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

Question

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.

Answer 1

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

Answer 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.

Answer 3

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