Solution Verification - find values of $a,b,c$ such that $\lim_{h\to 0} \frac{a\cdot f(x+h)+b\cdot f(x) + c\cdot f(x-h)}{h^2}$ exists 
Let $f$ be any twice differentiable function. Find values of $a,b,c$
such that  $$\lim_{h\to 0} \frac{a\cdot f(x+h)+b\cdot f(x) + c\cdot f(x-h)}{h^2}\tag{1}$$
exists for all $x$.

For $(1)$ to exist, we require
$$\lim_{h\to 0} 
\left(a\cdot f(x+h)+b\cdot f(x) + c\cdot f(x-h)\right)=0\tag{$*$}$$
(I've used this statement intuitively. I do not have a rigorous/proper proof for this statement.)
Which gives $a+b+c=0$.
Hence applying L'Hôpital's rule on $(1)$, we get$$\lim_{h\to 0} \frac{a\cdot f(x+h)+b\cdot f(x) + c\cdot f(x-h)}{h^2}=\lim_{h\to 0} \frac{a\cdot f'(x+h) - c\cdot f'(x-h)}{2h}\tag{2}$$
For $(2)$ to exist, we require $$\lim_{h\to 0} \left(a\cdot f'(x+h) - c\cdot f'(x-h)\right)=0\tag{$**$}$$
(I've used this statement intuitively. I do not have a rigorous/proper proof for this statement.)
Which gives $a-c=0$.
Hence applying L'Hôpital's rule on $(2)$, we get$$\lim_{h\to 0} \frac{a\cdot f'(x+h) - c\cdot f'(x-h)}{2h}=\lim_{h\to 0} \frac{a\cdot f''(x+h) + c\cdot f''(x-h)}{2}  \tag{3}$$
$$\implies\lim_{h\to 0} \frac{a\cdot f''(x+h) + c\cdot f''(x-h)}{2} = \frac{a+ c}{2}\cdot f''(x) \tag{4}$$
My questions are:

*

*Is the above solution correct?

*Can you provide a proper proof for the two statements that I have marked with asterisks? i.e  if it is given that $\lim_{x\to 0}\phi_2(x)=0$, then for $\lim_{x\to 0}\frac{\phi_1(x)}{\phi_2(x)}$ to exist is it necessary that $\lim_{x\to 0}\phi_1(x)=0$
Edit:-
 A: For your second question, notice
$$
    \phi_1(x) = \frac{\phi_1(x)}{\phi_2(x)} \cdot \phi_2(x)
$$
So you can apply the product rule for limits.  If the quotient and the denominator tend to zero at some point, then the numerator must also tend to zero at that point.  This is the trick used to prove that differentiability implies continuity.
A: Your reasoning behind the method you employ is correct. The answer to question $(2)$ is a consequence of the proof of the L'Hopital's rule in the $0/0$ case. You can also Taylor expand the numerator to solve your problem:
Expanding the numerator in $(1)$ about $h=0$, we get (because $f$ is twice-differentiable)
$$a\cdot f(x+h)+b\cdot f(x)+c\cdot f(x-h)=(a+b+c)f(x)+(a-c)hf'(x)+\left(\frac{a+c}{2}\right)h^2f''(x).$$ from which it must be that $a+b+c=0$ and $a-c=0$.
As a consequence, we get the formula $$f''(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2}.$$
A: As mentioned by Paramanand Singh in the comments, the solution is correct up to equation $(3)$
The solution breaks down at equation $(4)$ because, $f''(x)$ need not be continuous.
$$\text{hence, }\space\space \lim_{n\to 0}f''(x+h) \space\space\text{   and   }\space\space \lim_{h\to0}f''(x-h)  \space\space \text{ need not exist} $$
And so, we need to go back to equation $(2)$.
From equation $(2)$, we get  $$\lim_{h\to 0} \frac{a\cdot f'(x+h) - c\cdot f'(x-h)}{2h}=\lim_{h\to 0}\frac{a}{2} \cdot\frac{f'(x+h) - f'(x-h)}{h}$$
$$=\frac{a}{2}\lim_{h\to 0}\Bigg[\frac{\Big(f'(x+h)-f'(x)\Big)}{h}-\frac{\Big(f'(x-h)-f'(x)\Big) }{h}\Bigg]$$
$$=\frac{a}{2}\lim_{h\to 0}\Bigg[\frac{\Big(f'(x+h)-f'(x)\Big)}{h}+\frac{\Big(f'(x-h)-f'(x)\Big) }{-h}\Bigg]=\frac{a}{2}\Big[f''(x)+f''(x)\Big]=a\cdot f''(x)$$
