Limits to directly check second order differentiability

[Kumar8434]

Sorry, I mistakenly reported my own post last time. But later I realized that these limits do work. So, I'm posting this again.
I'm using these limits to check second-order differentiability:
$$\lim_{h\rightarrow 0}\frac{f(x+2h)-2f(x+h)+f(x)}{h^2}$$
And,
$$\lim_{h\rightarrow 0}\frac{f(x-2h)-2f(x-h)+f(x)}{h^2}$$
I think that when these limits are equal, then the function is twice differentiable.
Now, I checked it for this function:
$$f(x)=\frac{x^2}{2},x\geq0$$
$$=\frac{-x^2}{2},x<0$$.
Now, this function is differentiable once at $x=0$. But after the first differentiation, its derivative comes out to be $|x|$ which isn't differentiable at $x=0$. So, this function is not twice differentiable at $x=0$. So, those limits should give different values.
Now,$$\lim_{h\rightarrow 0}\frac{f(x+2h)-2f(x+h)+f(x)}{h^2}$$ At $x=0$,
$$=\lim_{h\rightarrow 0}\frac{f(2h)-2f(h)+f(0)}{h^2}$$
$$=\lim_{h\rightarrow 0}\frac{\frac{(2h)^2}{2}-2\frac{h^2}{2}+0}{h^2}$$
$$=2-1=1$$
And, at $x=0$,
$$\lim_{h\rightarrow 0}\frac{f(x-2h)-2f(x-h)+f(x)}{h^2}$$
$$=\lim_{h\rightarrow 0}\frac{f(-2h)-2f(-h)+f(0)}{h^2}$$
$$=\lim_{h\rightarrow 0}\frac{\frac{-(-2h)^2}{2}+2\frac{-h^2}{2}+0}{h^2}$$
$$=-2+1=-1$$
So, these limits are not equal, which means $f(x)$ is not twice differentiable at $x=0$.
So, I think this limit works. Is there some way to prove that a function is twice differentiable only if these two limits are equal?

 

[Kumar8434]

In a similar way, I got the general rule:
A function $f(x)$ is differentiable $n$ times only if these two limits are equal:
1.$$\lim_{h\rightarrow 0}\frac{\sum_{r=0}^n(-1)^r\cdot ^nC_r\cdot f(x+(n-r)h)}{h^n}$$
2.$$\lim_{h\rightarrow 0}\frac{\sum_{r=0}^n(-1)^r\cdot ^nC_r\cdot f(x-(n-r)h)}{(-h)^n}$$
Please note that $h$ itself is assumed to be positive in both the limits. $x$ can be replaced with $c$ to check the differentiability at $x=c$.
Have I got this result correct or Can counter examples be given against it?