As I've said fairly often in the last few days (for some reason), one of my favorite equations is:
$$e^x\ge x+1$$
The reason, partly, is that it uniquely defines $e$ without calculus. Hint for a proof: use this. (By the way, do equations need equals signs? Or is it equalities that need equals signs?)
Now, replacing $x$ by $-x$, we get $e^{-x}\ge1-x$, so:
$$e^x\le\frac1{1-x}$$
(The inequality gets reverse for $x>1$, as the right-hand side is negative there. But we only care about when $x$ is near zero.)
Thus:
\begin{align}
x+1\le{}&e^x\le\frac1{1-x}\\
x\le{}&e^x-1\le\frac x{1-x}\\
1\le^*{}&\frac{e^x-1}x\le^*\frac1{1-x}
\end{align}
*Since we just divided by $x$, the inequalities get reversed if $x$ is negative. It doesn't affect the argument.
Let $x$ tend to zero. By the squeeze theorem:
$$1=\lim_{x\to0}\frac{e^x-1}x$$
Your expansion of $f''(a)$ as a double limit is indeed correct based on the definition of the derivative. However, that definition requires $f'$ to be continuous at $a$. It turns out that the theorem you ask about actually holds under weaker conditions.
$
\def\lfrac#1#2{{\large\frac{#1}{#2}}}
$
Theorem 1. Take any open subset $D$ of $ℝ$ and any function $f : D→ℝ$ and any $x∈D$ such that $f(x+h) ∈ f(x) + c_1·h + c_2·h^2 + o(h^2)$ as $h → 0$. Then $\lfrac{f(x+h)-2·f(x)+f(x-h)}{h^2} → c_2$ as $h → 0$.
Proof. As given in this post, which is the second answer to one of the posts you cited.
Remark. This theorem is more general because it does not require $f$ to be differentiable at $x$, nor does it require $f$ to be continuous in some open interval around $x$. For example, let $f : ℝ→ℝ$ such that $f(0) = 0$ and $f(x) = \lfrac{x}{\lfloor 1/x \rfloor}$ for every $x∈ℝ_{≠0}$. Then $f(h) = \lfrac{h}{1/h+O(1)}$ $∈ h^2·(1+O(h))$ $⊆ h^2+o(h^2)$ as $h → 0$, so the theorem applies to $f$.
Theorem 2. Take any open subset $D$ of $ℝ$ and any function $f : D→ℝ$ and any $x∈D$ such that $f$ is differentiable on $D$ and $f'$ is differentiable at $x$. Then there are constants $c_1,c_2$ such that $f(x+h) ∈ f(x) + c_1·h + c_2·h^2 + o(h^2)$ as $h → 0$.
Proof. Essentially the same as in that same linked post.
Remark. Together with Theorem 1, this gives a much stronger theorem than the one you cited from Abbott, since we do not need $f$ to be twice differentiable on an open interval around $x$, much less that its second derivative is continuous at $x$!
Best Answer
What happens when you multiply the numerator and denominator by the (clearly non-zero) conjugate $$\sqrt{(1-h)^2+(2+h)^2} + \sqrt{5}?$$