Let f be a function twice differentiable and with derivatives continuous on an interval $[a,b]$ containing $0$. Prove the following statement: $$\lim_{x\to0}  \frac{f'(x) - \frac{f(x)-f(0)}{x}}{x} = \frac{f''(0)}{2}$$
I have been thinking about this for a bit of time now, but I'm not getting anything. What I have done: write $f'(x)$ as it's defined:
$$f'(x) = \lim_{h\to x}\frac{f(h) - f(x)}{h-x}$$
Now we can write: 
$$\lim_{x\to0}  \frac{f'(x) - \frac{f(x)-f(0)}{x}}{x} = \lim_{x\to0}  \frac{\lim_{h\to x}\frac{f(h) - f(x)}{h-x} - \frac{f(x)-f(0)}{x}}{x}  $$
I have tried lots of algebraic manipulation after this, but nothing has come out so far. Could someone steer me in the right direction?
EDIT: I just opened Rudin, which has a similar question that indicates that l'Hopital's should be used. Indeed:
$$\lim_{x\to0} \frac{f'(x) - \frac{f(x)-f(0)}{x}}{x} = \lim_{x\to0} \frac{xf'(x) - f(x)-f(0)}{x^2}$$
We can apply l'Hôpital:
$$ \lim_{x\to0} \frac{xf'(x) - f(x)-f(0)}{x^2} = \lim_{x\to0}\frac{f'(x) + xf''(x) - f'(x)}{2x} = \frac{f''(0)}{2}$$
EDIT #2Using Taylor's theorem, as suggested in the comments.
We choose to perform the Taylor expansion of $f(0)$ at $x_0 = x $. Then:
$\exists c $ between $x$ and $0$ such that:
$$f(0) = f(x) - xf'(x) + \frac{x^2f''(c)}{2} \implies \frac{f''(c)}{2} = \frac{f'(x) - \frac{f(x)-f(0)}{x}}{x}   $$
As $x\to 0$, $c\to 0 $ and we have 
$$\lim_{x\to0}  \frac{f'(x) - \frac{f(x)-f(0)}{x}}{x} = \frac{f''(0)}{2}$$
 A: Assuming only $f''(0)$ exists, we have by Taylor (aka the MVT applied twice)
$$f(x)=f(0)+f'(0)x+ (f''(0)/2)x^2 + o(x^2).$$
Thus $(f(x)-f(0))/x = f'(0) +(f''(0)/2)x + o(x).$ Therefore
$$\frac{f'(x) - (f(x)-f(0))/x}{x} = \frac{f'(x) - f'(0) -(f''(0)/2)x + o(x))}{x}$$ $$ = \frac{f'(x) - f'(0)}{x} -f''(0)/2 + o(1).$$
As $x\to 0,$ this $\to f''(0)-f''(0)/2 = f''(0)/2,$ and we're done.
A: The answer by user "zhw." is my favorite but here is an approach which uses L'Hospital and assumes only the existence of $f''(0)$ (and not the continuity of $f''$).
We have
\begin{align}
L &= \lim_{x \to 0}\dfrac{f'(x) - \dfrac{f(x) - f(0)}{x}}{x}\notag\\
&= \lim_{x \to 0}\frac{xf'(x) - f(x) + f(0)}{x^{2}}\tag{*}\\
&= \lim_{x \to 0}\frac{xf'(x) - xf'(0) + xf'(0) - f(x) + f(0)}{x^{2}}\notag\\
&= \lim_{x \to 0}\frac{f'(x) - f'(0)}{x} + \frac{xf'(0) - f(x) + f(0)}{x^{2}}\notag\\
&= f''(0) + \lim_{x \to 0}\frac{xf'(0) - f(x) + f(0)}{x^{2}}\notag\\
&= f''(0) + \lim_{x \to 0}\frac{f'(0) - f'(x)}{2x}\text{ (via L'Hospital's Rule)}\notag\\
&= f''(0) - \frac{1}{2} \lim_{x \to 0}\frac{f'(x) - f'(0)}{x}\notag\\
&= f''(0) - \frac{f''(0)}{2}\notag\\
&= \frac{f''(0)}{2}\notag
\end{align}
The direct use of L'Hospital's Rule just after the step marked $(*)$ above also gives the answer but only if we assume that $f''$ is continuous at $0$ (this is the approach given in question also). Moreover note that the existence of $f''$ in a neighborhood of $0$ is essential to apply L'Hospital's Rule after step marked $(*)$. The approach above avoids these unnecessary conditions and relies only on existence of $f''(0)$ and nothing more.
Also note that we have $$\lim_{x \to 0}\frac{f(x) - f(0)}{x} = f'(0)$$ but we must resist the temptation to replace the expression $(f(x) - f(0))/x$ with $f'(0)$ (this leads to wrong answer $f''(0)$) because in general while evaluating the limit of an expression we can not replace a sub-expression by its limit.
A: From the extended law of the mean, if $f$ is twice differentiable in the neighborhood of $x$, then for $h$ sufficiently small, there exists a number $\theta\in (0,1)$ such that 
$$f(x-h)=f(x)-f'(x)h+\frac12 f''(x+\theta h)h^2 \tag 1$$
If $f''$ is continuous in a neighborhood of $x$, then we have from $(1)$
$$\frac12f''(x)=\lim_{h\to 0}\frac12f''(x+\theta h)=\lim_{h\to 0}\frac{f'(x)h+f(x-h)-f(x)}{h^2}\tag 2$$
We can rearrange the limit on the right-hand side of $(2)$ to obtain
$$\frac12f''(x)=\lim_{h\to 0}\frac{f'(x)-\frac{f(x)-f(x-h)}{h}}{h} \tag 3$$
We can replace $x$ in the limit expression of $(3)$ with $x+h$.  Proceeding and letting $x=0$ yields
$$\frac12f''(0)=\lim_{h\to 0}\frac{f'(h)-\frac{f(h)-f(0)}{h}}{h} \tag 4$$
Since $h$ is a "dummy" parameter in $(4)$ we have
$$\bbox[5px,border:2px solid #C0A000]{\frac12f''(0)=\lim_{x\to 0}\frac{f'(x)-\frac{f(x)-f(0)}{x}}{x} }\tag 5$$
as was to be shown!

NOTES:
It is of interest to note that we could have written 
$$f(x+h)=f(x)+f'(x)h+\frac12 f''(x+\eta h)h^2 $$
for some $\eta\in (0,1)$.  If $f''$ is continuous in a neighborhood of $x$, then 
$$\frac12f''(x)=\lim_{h\to 0}\frac{\frac{f(x+h)-f(x)}{h}-f'(x)}{h}$$
whereupon letting $x=0$ (and changing the "dummy" index $h$ to $x$) yields
$$\frac12f''(0)=\lim_{x\to 0}\frac{\frac{f(x)-f(0)}{x}-f'(0)}{x} \tag 6$$
The expression in $(6)$ is an alternative to that in $(5)$.

