$\lim\limits_{h \to 0}{f(x+h)-f(x)-f'(x)h \over h^2} $ $f$ is twice-differentiable function at the domain of $\Bbb R$
\begin{align}
\lim_{h \to 0}{f(x+h)-f(x)-f'(x)h \over h^2} 
&= \lim_{h \to 0}\left[{f(x+h)-f(x) \over h^2} -{f'(x)h \over h^2}\right] \\
&= \lim_{h \to 0}{f(x+h)-f(x) \over h}{1\over h} -\lim_{h \to 0}{f'(x) \over h} \\
&= \lim_{h \to 0}{f'(x)\over h} -\lim_{h \to 0}{f'(x) \over h} \\
&= \lim_{h \to 0}\left[{f'(x)\over h}-{f'(x)\over h}\right] \\
& = 0
\end{align}
Is the above reasoning correct? I especially concerns my repeating distribution of limits back and forth. 
 A: You made an error at this step: 
$\lim\limits_{h \to 0}{f(x+h)-f(x) \over h}{1\over h}=\lim\limits_{h \to 0}{f'(x) \over h} $
Because by doing so you assume that $\lim\limits_{h \to 0}{f(x+h)-f(x) \over h}$ and $\lim\limits_{h \to 0}{1 \over h}$ both exists which is clearly false for the latter. 
Now since $f(x)$ is twice differentiable, then $f$ and $f'$ are continuous. Substituting $h=0$ into $\lim\limits_{h \to 0}{f(x+h)-f(x)-f'(x)h \over h^2}$ gives us ${0 \over 0}$. 
(Note that we can directly let $h=0$ due to continuity!)
Now applying L'Hopital's rule by taking derivatives on the numerator and denominator with respect to $h$ yields: $\lim\limits_{h \to 0}{f'(x+h)-f'(x) \over 2h}$.
Then it is clear that this expression is equivalent to ${1 \over 2} f''(x)$.
A: By the Taylor formula
$$f(x+h) = f(x) + f'(x)h + \frac{1}{2}f''(x) h^2 + o(h^2)$$
Now substitute and get
$$\lim_{h \to 0} \frac{1}{h^2} (f(x+h) -f(x) - f'(x)h) = \lim_{h \to 0} \frac{1}{h^2} (\frac{1}{2}f''(x) h^2 + o(h^2)) = \frac{1}{2}f''(x)$$
A: The first and main error is in
$$
\lim_{h \to 0}\left[{f(x+h)-f(x) \over h^2} -{f'(x)h \over h^2}\right]
= \lim_{h \to 0}{f(x+h)-f(x) \over h}{1\over h} -\lim_{h \to 0}{f'(x) \over h}
$$
because neither limit in the right-hand side exists (unless you're in a very very special situation, which is not assumed in the hypotheses).
The second mistake is in
$$
\lim_{h \to 0}{f(x+h)-f(x) \over h}{1\over h}= 
\lim_{h \to 0}{f'(x)\over h}
$$
When computing a limit, you cannot just take one part and substitute it with its limit; this is essentially the same as doing
$$
\lim_{h\to0}\frac{h}{h}=\lim_{h\to0}\frac{0}{h}
$$
