TITLE: Proof verification: $\lim_{n\to\infty}(\sqrt{n^2+1}-n)=0$
QUESTION [7 upvotes]: I'm having issues forming the discussion part of the proof because I am not sure if I am coming up with the right estimation. Is this an appropriate way of coming up with an estimation?
I wrote:
We want to show that $\forall \epsilon >0$, $\exists N>0$, $N\in \mathbb{N}$ s.t. $n>N \Longrightarrow |(\sqrt{n^2+1}-n)-0|<\epsilon$. Then we will proceed by simplifying 
$$
\begin{split}
\sqrt{n^2+1}-n
 &= \left(\sqrt{n^2+1}-n\right) \times
    \frac{\sqrt{n^2+1}+n}{\sqrt{n^2+1}+n}\\
 &=\frac{n^2+1-n^2}{\sqrt{n^2+1}+n}\\
 &=\frac{1}{\sqrt{n^2+1}+n}
\end{split}
$$
by using the conjugate. 
Now we will proceed by making an estimation, we see that $$
\frac{1}{\sqrt{n^2+1}+n} \leq \frac{1}{n+1},
\quad \text{where } n > 1.
$$
So let $\frac{1}{n+1} < \epsilon$ Then by multiplying both sides by $(n+1)$ and dividing both sides by $\epsilon$ we have $\frac{1}{\epsilon}< n+1$. Now we want to subtract 1 from both sides and we arrive at $\frac{1}{\epsilon}-1 < n$. We will choose $N=\frac{1}{\epsilon}-1$ for when $n>1$
I'm new to formulating proofs with rigor. Thanks for your help.

REPLY [3 votes]: You did well.
A perhaps simpler way is to observe that $\sqrt{n^2+1}+n>2n$, so
$$
\frac{1}{\sqrt{n^2+1}+n}<\frac{1}{2n}
$$
and we just need to take as $N$ any integer such that
$$
N>\frac{1}{2\varepsilon}
$$
(which exists by the Archimedean property). As soon as $n>N>1/(2\varepsilon)$ we have
$$
\frac{1}{\sqrt{n^2+1}+n}<\frac{1}{2n}<\frac{1}{2N}<\varepsilon
$$