# real analysis – Necessity of selecting value for \$delta\$ in \$epsilon\$-\$delta\$ proofs

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|

$$|x+3||x-3|

$$text{Let } delta = 1, text{so } |x-3|<1$$

$$-1

$$5

$$therefore |x-3|

$$delta = min { 1, frac{epsilon}{7} }$$

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 $$left(0,1right)$$, for example? Otherwise, the proof seems incomplete because it doesn’t take into account any $$delta>1$$.