<s>[INST] How do I prove that $\lim _{n\to \infty} (\sqrt{n^2 + 1} - n) = 0$? As you can see from the title, I need help proving that $\lim _{n\to \infty} (\sqrt{n^2 + 1} - n) = 0$. I first looked for $N$ by using $|\sqrt{n^2 + 1} - n - 0| = \sqrt{n^2 + 1} - n < \varepsilon$ and solving for $n$. I got $n > \frac{1 - \varepsilon ^2}{2\varepsilon}$. I then set $N = \frac{1 - \varepsilon ^2}{2\varepsilon}$. However, I afterwards realized that I had squared an inequality to get this. The following is what my proof looked like before I realized this mistake:
Let $\varepsilon > 0$. Choose $N = \frac{1 - \varepsilon ^2}{2\varepsilon}$ and let $n > N$. Then $n > \frac{1 - \varepsilon ^2}{2\varepsilon}$. Thus $2n\varepsilon > 1 - \varepsilon^2$, $\varepsilon^2 + 2n\varepsilon > 1$, $\varepsilon^2 + 2n\varepsilon + n^2 = (\varepsilon + n)^2 > 1 + n^2$, $\varepsilon + n > \sqrt{1 + n^2}$, and $\varepsilon > \sqrt{n^2 + 1} - n$. Hence $| \sqrt{n^2 + 1} - n - 0 | = \sqrt{n^2 + 1} - n < 
\varepsilon.$ Therefore, $\lim _{n\to \infty} (\sqrt{n^2 + 1} - n) = 0$.
How should I find $N$? Also, are there any other errors in this proof?
 [/INST] $$\lim _{n\to \infty} (\sqrt{n^2 + 1} - n) =\lim _{n\to \infty} \frac{(\sqrt{n^2 + 1} - n)(\sqrt{n^2 + 1} + n)}{\sqrt{n^2 + 1} + n}\\ = \lim _{n\to \infty} \frac{n^2 + 1-n^2}{\sqrt{n^2 + 1} + n}=\lim _{n\to \infty} \frac{1}{\sqrt{n^2 + 1} + n}=0.$$
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