Epsilon-delta proof - minimum function

Prove that $\displaystyle\lim_{x \to 2} \dfrac{x+2}{(x-2)^4} = \infty$ using epsilon-delta.

The following proof is provided:

Note that if $x>0$, then $x+2>1$, in which case $\dfrac{x+2}{(x-2)^4} > \dfrac{1}{(x-2)^4}$. If furthermore, $|x-2| < \delta$, then $\dfrac{x+2}{(x-2)^4} > \dfrac{1}{(x-2)^4} > \dfrac{1}{\delta ^4}$. This is larger than a given $N>0$, if $\delta \leq \dfrac{1}{\sqrt[4]{N}}$.

Then it goes on to define delta as $\delta= \min(2,\dfrac{1}{\sqrt[4]{N}}).$ Now I don't understand why the $2$ is there in particular instead of some other number. Usually I understand the necessity of the minimum function, but in this case I cannot grok it for the life of me. Can anyone enlighten me, perhaps by using an example where it goes wrong if there is no $\min(2,..)$?

It is there to guarantee that $x>0$ which is used on the first line.
• I still don't quite get why. How does your choice of $\delta$ have bearing on $x$? – YakSal Tafri Feb 21 '17 at 22:50
• It determines the maximum distance $x$ can be away from 2. Since $\delta\leq 2$, $0\leq x\leq 4$. – user416426 Feb 21 '17 at 22:54