= Solution
Set $F(h)=f(a+h)+f(a-h)-2f(a)$ and $G(h)=h^2$. Both vanish at zero. For $h\ne0$, their derivative ratio is
$$
\frac{F'(h)}{G'(h)}=\frac{f'(a+h)-f'(a-h)}{2h}
=\frac12\left(\frac{f'(a+h)-f'(a)}h+\frac{f'(a)-f'(a-h)}h\right).
$$
Twice <differentiability> at $a$ makes both quotients tend to $f''(a)$. Apply the endpoint <L'Hôpital's rule> proved in part (iii) for $h\downarrow0$. For the other side use the reflected positive variable $t=-h$; $F$ and $G$ are even, so their quotient is unchanged. Thus the two-sided limit is
$$
\boxed{\lim_{h\to0}\frac{f(a+h)+f(a-h)-2f(a)}{h^2}=f''(a).}
$$
This is the <second-order central difference> limit. The proof only uses <differentiability> of $f'$ at $a$ and <differentiability> nearby; it does not incorrectly assume $f''$ is <continuous>.
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