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

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.

• why would someone downvote this legitimate question? SMH – James Sep 30 '18 at 21:32
• You did very well. – egreg Sep 30 '18 at 21:38

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

• I appreciate your clarity. In terms of what I can have as a constant multiple to my estimation, does it matter what value I choose? I'm assuming that I would choose a constant that is like yours one that is in $\mathbb{Q}$ or some value greater than $0$ that belongs to $\mathbb{N}$ right? And one that is fairly small? I'm wondering intuitively what qualities should this constant have. – dls Oct 1 '18 at 2:10
• @dls No, it doesn't matter what upper bound you choose, so long as it leads you to the end. Happily for you, this kind of exercises will stop early. – egreg Oct 1 '18 at 6:47

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

• May I ask for clarification for how you arrived at this inequality or why did you do what you did? I'm am interested in understanding how I can use this idea for other situations. – dls Oct 1 '18 at 2:46

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.