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!
 A: $\sqrt{n^4+n^2+20n+7}\geq \sqrt{n^4}\Rightarrow \dfrac{1}{\sqrt{n^4+n^2+20n+7}}\leq \dfrac{1}{\sqrt{n^4}}$
$\sqrt{n^4+n^2+1}\geq \sqrt{n^4}\Rightarrow \dfrac{1}{\sqrt{n^4+n^2+1}}\leq \dfrac{1}{\sqrt{n^4}}$
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!
A: 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.
A: 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.
A: 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 $n\geq3$).
A: 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 $h\to0,h\ne0,$
$$F=\lim_{h\to0}\frac{20h+6h^2}{(\sqrt{1+h^2+20h^3+7h^4}+\sqrt{1+h^2+h^4})}=\cdots$$
