Does the limit $\frac{f(x_0 + \varepsilon) - f(x_0 - \varepsilon) -2 \varepsilon f'(x_0)}{\varepsilon^3}$ exist? Let $f \in C^3(\mathbb{R})$, does the following limit exist
$$ \underset{\varepsilon 
\to 0}{\lim} \frac{f(x_0 + \varepsilon) - f(x_0 - \varepsilon) - 2 \varepsilon f'(x_0)}{\varepsilon^3}?
$$
It's been a while since I took Calculus and somehow I fail to see how I should approach this.
 A: By applying Taylor's Theorem with Peano's Form of Remainder, we get that
$f(x_0\!+\!\varepsilon)=f(x_0)\!+\!\varepsilon f’(x_0)\!+\!\varepsilon^2\dfrac{f’’(x_0)}{2!}\!+\!\varepsilon^3\dfrac{f’’’(x_0)}{3!}\!+\!o\!\left(\epsilon^3\right)$
$f(x_0\!-\!\varepsilon)=f(x_0)\!-\!\varepsilon f’(x_0)\!+\!\varepsilon^2\dfrac{f’’(x_0)}{2!}^\!-\!\varepsilon^3\dfrac{f’’’(x_0)}{3!}\!+\!o\!\left(\epsilon^3\right)$
where $\,o\!\left(\varepsilon^3\right)\,$ represents a function such that $\,\lim_\limits{\varepsilon\to0}\dfrac{o\!\left(\varepsilon^3\right)}{\varepsilon^3}=0\,.$
Therefore ,
$\lim_\limits{\varepsilon\to0}\dfrac{f(x_0+\varepsilon)-f(x_0-\varepsilon)-2\varepsilon f'(x_0)}{\varepsilon^3}=$
$=\lim_\limits{\varepsilon\to0}\dfrac{2\varepsilon^3\dfrac{f’’’(x_0)}{3!}+o\!\left(\varepsilon^3\right)}{\varepsilon^3}=$
$=\lim_\limits{\varepsilon\to0}\left[\dfrac{f’’’(x_0)}3+\dfrac{o\!\left(\varepsilon^3\right)}{\varepsilon^3}\right]=\dfrac{f’’’(x_0)}3\,.$
It is not necessary that $\,f\in C^3(\mathbb{R})\,,\,$ but it is sufficient that $\,f’’’(x_0)\,$ exists.
It means that if $\,f’’’(x_0)\,$ exists, then
$\lim_\limits{\varepsilon\to0}\dfrac{f(x_0+\varepsilon)-f(x_0-\varepsilon)-2\varepsilon f'(x_0)}{\varepsilon^3}=\dfrac{f’’’(x_0)}3\,.$
