Suppose $h(x)=x^2\sin(1/x)$ for $x\neq 0$, $h(0)=0$. Then it's derivative exists everywhere and is finite, but $$\frac{h(0+\varepsilon)-2h(0)+h(0+\varepsilon)}{\varepsilon^2}=\frac{2\varepsilon^2\sin(\frac{1}{x})}{\varepsilon^2}=2\sin(\frac{1}{\varepsilon})$$$$\frac{h(0+\varepsilon)-2h(0)+h(0-\varepsilon)}{\varepsilon^2}=\frac{2\varepsilon^2\sin(\frac{1}{\varepsilon})}{\varepsilon^2}=2\sin(\frac{1}{\varepsilon})$$ has no limit as $\varepsilon\rightarrow 0$.
