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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?

Martin Sleziak
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Connor
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3 Answers3

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$$\lim _{n\to \infty} (\sqrt{n^2 + 1} - n) =\lim _{n\to \infty} \frac{(\sqrt{n^2 + 1} - n)(\sqrt{n^2 + 1} + n)}{\sqrt{n^2 + 1} + n}\\ = \lim _{n\to \infty} \frac{n^2 + 1-n^2}{\sqrt{n^2 + 1} + n}=\lim _{n\to \infty} \frac{1}{\sqrt{n^2 + 1} + n}=0.$$

mfl
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Another way using Taylor series $$A=\sqrt{n^2 + 1} - n=n\Big(\sqrt{1+\frac{1}{n^2}}-1\Big)$$ Remembert that, for samm values of $x$, $$\sqrt{1+x} =1+\frac{x}{2}-\frac{x^2}{8}+O\left(x^3\right)$$ Replace $x$ by $\frac{1}{n^2}$ so $$\sqrt{1+\frac{1}{n^2}}-1=1+\frac{1}{2 n^2}-\frac{1}{8} \left(\frac{1}{n^2}\right)^2+\cdots-1=\frac{1}{2 n^2}-\frac{1}{8} \left(\frac{1}{n^2}\right)^2+\cdots$$ Finally, $$A=\frac{1}{2 n}-\frac{1}{8n^3}+\cdots$$ which shows the limit and how it is approched.

Claude Leibovici
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For any positive $c$, $(n+\frac{c}{2n})^2 =n^2+c+\frac{c^2}{4n^2} > n^2+c $, so $\sqrt{n^2+c} <n+\frac{c}{2n} $.

Therefore, if $c > 0$, $n <\sqrt{n^2+c} <n+\frac{c}{2n} $ so $\lim_{n \to \infty} \sqrt{n^2+c}-n = 0 $.

You can prove this when $c < 0$ similarly.

marty cohen
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