Show that $\sqrt{n^2+1}-n$ converges to 0

Hint:

$$\sqrt{n^2+1}-n=\left(\sqrt{n^2+1}-n\right)\frac{\sqrt{n^2+1}+n}{\sqrt{n^2+1}+n}=\frac1{\sqrt{n^2+1}+n}\xrightarrow[n\to\infty]{}\ldots$$

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.

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ $$ \verts{\root{n^{2} +1} - n - 0} = {1 \over \root{n^{2} +1} + n} < {1 \over 2n} $$ Given $\epsilon > 0$ $$ n > N \equiv \floor{1 \over 2\epsilon} + 1 \quad\imp\quad \verts{\root{n^{2} +1} - n} < \epsilon \quad\imp\quad \lim_{n \to \infty}\pars{\root{n^{2} +1} - n} = 0 $$