Explanation Regarding Limits from Spivak's Calculus

Question

I'm studying Spivak's Calculus. In the limits chapter, there is a particular expression I struggle to understand. I shall first quote the book then explain my thought process and where I struggle so hopefully someone can point my mistakes and even give me some insight. Quoting from the book:

For the function: $$ f(x) ={\begin{cases} -1, & x<0 \\ 1, & x>0 \end{cases}} $$

If $a>0$ then $f$ approaches $1$ near $a$: indeed, to ensure that $|(f) - 1| < \epsilon$, it certainly suffices to require that $|x-a| <a$ , since this implies $-a < x-a$ which further implies that $0<x$ so that $f(x) = 1$.

If $a>0$ then $f$ approaches $1$ near $a$:

I assume here that based on the function, it should be intuitively apparent that the value of the limit is 1.

indeed, to ensure that $|(f) - 1| < \epsilon$,

Here begins the proof:

it certainly suffices to require that $|x-a| <a$ , since this implies $-a < x-a$ which further implies that $0<x$

Now I know $|x-a|$ is the distance between $x$ and $a$ and if $|x-a|<a$ then that implies x is positive. My trouble begins here. If his goal was to show that $x>0$, wouldn't requiring that $(x+a) > a$ be simpler? Using $|x-a|<a$ creates another condition $x<2a$ which can make the expression false for some $x$ even though it is positive.

I then realized that $|x-a|<a$ may not be to show $x>0$ but to show that as $x$ gets closer to $a$ it must be positive, similar to $|f(x) - l| < \epsilon$. If this is the case, why did he choose $a$ as the $\epsilon$. Which leads to another question:

When choosing a value of $\epsilon$ for $|f(x) - l| < \epsilon$, what should we pay attention to?

Answer 1

When choosing a value of $\varepsilon$ for $\bigl|f(x)-l\bigr|<\varepsilon$, we should pay attention to nothing, since $\varepsilon$ can be any number greater than $0$. There nothing to choose, other than the fact that $\varepsilon>0$.

We choose $|x-a|<a$ because that's part of the definition of limite: we must find a $\delta>0$ such that $|x-a|<\delta\implies\bigl|f(x)-l\bigr|<\varepsilon$. That is, we have to prove that, if $x$ is close enough to $a$, then $\bigl|f(x)-l\bigr|<\varepsilon$.

Answer 2

You have a couple of questions here

If his goal was to show that $x>0$, wouldn't requiring that $(x+a)>a$ be simpler?

But remember that for a limit we want to show that $|x-a| < \delta$ implies $|f(x) - a| < \epsilon$? So he starts with

$$ \begin{align} |x-a| &< \delta = a\\ \implies -a < x-a &< a \end{align} $$

From which, he only uses the inequality to the left $(-a < x-a)$.

For your more general question

When choosing a value of $\epsilon$ for $|f(x)−l|<\epsilon$, what should we pay attention to?

I assume you mean choosing values of $\delta$ that will lead to the inequality $|f(x)−l|<\epsilon$. Usually you play around with the inequalities to figure out what would work. (If you're using this book as part of a class, your teacher should be demonstrating the process.) Often setting $\delta$ to something "like" $f^{-1}(\epsilon)$ will get you going in the right direction. (I realize this is vague. You will get more comfortable with the process the more problems you solve.)