# Prove that $\lim\limits_{n\to\infty}\left(\sqrt{n^4+n^2+20n+7}\,-\,\sqrt{n^4+n^2+1} \,\right)=0$

Prove that:

$$\lim_{n\to\infty}\left(\sqrt{n^4+n^2+20n+7}\,-\,\sqrt{n^4+n^2+1} \,\right)=0$$

Epsilon>0. According to the Archimedean Property of reals, we have n1 element N with epsilon*n1>11. (Why 11? Seems so random...)

n0:=max{3,n1} (What's the point of that? It doesn't appear anywhere in the proof...). For each n element N with n>=n0 we have:

\begin{align} 0&\lt\sqrt{n^4+n^2+20n+7}-\sqrt{n^4+n^2+1} \\\,\\ &=\dfrac{\left(\sqrt{n^4+n^2+20n+7}\,-\,\sqrt{n^4+n^2+1}\right)\cdot\left(\sqrt{n^4+n^2+20n+7}+\sqrt{n^4+n^2+1}\right)}{\sqrt{n^4+n^2+20n+7}\,+\,\sqrt{n^4+n^2+1}} \\\,\\ &=\dfrac{20n+6}{\sqrt{n^4+n^2+20n+7}\,+\,\sqrt{n^4+n^2+1}}\leqslant\dfrac{20n+2n}{\sqrt{n^4}+\sqrt{n^4}}=\dfrac{11}n\leqslant\dfrac{11}{n_1}\lt\epsilon \end{align}

(I don't get the circled part. Why 20n+2n, why "cut off" the roots? And why should it equal 11/n?)

From this we get

$$\left|\sqrt{n^4+n^2+20n+7}\,-\,\sqrt{n^4+n^2+1} \,\right|\lt\epsilon$$

, thus the proof is complete.

Thanks for the clarifications!

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In addition to Peter's answer, the importance of having chosen the number $11$ is to ensure the final inequality in that block. Whoever did probably did the calculation with no $n_1$, then realized at the end that such a thing had to be added, and to be as mathematically sound as possible, it's typically added at the beginning. This you will see happening in essentially any proof involving $\varepsilon$s. – fuglede Jan 9 '14 at 13:57
A superb observation. Is is often very difficult to predict exactly, what is needed, but in many cases, it can be adjusted, so that everything works well. Perhaps, the proof for the product rule with limits (where a zero-term is added) was found in such a way. – Peter Jan 9 '14 at 14:01
What often helps, is to go backwards : Starting with the desired inequality and manipulating it until it can be easily proved. – Peter Jan 9 '14 at 14:05

$\sqrt{n^4+n^2+20n+7}\geq \sqrt{n^4}\Rightarrow \dfrac{1}{\sqrt{n^4+n^2+20n+7}}\leq \dfrac{1}{\sqrt{n^4}}$

$\sqrt{n^4+n^2+1}\geq \sqrt{n^4}\Rightarrow \dfrac{1}{\sqrt{n^4+n^2+1}}\leq \dfrac{1}{\sqrt{n^4}}$

your $n$ is choosen such that $n\geq 3$ so : $6\leq 2n$

That choosen $n_1>11$ is not random and it is not choosen before solving that out...

After solving everything i got $\dfrac{11}{n}$ but then I want that $\dfrac{11}{n}$ to be so small so if i choose $n>11$ then I am done!

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Extarct n^4 from each radical. So, what is left inside the remaining radicals is (1 + 1/ n^2 + ...) and, since "n" is large, each remaining radical radical is almost 1.

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This is the intuitive way, which gives immediately the correct answer. – Peter Jan 9 '14 at 14:07
@Peter. I should not call that intuitive way. When the OP arrives at the last line, he is ready for the limit. This is the only thing I wanted to clarify for him. Cheers. – Claude Leibovici Jan 9 '14 at 14:11

The numerator was made bigger (or equal, if $n=3$) and the denominator was made smaller, so the fraction becomes bigger. This only works, when $n\geqslant3$ is assumed somewhere.

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You should, through looking at only the leading terms of
$$\dfrac{20 n + 6}{\sqrt{n^4+n^2+20n+7}+\sqrt{n^4+n^2+1}}$$ see $$\dfrac{20 n}{\sqrt{n^4}+\sqrt{n^4}}=\dfrac{20 n}{2 n^2}=\dfrac{10}{n}$$ which is clearly going to go to $0$ as $n\rightarrow \infty$. That's what we want to work towards.

So we try to get rid of the lower order terms but only by increasing the expression to preserve the inequality. So only to increase the denominator and reduce the numerator.

The numerator $\sqrt{n^4+n^2+20n+7}+\sqrt{n^4+n^2+1}$ can be replaced by the lesser $\sqrt{n^4}+\sqrt{n^4}$.

But we can only get rid of the $6$ term by increasing it up to a higher degree term. They could have just used $n$, but instead used $2n$, presumably so they could nicely cancel out the 2 in the denominator, and get the prettier $\dfrac{11}{n}$ rather than $\dfrac{21}{2n}$.

So the $11$ is arbitrary, and so is the $3$. It could have been $10.5$ and $6$ if we went with $6 \leq n$ (for $n\geq6$) rather than $6 \leq2n$ (for $n\geq3$).

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Setting $\frac1n=h$

$$F=\lim_{n\to\infty}\left(\sqrt{n^4+n^2+20n+7}\,-\,\sqrt{n^4+n^2+1} \,\right)$$ $$=\lim_{h\to0}\frac{\sqrt{1+h^2+20h^3+7h^4}-\sqrt{1+h^2+h^4}}{h^2}$$

Rationalizing the numerator,

$$F=\lim_{h\to0}\frac{(1+h^2+20h^3+7h^4)-(1+h^2+h^4)}{h^2(\sqrt{1+h^2+20h^3+7h^4}+\sqrt{1+h^2+h^4})}$$

As $h\to0,h\ne0,$

$$F=\lim_{h\to0}\frac{20h+6h^2}{(\sqrt{1+h^2+20h^3+7h^4}+\sqrt{1+h^2+h^4})}=\cdots$$

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