# Limit exercise from Rudin

This is Chapter 3, Exercise 2 of Rudin's Principles.

Calculate $\lim\limits_{n \to \infty} \sqrt{n^2+n} -n$.

Hints will be appreciated.

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Hint: $$\frac{\sqrt{n^2+n}-n}{1} = \frac{\sqrt{n^2+n}-\sqrt{n^2}}{1}\times \frac{\sqrt{n^2+n}+\sqrt{n^2}}{\sqrt{n^2+n}+\sqrt{n^2}} = \cdots$$ I will expand more if needed.

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So $1/2$ is the final answer, huh? Thanks. – BSNL-IN Jun 11 '12 at 13:21
Yes, 1/2 is correct. – Old John Jun 11 '12 at 13:54

$\sqrt{n^2+n} - n= \frac{n}{\sqrt{n^2+n} + n} = \frac{1}{\sqrt{1+\frac 1n} + 1}$

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Or $\sqrt{n^2+n}=n \left(1+\frac 1n \right)^{\frac 12}=n \left(1 + \frac{1}{2n} -\frac{1}{8n^2} + o \left(\frac{1}{n^2} \right)\right)$ – Mohamed Jun 11 '12 at 13:05

When you have a sequence that involves difference between two roots, it's often a good idea to try using the identity $(a-b)(a+b) = a^2-b^2$ to git rid of the root difference: $$\sqrt{a} - \sqrt{b} = \left(\sqrt{a} - \sqrt{b}\right)\frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} + \sqrt{b}} = \frac{\left(\sqrt{a} - \sqrt{b}\right)\left(\sqrt{a} + \sqrt{b}\right)}{\sqrt{a} + \sqrt{b}} = \frac{a - b}{\sqrt{a} + \sqrt{b}}$$

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@ Ayman Hourieh : Thanks. In general the identity $$(a-b) \: \sum_{k=0}^{p-1} a^{p-1-k}b^k =a^p-b^p$$ where $p$ is a positive integer is used when we have limits with $\sqrt[p]{\text{expression 1}} - \sqrt[p]{\text{expression 2}}$ . – Mohamed Jun 11 '12 at 13:18

To get an idea of what the answer might be, I tried inputting values: n=100, n=1000, n=1000000. e.g. for n = 1000, sqrt(1000000 + 1000) ~ 1000 + 1/2 because (1000 + 1/2)^2 = 1000000 + 2*1000*(1/2) + 1/4 ~ 1000000 + 1000.

If you can spot this, then the rest is easy.

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You can also do this by inductively showing it is increasing and bounded, but their ideas are probably a better route.

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This will show that the limit exists; OP asked to compute it. – sdcvvc Jun 11 '12 at 19:52
You can also show the least upper bound is 1/2 however... – toypajme Jun 21 '12 at 16:28