Answer
See proof below.
Work Step by Step
If we are given an $\epsilon\gt 0$,
we want to find an N such that, for $n\gt N,$
$|(1-\displaystyle \frac{1}{n})-1|\lt \epsilon.$
\begin{align*}
|(1-\displaystyle \frac{1}{n^{2}})-1|&=|-\displaystyle \frac{1}{n^{2}}| \\
& =\displaystyle \frac{1}{n^{2}}\lt \epsilon\\
& \text{... so, we take}\\
N^{2}&\displaystyle \gt\frac{1}{\epsilon} \\
N&\displaystyle \gt\frac{1}{\sqrt{\epsilon}} \end{align*}
Since for any $\epsilon\gt 0$, we can find N= any integer greater than $\displaystyle \frac{1}{\sqrt{\epsilon}}$,
by the definition, $\displaystyle \lim_{n\rightarrow\infty}(1-\frac{1}{n})=1$
