} \right| < \varepsilon $ for all $n>M$.
We have
$\frac{1}{{{n^2}}} < \varepsilon $,
$n > \frac{1}{{\sqrt \varepsilon }}$.
We can choose $\varepsilon > 0$ and $M = \frac{1}{{\sqrt \varepsilon }}$ so that equation (1) is valid. By limit definition, this proves that $\mathop {\lim }\limits_{n \to \infty } {n^{ - 2}} = 0$.
