Why do we need min to choose $\delta$? On this thread:
Problem:
The person uses 
$\delta = \text {min} (\frac{\epsilon}{2}, \frac{1}{2})$
why do we need the min function to determine $\delta$?
Thanks!
 A: Note that along the way he assumed $\delta \lt \frac 12$  This allowed the following calculations to go through.  As long as $\epsilon$ is small, $\frac \epsilon 2 \lt \frac 12$, but a nasty opponent (who knew our proof, say)could give us $\epsilon =5$, say.  If we just say $\delta = \frac \epsilon 2$ our opponent could say "Look, it doesn't work at $x=0$ and $|0-1| \lt \frac 52$
A: Is is just for convenience. 
Suppose (for example) that you want $\delta^2 + \delta < \epsilon$. You can factor the left-hand side to write $\delta(\delta + 1) < \epsilon$. Let $m$ be any positive number. As long as $\delta < m$ you have $\delta(\delta + 1) < \delta (m+1)$, and if in addition you have $\delta < \dfrac{\epsilon}{m+1}$ you arrive at $\delta(\delta + 1) < \delta(m+1) < \dfrac{\epsilon}{m+1}(m+1) = \epsilon$. 
What this means is that if both $\delta < m$ and $\delta < \dfrac{\epsilon}{m+1}$, then $\delta^2 + \delta < \epsilon$.  That is,
$$ \delta < \min \left\{ m,\dfrac{\epsilon}{m+1} \right\} \implies \delta^2 + \delta < \epsilon.$$
You don't need the min function to do this, in general. You can solve the inequality by other means. For instance, you can add $\dfrac 14$ to both sides of the example inequality to get
$$ \delta^2 + \delta + \frac 14 < \epsilon + \frac 14$$ so that $$\left( \delta + \frac 12 \right)^2 < \epsilon + \frac 14.$$ The solution to this is an interval:
$$
- \sqrt{ \epsilon + \frac 14} < \delta + \frac 12 < \sqrt{ \epsilon + \frac 14}
$$
so for positive $\delta$, 
$$
\delta <  \sqrt{ \epsilon + \frac 14} - \frac 12 \implies \delta^2 + \delta < \epsilon.
$$
In almost all cases it is simpler to impose conditions on $\delta$ and use a min.
A: Here's a general answer:
The definitions of analysis are formulated in terms of conditions depending on a positive real number $\delta$ that "remain true if $\delta$ is made smaller". For example, the precise definition of the statement $\lim\limits_{x \to a} f(x) = L$ includes the condition
$$
\text{If $|x - a| < \delta$, then $|f(x) - L| < \varepsilon$,}
$$
which we might denote $P(\delta)$, regarding $f$, $a$, $L$, and $\varepsilon$ as given/known.
If the condition $P(\delta)$ is true for some $\delta > 0$, and if $0 < \delta' < \delta$, then $P(\delta')$ is also true, because its hypothesis is logically more strict.
Now suppose you have finitely many such conditions satisfied by positive numbers $\delta_{1}, \dots, \delta_{k}$, and you want a single $\delta > 0$ that satisfies all your conditions. It suffices to take a positive $\delta$ that does not exceed $\delta_{1}, \dots, \delta_{k}$. The standard idiom of analysis is to take
$$
\delta = \min(\delta_{1}, \dots, \delta_{k}).
$$
To be picky, it's not that we need to use the minimum, but it's sufficient or enough to use the minimum.
A: If challenged with any $\varepsilon \gt 0$, by setting
$\tag 1 \Large{\delta =  1-\frac{1}{\varepsilon+1}}$
the statement
$\tag 2 \Large{|x - 1| \lt \delta \text{ implies } |\frac{1}{x} - 1| \lt \varepsilon}$
will always be true (and makes sense since $\delta$ is always less than $1$).
We won't bother proving this, but we can demonstrate it by plugging in some positive values for $\varepsilon$ into this Wolfram Calculation.
A: Because here $\epsilon$ is an arbitrary positive number, it could be 2. Hence we use minimum so that $|x|> \frac{1}{2}$ (bounded away from 0) to control the denominator.
