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Suppose a real differentiable function $h(x)$ with its derivative not infinity, it is sure that its second symmetric derivative $\lim_{\epsilon->0}\frac{h(x+\epsilon)-2h(x)+h(x-\epsilon)}{\epsilon^2}$ should exist?

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1 Answer 1

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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 $$\frac{h(0+\varepsilon)-2h(0)+h(0-\varepsilon)}{\varepsilon^2}=\frac{2\varepsilon^2\sin(\frac{1}{\varepsilon})}{\varepsilon^2}=2\sin(\frac{1}{\varepsilon})$$ has no limit as $\varepsilon\rightarrow 0$.

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