<s>[INST] Proving limit using the definition of a limit Heres the question (provided for context) about proving a limit by using the definition of a limit:
$\mathop {\lim }\limits_{x \to 4} \left({x^2} + x - 11\right) = 9$
So, let’s get started. Let $\varepsilon  > 0$ be any number then we need to find a number  $\delta  > 0$ so that the following will be true.
$\left| {\left( {{x^2} + x - 11} \right) - 9} \right| < \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - 4} \right| < \delta$
simplifying a bit
$\left| {\left( {{x^2} + x - 11} \right) - 9} \right| = \left| {{x^2}
+ x - 20} \right| = \left| {\left( {x + 5} \right)\left( {x - 4} \right)} \right| = \left| {x + 5} \right|\left| {x - 4} \right| < \varepsilon$
if, by some chance, we can show that $\left| {x + 5} \right| < K$ for some number $K$ then, we’ll have the following
$\left| {x + 5} \right|\left| {x - 4} \right| < K\left| {x - 4} \right|$
If we now assume that what we really want to show is $K\left| {x - 4} \right| < \varepsilon$ instead of $\left| {x + 5} \right|\left| {x - 4} \right| < \varepsilon$ we get the following,
$\left| {x - 4} \right| < \frac{\varepsilon }{K}$
All this is based on the assumption that we can show $\left| {x + 5} \right| < K$ for some $K$. To do this we assume that whatever $x$ is it must be close to $x=4$ since we are working with a limit. So lets assume that $x$ is within a distance of one of $x=4$. In terms of inequality we can assume
$\left| {x - 4} \right| < 1$
Starting with removing the absolute value bars we have
$- 1 < x - 4 < 1\hspace{0.5in} \Rightarrow \hspace{0.5in}3 < x < 5$
If we now add 5 to all parts of this inequality we get,
$8 < x + 5 < 10$
Now, since $x + 5 > 8 > 0$ (the positive part is important here) we can say that, provided $\left| {x - 4} \right| < 1$ we know that $x + 5 = \left| {x + 5} \right|$. Or, if take the double inequality above we have,
$8 < x + 5 = \left| {x + 5} \right| < 10$ $\hspace{0.5in} \Rightarrow \hspace{0.25in}\,\,\,\,\left| {x + 5} \right| < 10\hspace{0.5in} \Rightarrow \hspace{0.5in}K = 10$
So, provided $\left| {x - 4} \right| < 1$ we can see that $\left| {x + 5} \right| < 10$ which in turn gives us,
$\left| {x - 4} \right| < \frac{\varepsilon }{K} = \frac{\varepsilon }{{10}}$
source
How did we go from this double inequality $8 < x + 5 < 10$ to this $8 < x + 5 = \left| {x + 5} \right| < 10$. From what I understand $|{x + 5}| < 10 $ can also be written as $-10<x + 5<10$, then the authors statement:
$8 < x + 5 = \left| {x + 5} \right| < 10$
should not be true since it is excluding a part of the interval that $|{x + 5}| < 10 $ includes (the interval which can be seen when the absolute value inequality is expanded into a double inequality i.e. $-10<x + 5<10$)
 [/INST] You are working with the assumption $|x-4|<1$, that means $3<x<5$.
So $x+5=|x+5|$ because $x+5$ is always positive when $x \in (3,5)$.
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