calculus - Proving Schwarz derivative $frac{f''(0)}{2} =limlimits_{xto 0} frac{frac{f(x) -f(0)}{x}-f'(0)}{x}$ without Taylor expansion or L'Hopital rule?

Assume that $f$ is $C^2$ near 0. I would like to show the following Schwartz derivative

$$\frac{f''(0)}{2} =\lim_{x\to 0} \frac{\frac{f(x) -f(0)}{x}-f'(0)}{x}$$

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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.

Answer

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
$$\lim_{x\to0}\frac{\frac{f(x)-f(0)}{x}-f'(0)}{x}=\frac12f''(0).$$