Proving Schwarz derivative $\frac{f''(0)}{2} =\lim\limits_{x\to 0} \frac{\frac{f(x) -f(0)}{x}-f'(0)}{x}$ without Taylor expansion or L'Hopital rule? Assume that $f$ is $C^2$ near 0. I would like to show the following Schwartz derivative 
$$
\frac{f''(0)}{2} =\lim_{x\to 0} \frac{\frac{f(x) -f(0)}{x}-f'(0)}{x}
$$ 


I am able to do this by using the Taylor expansion and L'Hopital rule. I am wondering how one can prove it without using Taylor expansion or L'Hopital rule. 


Suppose $x>0$. Note 
\begin{eqnarray} 
&&\frac{\frac{f(x)-f(0)}{x}-f'(0)}{x}\\
&=&\frac{f(x)-f(0)-xf'(0)}{x^2} \\
&=&\frac{\int_0^xf'(t)dt-\int_0^xf'(0)dt}{x^2} \\
&=&\frac{\int_0^x[f'(t)-f'(0)]dt}{x^2} \\
&=&\frac{\int_0^x\bigg[\int_0^tf''(s)ds\bigg]dt}{x^2} \\
&=&\frac{\int_0^x\bigg[\int_s^xf''(s)dt\bigg]ds}{x^2} \\
&=&\frac{\int_0^x(x-s)f''(s)ds}{x^2} 
\end{eqnarray}
and
$$ \int_0^x(x-s)ds=\frac12x^2. $$
So
\begin{eqnarray} 
&&\bigg|\frac{\frac{f(x)-f(0)}{x}-f'(0)}{x}-\frac12f''(0)\bigg|\\
&=&\bigg|\frac{\int_0^x(x-s)[f''(s)-f''(0)]ds}{x^2}\bigg|\\
&\le&\bigg|\frac{\int_0^x(x-s)|f''(s)-f''(0)|ds}{x^2}\bigg|
\end{eqnarray}
Since $f\in C^2$, for $\forall \varepsilon>0$, $\exists \delta>0$ such that
$$ |f''(x)-f''(0)|<\varepsilon \forall x\in(0,\delta). $$
Thus for $x\in(0,\delta)$,
\begin{eqnarray} 
&&\bigg|\frac{\frac{f(x)-f(0)}{x}-f'(0)}{x}-\frac12f''(0)\bigg|\\
&\le&\frac{\int_0^x(x-s)|f''(s)-f''(0)|ds}{x^2}\\
&\le&\bigg|\frac{\int_0^x(x-s)\varepsilon ds}{x^2}\bigg|\\
&=&\frac12\varepsilon.
\end{eqnarray}
So
$$ \lim_{x\to0}\frac{\frac{f(x)-f(0)}{x}-f'(0)}{x}=\frac12f''(0). $$
