$\epsilon$-$\delta$ definition of the limit applied to nonlinear functions (teaching) When I teach the $\epsilon$-$\delta$ definition of the limit, I usually start with a linear function and a table of values to intuitive show the idea where the 'guesses' for $\delta$ in terms of $\epsilon$ is taken from. 
For example, $\displaystyle\lim_{x \to 3} (2x-4) = 2 $, I use a table of values for $x = 3, 3.01, 3.1 ... $ with a corresponding $f(x) = 2, 2.02, 2.2, ...$. 
Here we see that if the distance from $x$ is 0.01 (from 3 to 3.01), the corresponding 'distance' from $f(x)$ is 0.02. And so $\delta = \epsilon /2$, which nicely fits in the proof. 
So my question is, when we move to quadratic functions, $\displaystyle\lim_{x \to 2} x^2 = 4.$ How do I use the same illustration to intuitively explain my choice for epsilon? 
 A: You do not choose the $\epsilon>0$. This tolerance level is prescribed by your client, and you have to exhibit a $\delta>0$ that guarantees a value error $<\epsilon$. Such a $\delta$ is not uniquely determined; it just has to work.
For the function $g(x):=x^2$ we have to show that $g(x)$ is near $4$ when $x$ is near $2$. Now $$g(x)-4=x^2-4=(x+2)(x-2)$$
and therefore
$$|g(x)-4|= |x+2|\ |x-2|\leq 5 |x-2|\qquad(1\leq x\leq 3)\ .\tag{1}$$
Note that we have restricted the $x$ we are looking at to a distance of $1$ from the given limiting point $2$. This measure was taken in order to come up with a "worst case" factor $5$. The relation $(1)$ says: If $|x-2|\leq1$ then the output error $|g(x)-4|$ is at most five times the input error $|x-2|$.
Given a tolerance $\epsilon>0$ we are therefore tending to put $\delta:={\epsilon\over 5}$. For an  $\epsilon\ll1$ this would be fine. But if the given $\epsilon$ is $6$ (a highly improbable datum, but anyway) then we are in error. We therefore put 
$$\delta:=\min\left\{{\epsilon\over5},1\right\}>0\ .$$
