Proof verification: $\lim_{n\to\infty}(\sqrt{n^2+1}-n)=0$

$$ n^2 < n^2 + 1 < \left( n + \frac{1}{2n} \right)^2 $$

Here is the template:

Let $\epsilon > 0 $ be given.

Choose $N= .... $ (Here is you choice of $N$ that depends on $\epsilon)$

Now, for $n > N $, show that your $N$ satisfies

$$ |a_n - L | < \epsilon $$

Just show your work backwards.

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 $$ \frac{1}{\sqrt{n^2+1}+n}<\frac{1}{2n}<\frac{1}{2N}<\varepsilon $$