question about epsilon delta proof I just had a few questions concerning the epsilon delta proof of limits. To be more precise, I always get lost at the part where we take $\delta$ to be the minimum of two real numbers. Let me provide a concrete example to work with. 
Consider $\displaystyle\lim_{x \to 2}x^2 = 4$
Given $\epsilon > 0$, let $\delta = \min\{1,\displaystyle\frac{\epsilon}{5}\}$, and assume $0 < |x-2|< \delta$
Then, $|x^2 - 4| = |x+2||x-2| < 5|x-2| < 5 \cdot \displaystyle\frac{\epsilon}{5} = \epsilon$
I know we are taking $|x-2| < 1$ AND $|x-2| < \displaystyle\frac{\epsilon}{5}$ however I do not understand the following:


*

*What if it turned out that the minimum is $1$. I do not quite understand why $5|x-2| < 5 \cdot \displaystyle\frac{\epsilon}{5} = \epsilon$ would hold still.

*Why does it matter that we take the minimum? If one is bigger then the other wouldn't it still incorporate both conditions:  $|x-2| < 1$ AND $|x-2| < \displaystyle\frac{\epsilon}{5}$? I am just imagining two open intervals where one interval is inside the other so choosing the bigger one would count towards both of them.
I don't know maybe I am just not seeing something obvious. I apologize if my questions are easy or not the best. It has been bugging me though.
 A: If the minimum turned out to be $1$, then $1 < \dfrac{\epsilon}{5}$, so that $\epsilon > 5$.
Then 
$$5|x-2| < 5(1) = 5 < \epsilon\text{.} $$
See also $\epsilon$-$\delta$ proof that $\lim\limits_{x \to 1} \frac{1}{x} = 1$. for a similar problem, and Why do we need min to choose $\delta$? for an explanation of the minimum. 
A: A typical epsilon-delta proof appeared in textbooks is arranged in an arguably unnatural way; I provide a more natural arrangement for your reference, I believe by which you can figure out why Minimum usually appears in such a proof:
If $x \in \Bbb{R}$, then
$$
|x^{2}-4| = |x-2||x+2|;
$$
if in addition $|x-2| < 1$ (this is to get rid of $|x+2|$, with the bound $1$ chosen for convenience),
then $|x| - 2 \leq |x-2| < 1$, implying $|x+2| \leq |x|+2 < 5$, implying $|x-2||x+2| < 5|x-2|$;
given any $\varepsilon > 0$, we have $5|x-2| < \varepsilon$ if in addition $|x-2| < \varepsilon/5$. Hence, for every $\varepsilon > 0$ it holds that $|x-2| < \min \{ 1, \varepsilon/5 \}$ (which just says that $|x-2| < 1$ and $|x-2| < \varepsilon/5$) implies $|x^{2}-4| < \varepsilon$; we have proved that $\lim_{x \to 2}x^{2}=4$.
A proof like the above shows clearly how a choice of $\delta$ is made; on the contrary, if a choice of $\delta$ is written in the first place then it is not "healthy" for beginners to understand the nature of such a proof.
Note that, because the map $x \mapsto x^{2}$ can be defined at $2$, no need to put the condition $0 < |x-2|$.
