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!

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.
