If $\lim_{h \rightarrow 0}$, then where does $|h|$ spring from? To avoid typos, please see my screen captures below, and the red underline. The question says $h \rightarrow 0$, thus why $|h|$ in the solution? Mustn't that $|h|$ be $h$?

Spivak, Calculus 2008 4 edn.  His website's errata lists no errata for these pages.
 A: Writing
$$
0 < |h| < \delta
$$
is easier than writing
$$
-\delta  < h < \delta   \text{ and } h \ne 0 .
$$
A: Note that the definition of a limit involves the inequality $0 < |x - a| < \delta$. Substituting $a = 0$ and $x = h$ yields
$$
0 < |h| < \delta,
$$
which is the inequality that you've underlined.
A: $h$ might tend to zero from either side, so the statement $$-\delta \lt h \lt \delta \iff 0\lt |h| \lt \delta$$ is required.
A: Don't worry. It takes a while and some practice to wrap your head around this stuff. It's easy to get confused by all the different conditions and variables and what depends on what.
Suppose $\lim_{x \to a}f(x) = \ell$
This means that for any $\varepsilon > 0$ there exists some $\delta > 0$ such that for all $x$, if
$$0 < |x - a| < \delta \text{ then } |f(x) - \ell| < \varepsilon$$
Now, for this same $\delta$, suppose you have a number $h$ such that
$$0 < |h| < \delta$$
We can make this look like the previous condition by noting that $h = (a + h) - a$.
Therefore, for all $h$ if
$$0 < |h| < \delta$$
then
$$0 < |(a + h) - a| < \delta$$
Compare this with the initial $\delta$-requirement placed on $x$, namely $0 < |x - a| < \delta$.
The number $a+h$ satisfies the $x$ requirement from the first limit, so we have $|f(a + h) - \ell| < \varepsilon$.
Putting it all together, if $\lim_{x \to a}f(x) = \ell$ then for any $\varepsilon > 0$ there exists some $\delta > 0$ such that for all $h$, if
$$0 < |h| < \delta \text { then } |f(a + h) - \ell| < \varepsilon$$
Another way of writing this is $\lim_{h \to 0} f(a+h) = \ell$. If the first limit exists, then so too does the second, and they are equal.
To complete the proof we need to begin with the second limit $\lim_{h \to 0} f(a+h) = \ell$ and use similar arguments to show how it leads to the first limit.
Suppose $\lim_{h \to 0} f(a+h) = \ell$. This tells us that for any $\varepsilon > 0$ there is a $\delta > 0$ such that for all $h$, if
$$0 < |h| < \delta \text { then } |f(a + h) - \ell| < \varepsilon$$
Suppose for this same $\delta$ we have numbers $x$ such that
$$0 < |x - a| < \delta$$
Compare this with the above $\delta$-requirements for $h$. These numbers $x-a$ fulfill the $\delta$-requirements for $h$, therefore
$$|f(a + (x - a)) - \ell| < \varepsilon$$
but this is just
$$|f(x) - \ell| < \varepsilon$$.
Putting it all together, we know that if $\lim_{h \to 0} f(a+h) = \ell$ then for any $\varepsilon > 0$ there is a $\delta > 0$ such that for all $x$, if
$$0 < |x - a| < \delta \text{ then } |f(x) - \ell| < \varepsilon$$
We can of course write this as $\lim_{x \to a}f(x) = \ell$. Again, existence of one limit guarantees existence of the other, and they are equal.
