In a specific part of the solution to an $\varepsilon - \delta$ limit problem (substituting for some number $k$ one of the factors of the quadratic to find $\delta$ correspondence), the author establishes that
$-3<x-1<-1$ , therefore $|x-1|<3$
He supports it with the following comment:
"The smallest x can be is -1.9999..., and that number minus 1 is -2.9999..., which is less than 3.
I'm trying to wrap my head around how he establishes equivalency between that inequality and that absolute value, and the supporting comment confuses me even more. What I'm trying to understand is what he's doing or how to properly go from a compound inequality into an absolute value, and to do so in a way that allows me to apply whatever technique he used there on other problems that require me to do so.
If it helps in any way, the problem is as follows:
Identify a value of $\delta$ (accurate to two decimal places) that corresponds to $\varepsilon = 0.01$ given $\lim_{x \to -1}(x^2+3)=4$, according to the definition of limits
edit: spelling and readability
Well $-1 < 3$ right? Therefore $-3 < x-1 < -1 < 3$. That is to say $-3 < x-1 <3$ which is equivalent to $|x-1| <3$.
The justification the author is making is confusing and unnecessary, it's much simpler than that.
edit: Its a known property of absolute values that $|x-a|<b \Leftarrow \Rightarrow -b<x-a<+b$