Multiple possible delta in epsilon-delta In most textbooks and websites there is often this picture:

where delta is often presented as $\min\{f^{-1}(L+\epsilon) -f^{-1}(L),f^{-1}(L) - f^{-1}(L-\epsilon)\}$ but for most limits, it's not that and there are also multiple $\delta$, for example:
$\lim\limits_{x \to 4} \sqrt{x} = 2$
\begin{align}
\delta &= \min\{f^{-1}(L+\epsilon) -f^{-1}(L),f^{-1}(L) - f^{-1}(L-\epsilon)\} \\ &= \min\{(L+\epsilon)^2 - L^2,L^2 - (L^2 -2L\epsilon + \epsilon^2)\}\\ 
&=  \min\{2L\epsilon + \epsilon^2,2L\epsilon + \epsilon^2\} \\
&= 4\epsilon - \epsilon^2\\
&\text{edit:}\\
& \epsilon \leq 2 \text{ since } f^{-1}(a) \text{ is only defined for } a \geq 0  \\
&\text{end of edit}
\end{align}
Another way:
$|x-4| = |\sqrt{x}-2|\cdot|\sqrt{x}+2| < \delta $
$|\sqrt{x}-2| < \frac{\delta}{|\sqrt{x}+2|}$
let $\delta = 1$:
$3 < x < 5$
$\sqrt{3} < \sqrt{x} < \sqrt{5}$
$\sqrt{3} + 2 < \sqrt{x} + 2 < \sqrt{5} + 2$
$\frac{1}{|\sqrt{5}+2|} < \frac{1}{|\sqrt{x}+2|} < \frac{1}{|\sqrt{3}+2|}$
so:
$|\sqrt{x}-2| <\frac{\delta}{|\sqrt{x}+2|} < \frac{\delta}{(\sqrt{3}+2)}$
$\min\{1,\epsilon(\sqrt{3}+2)\} = \delta$
Another question I have is:
How do you find the smallest delta for any given epsilon and how do you find the max delta for any epsilon.
 A: The definition of $\lim_{x \to x_0} f(x) = L$ is that for every $\epsilon > 0$ there exists $\delta > 0$ such that $|f(x) - L| < \epsilon$ whenever $0 < |x - x_0| < \delta$.
The thing is, once you know such a $\delta$ exists you don't really care  about what it is. There's no need to find the largest possible $\delta$ (there is never a "smallest possible $\delta$", because if $\delta_1$ works you could take any $\delta_2$ with $0 < \delta_2 < \delta_1$).
Moreover, an "$\epsilon$-$\delta$ proof"
is generally only used when there is no alternative available: more often you use various results that say how limits behave when you combine functions in various ways, e.g. if $\lim_{x \to x_0} f(x) = A$ and $\lim_{x \to x_0} g(x) = B$ then $\lim_{x \to x_0} (f(x) + g(x)) = A + B$.
If your function $f$ is increasing on an interval $[a,b]$ around $x_0$ such that $f(a) < f(x_0)-\epsilon$ and $f(b) > f(x_0) + \epsilon$, then you can the largest possible $\delta$ is $$\min(f^{-1}(L+\epsilon)-x_0, x_0 - f^{-1}(L-\epsilon))$$
So for $f(x) = \sqrt{x}$ and $x_0 = 4$, since $f$ is increasing on $[0,\infty)$ and $(2+\epsilon)^2 - 4 = 4 \epsilon + \epsilon^2 > 4 \epsilon - \epsilon^2 = 4 - (2-\epsilon)^2$, the largest possible $\delta$ is $4 \epsilon - \epsilon^2$ for $\epsilon < 4$.
A: You should not worry too much about maximum and even less about minimum (there is no minimum !). What really matters is to find a $\delta$ that fits ! In practice, you often use gross lower bounds.
E.g., we want to establish
$$|x-4|<\delta\implies|\sqrt x-2|<\epsilon.$$
We can use
$$x-4=(\sqrt x-2)(\sqrt x+2)$$ and multiplying by $|\sqrt x+2|$ write
$$|x-4|<\delta\implies|x-4|<\epsilon|\sqrt x+2|$$
Then, whatever $x$, we have
$$2\epsilon\le \epsilon|\sqrt x+2|$$
and if we set $$\delta=\dfrac\epsilon2$$ we do have
$$|x-4|<\delta=\frac\epsilon2\implies|x-4|<\frac\epsilon2|\sqrt x+2|\le\epsilon.$$

The trick is to find a lower bound of
$$\frac{|f(x)-L|}{|x-x_0|}.$$
