How to use the $\varepsilon$-$\delta$ definition.

Question

I was trying to understand the demonstration of a limit existing or not using the $\varepsilon$-$\delta$ definition. I've came across this derivation:

I understand it correctly until this phrase:

Finding such $x$ isn't that difficult, just take $0<r<\min(\delta,1/2)$ and consider $x=4+r$.

Why finding and $x$ that obeys both inequallities has to satisfy that $x = 4 + r$ and $0 < r < \min(\delta,1/2)$?

Answer 1

The logic is that each of $\delta$ and $1/2$ are strictly positive, and thus their minimum is strictly positive. Then, strictly between any two distinct reals (here zero and the minimum) we can find a third real. We label this $r$.

We then define $x$ to be $4+r$. Note $x$ depends on the arbitrary $\delta$, i.e. $x=x(\delta)$.

Then Julio Cáceres directly shows that both inequalities are satisfied:

$$|x-4|<\delta,\;\;\;\;\;|9-x-4|\geq 1/2.\quad\quad\quad\quad\quad\quad\quad\quad(*)$$

It is not that finding an $x$ that obeys both inequalities has to be this particular $x$; we just need one $x$ that obeys $(*)$. I think $x=4-r$ would work as well.

Answer 2

If $0<r<\min\left\{\delta,\frac12\right\}$, then $0<r<\delta$, and therefore$$|x-4|=|4+r-4|=|r|=r<\delta.$$And$$9-x-4=5-(4+r)=1-r>\frac12,$$since $r<\frac12$.