Thread: Use definition of limit to prove lim (sqrt(1 + n^(-1))) = 1?

2. Limits

Hello qtpipi
Originally Posted by qtpipi
We need to show that, for any $\epsilon > 0, \exists N \in \mathbb{N}: \, \forall n>N,\, \sqrt{1+n^{-1}}<1+\epsilon$

So, given $\epsilon >0$, let $\sqrt{1+x^{-1}}= 1+\epsilon$

$\Rightarrow 1 + x^{-1} = 1+2\epsilon + \epsilon^2$

$\Rightarrow x = \frac{1}{2\epsilon + \epsilon^2}$

So for $n > \lfloor{x}\rfloor, n > \frac{1}{2\epsilon + \epsilon^2}$

$\Rightarrow n^{-1} < 2\epsilon+\epsilon^2$

$\Rightarrow \sqrt{1+n^{-1}} < 1 + \epsilon$