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Can the equality of these two be used as a second-order differentiability check?:

$$\lim_{h\rightarrow 0}\frac{f(x+2h)-2f(x+h)+f(x)}{h^2}$$

And,

$$\lim_{h\rightarrow 0}\frac{f(x-2h)-2f(x-h)+f(x)}{h^2}$$

I think that just like when the limits ##\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}## and ##\lim_{h\rightarrow 0}\frac{f(x-h)-f(x)}{-h}## are equal, then it means that the function is differentiable, the equality of these two limits should mean that the function is twice differentiable.

$$\lim_{h\rightarrow 0}\frac{f(x+2h)-2f(x+h)+f(x)}{h^2}$$

And,

$$\lim_{h\rightarrow 0}\frac{f(x-2h)-2f(x-h)+f(x)}{h^2}$$

I think that just like when the limits ##\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}## and ##\lim_{h\rightarrow 0}\frac{f(x-h)-f(x)}{-h}## are equal, then it means that the function is differentiable, the equality of these two limits should mean that the function is twice differentiable.

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