Show that $\sqrt{n^2+1}-n$ converges to 0 I want to use the definition of the limit to show that $\sqrt{n^2+1}-n$ converges to 0.
The definition is as follows: if $\sqrt{n^2+1}-n$ converges to 0, then $\forall \epsilon>0$, there exists an $N>0$ such that $n\ge N \implies \mid\sqrt{n^2+1}-n\mid<\epsilon$.
Now I want to start backwords in order to figure out how to pick N. I know: 
$\mid\sqrt{n^2+1}-n\mid=\sqrt{n^2+1}-n$ 
since $n$ is a natural number and $\sqrt{n^2+1}>n$. So I need to pick an N such that $\sqrt{n^2+1}-n<\epsilon$. I tried multiplying $\sqrt{n^2+1}-n$ by $\frac{\sqrt{n^2+1}+n}{\sqrt{n^2+1}+n}$ but that didn't really seem to help. 
Do you have any ideas on how to find this N?
Thanks
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$$
\verts{\root{n^{2} +1} - n - 0} = {1 \over \root{n^{2} +1} + n} < {1 \over 2n}
$$
Given $\epsilon > 0$
$$
n > N \equiv \floor{1 \over 2\epsilon} + 1
\quad\imp\quad
\verts{\root{n^{2} +1} - n} < \epsilon
\quad\imp\quad
\lim_{n \to \infty}\pars{\root{n^{2} +1} - n} = 0
$$
A: If we do not wish to rationalize the numerator, note that for positive $n$ we have $\sqrt{n^2+1}\lt n+\frac{1}{2n}$, since $\left(n+\frac{1}{2n}\right)^2=n^2+1+\frac{1}{4n^2}\gt n^2+1$. 
Thus $\left|\sqrt{n^2+1}-n\right|\lt \frac{1}{2n}$, and now $\epsilon$-$N$ works nicely. 
A: Note $$ - \frac{1}{n} \leq\frac{1}{\sqrt{n^2 + 1 } + n} \leq \frac{1}{n} $$
we know $\frac{1}{n} \to 0 $. Apply squeeze rule now.
A: Hint:
$$\sqrt{n^2+1}-n=\left(\sqrt{n^2+1}-n\right)\frac{\sqrt{n^2+1}+n}{\sqrt{n^2+1}+n}=\frac1{\sqrt{n^2+1}+n}\xrightarrow[n\to\infty]{}\ldots$$
