Calculate $\lim_{n\to\infty}(\sqrt{n^2+n}-n)$.

Question

Introduction:

An exercise from "Principles of mathematical Analysis, third edition" by Rudin, page 78.

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Exercise:

Calculate $\lim_{n\to\infty}(\sqrt{n^2+n}-n)$.

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Explanation:

I have a hard time to grasp how to handle limits like this. I don't know how to start and what to look for. I've checked with mathematica and the answer should be $\frac{1}{2}$, and of course i've got the wrong answer. I find limits unintuitive. In the book, they proved "the limits of some sequences which occur frequently". The limits they proved were:

(a) If $p>0$ then $\lim_{n\to\infty}\frac{1}{n^p}=0$

(b) If $p>0$ then $\lim_{n\to\infty}\sqrt[n]{p}=1$

(c) $\lim_{n\to\infty}\sqrt[n]{n}=1$

(d) If $p>0$ and $\alpha$ is real, then $\lim_{n\to\infty}\frac{n^\alpha}{(1+p)^n}=0$

(e) If $|x|<1$, then $\lim_{n\to\infty}x^n=0$.

When they proved all of the above theorems it felt like they used the fact that they knew the limits. For example:

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Proof of (b): If $p>1$, put $x_n=\sqrt[n]{p}-1$. Then, $x_n>0$ and by the binomial theorem,

$$1+nx_n\leq(1+x_n)^n=p$$

so that $$0<x_n\leq\frac{p-1}{n}.$$

Hence $x_n\to 0$. And so on...

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That is, I think they used the fact that the limit were 1 when they did put $x_n=\sqrt[n]{p}-1$. Before I compute a limit do I have to guess one? How can I do that when I don't think this is intuitive? Have you any tips how you do when you shall tackle a problem like this? How do you start when you want to compute a limit?

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Solution:

This is how I did it:

$\sqrt{n^2+n}-n=\sqrt{n}\sqrt{n-1}-n=\sqrt{n}(\sqrt{n-1}-\sqrt{n})$.

Since $$(\sqrt{n-1}-\sqrt{n})\to 0\text{ when }n\to\infty.$$

The product approaches $0$. Which is obviously not true. I did realise this after a while. Since one of the factor grows really big while the other gets really small and I guess they tend to take each other out, so it's pretty clear it shouldn't approach 0, but I don't think it's clear that it should approach $\frac{1}{2}$ either. Thanks for your help.

Answer 1

Hint: multiply and divide for $(\sqrt{n^2+n}+n)$.

Answer 2

Use $(a+b)(a-b) = a^2-b^2$ to modify the expression:

$$\lim_{n \to \infty} \sqrt{n^2+n} - n = \lim_{n \to \infty} {(\sqrt{n^2+n} - n) (\sqrt{n^2+n} + n )\over \sqrt{n^2+n} + n } = \lim_{n \to \infty} {n^2+n - n^2 \over \sqrt{n^2+n} + n } = \lim_{n \to \infty} {n \over \sqrt{n^2+n} + n } = \lim_{n \to \infty} {1 \over{ \sqrt{n^2+n} \over n} + 1 } = \lim_{n \to \infty} {1 \over{ \sqrt{n^2\left(1+ {1\over n}\right)} \over n} + 1 } = \lim_{n \to \infty} {1 \over{ \sqrt{1+ {1\over n}} } + 1 } = {1\over 2}$$

Answer 3

Set $1/n=h$

$$\sqrt{n^2+n}=\sqrt{\dfrac{1+h}{h^2}}=\dfrac{\sqrt{1+h}}{|h|}=\dfrac{\sqrt{1+h}}h$$ as $h>0$

$$\implies\lim_{n\to\infty}(\sqrt{n^2+n}-n)=\lim_{h\to0^+}\dfrac{\sqrt{1+h}-1}h$$

$$=\lim_{h\to0^+}\dfrac{{1+h}-1}h\cdot\dfrac1{\lim_{h\to0^+}(\sqrt{1+h}+1)}=?$$

Answer 4

Use $\sqrt{1+x}=1+{x\over 2}+o(x)$

$\sqrt{n^2+n}-n=n\sqrt{1+{1 \over n}}-n=n\left(1+{1 \over 2n} + o\left(1 \over n\right) - 1\right)={1\over 2}+o(1)$

So, $\lim\limits_{n\rightarrow\infty}(\sqrt{n^2+n}-n) = {1 \over 2}$

Answer 5

$$\begin{align*} \lim_{n\to\infty}(\sqrt{n^2+n}-n)&=\lim_{n\to\infty}\left(\frac{(\sqrt{n^2+n}-n)}{1}\times\frac{\sqrt{n^2+n}+n}{\sqrt{n^2+n}+n}\right)\\ &=\lim_{n\to\infty}\frac{n}{\sqrt{n^2+n}+n}=\lim_{n\to\infty}\frac{1}{\sqrt{1+0}+1}=\frac{1}{2} \end{align*}$$