I've been reviewing the epsilon-delta definition of a limit before beginning Analysis 1 and am slightly confused about the problems where $\delta$ is selected from a minimum of two values.
For example, a problem in my textbook asks to prove $$\lim_{x \to 3} x^2 = 9$$
using epsilon-delta. The process is explained as follows:
$$|x^2-9|<\epsilon$$
$$|x+3||x-3|<\epsilon$$
$$\text{Let } \delta = 1, \text{so } |x-3|<1$$
$$-1<x-3<1$$
$$5<x+3<7$$
$$\therefore |x-3|<\frac{\epsilon}{7}$$
$$\delta = \min \{ 1, \frac{\epsilon}{7} \}$$
I'm confused as to why we don't leave delta as delta, and simply solve for delta in terms of epsilon (i.e. $\delta = \sqrt{\epsilon+9}-3$). Are we able to simply pick a bound for $\delta$ because epsilon-delta proofs only require that there exists a positive delta that will work, and that there an infinite number of positive deltas that exist in $\left[0,1\right)$, for example? Otherwise, the proof seems incomplete because it doesn't take into account any $\delta>1$.
Best Answer
Yes, the definition only requires that, for every $\varepsilon>0$, there is some $\delta>0$ such that$$|x-a|<\delta\implies\bigl|f(x)-f(a)\bigr|<\varepsilon.$$So, the fact that, in that example, the $\delta$ is always smaller than or equal to $1$ is irrelevant.
And, yes, in some cases we can get a $\delta$ that works as a function of $\varepsilon$, but in most cases we cannot. Your example ($\delta=\sqrt{\varepsilon+9}-3$) is indeed correct, but try to do the same thing with $\lim_{x\to1}x^3+x^2+x=3$.