Find deltas algebraically for given epsilons The exercise below is from the book Calculus by Thomas / Finney (9th edition):

Prove that $\displaystyle \lim_{x\rightarrow 2}f(x) = 4$ if:
$f(x) = x^2$ when $x \neq 2$
$f(x) = 1$ when $x = 2$

The solution says that (I use "e" for "epsilon", "d" for "delta"):

Why was it all right to assume $\varepsilon < 4$? Because, in finding a $\delta$ such that for all $x$, $0 < | x - 2 | < \delta$ implied $| f(x) - 4 | < \varepsilon < 4$, we found a $\delta$ that would work for any larger $\varepsilon$ as well.

I have two questions:
First, it seems like they let $\varepsilon< 4$ because $f(x) > 0$, but why not including $\varepsilon = 4$ ($f(x) = 0$)? Or am I mistaking something?
Second, I don't understand the phrase "we found a $\delta$ that would work for any larger $\varepsilon$ as well"...?
Many thanks!
 A: This is often a cause for head scratching, but it's quite simple to explain.
You have to prove that, for any $\varepsilon>0$, the solution set of the inequality
$$
|f(x)-4|<\varepsilon
$$
contains an open interval around $2$ (except $2$, which is always excluded from considerations). You don't need to find the solution set, but have just to prove the condition above, that's usually expressed by the inequality with $\delta$.
Now, if you find that the solution set of $|f(x)-4|<\varepsilon$ contains an open interval around $2$, then this open interval will be contained in the solution set for any inequality of the form $|f(x)-4|<\varepsilon'$, where $\varepsilon'\ge\varepsilon$.
Thus it's always irrelevant to impose conditions such as $\varepsilon<4$ that simplify the computations without being restrictive.
I like to explain in this way. Suppose you're a producer of cogs for engines. If you have to produce those cogs for a ship engine, for which there is a rather high tolerance on measures, you can as well keep the same setup on your machines as the one used for racing car engine cogs.
The tolerance on measures is the $\varepsilon$, while the setup is the $\delta$: if you fix a setup that ensures fitting the required tolerance for racing car engines, you are of course sure that it will fit when ship engines are involved.
Of course, the cog producer will not use the same setup in those cases, because of higher cost; but numbers are free…
