Hello qtpipiWe need to show that, for any $\displaystyle \epsilon > 0, \exists N \in \mathbb{N}: \, \forall n>N,\, \sqrt{1+n^{-1}}<1+\epsilon$
So, given $\displaystyle \epsilon >0$, let $\displaystyle \sqrt{1+x^{-1}}= 1+\epsilon$
$\displaystyle \Rightarrow 1 + x^{-1} = 1+2\epsilon + \epsilon^2$
$\displaystyle \Rightarrow x = \frac{1}{2\epsilon + \epsilon^2}$
So for $\displaystyle n > \lfloor{x}\rfloor, n > \frac{1}{2\epsilon + \epsilon^2}$
$\displaystyle \Rightarrow n^{-1} < 2\epsilon+\epsilon^2$
$\displaystyle \Rightarrow \sqrt{1+n^{-1}} < 1 + \epsilon$
Grandad